Abstract Algebra — Chapter 6 Field theory and Galois theory

Chapter 6
Field theory and Galois theory

6.1. Extension fields

Definition 6.1.1. §

A field $E$ is an extension field (or extension) of $F$ if $F$ is a subfield of $E$ . We write $E ∕ F$ (which reads “ $E$ over $F$ ”) to denote that $E$ is an extension field of $F$ , and we say that $E ∕ F$ is a field extension, or an extension of fields.

Examples 6.1.2. §

We have that $ℝ$ is an extension field of $ℚ$ , and $ℂ$ is an extension of both $ℚ$ and $ℝ$ . We have that $ℚ (i)$ is an extension of $ℚ$ of which $ℂ$ , but not $ℝ$ , is an extension field.

We will often have cause to deal with the field $ℤ ∕ 𝑝ℤ$ , where $p$ is a prime. When we think of $ℤ ∕ 𝑝ℤ$ as a field, we make a change of notation.

Definition 6.1.3. §

For a prime $p$ , the field of $p$ elements, $𝔽_{p}$ , is $ℤ ∕ 𝑝ℤ$ .

Lemma 6.1.4. §

Let $F$ be a field. If $F$ has characteristic $0$ , then $F$ is an extension of $ℚ$ . If $F$ has characteristic equal to a prime $p$ , then $F$ is an extension of $𝔽_{p}$ .

Proof.

If $F$ has characteristic $0$ , we define $ι : ℚ \to F$ by $ι (a b^{- 1}) = (a \cdot 1) \cdot {(b \cdot 1)}^{- 1}$ , where $a, b \in ℤ$ and $b \neq 0$ . Since $ι$ is a ring homomorphism and $ℚ$ is a field, it is injective, so $ℚ$ sits isomorphically inside $F$ . If $F$ has characteristic $p$ , then we define $ι : 𝔽_{p} \to F$ by the same equation, where now $a, b \in ℤ$ and $b ≢ 0 mod p$ . Since $F$ has characteristic $p$ , this is a ring homomorphism, and again it is injective. □

Definition 6.1.5. §

An intermediate field of a field extension $E ∕ F$ is a subfield $E^{'}$ of $E$ containing $F$ . The extension $E^{'} ∕ F$ is said to be a subextension of $F$ in $E$ .

Definition 6.1.6. §

The ground field (or base field) of a field extension $E ∕ F$ is the field $F$ .

Definition 6.1.7. §

Let $E ∕ F$ be a field extension. Let $A \subset E$ . The field generated over $F$ by the set $A$ (or its elements) is the smallest subfield $K$ of $E$ containing $F$ and $A$ , often denoted $F (A)$ . We say that the elements of $A$ generate $K$ as an extension of $F$ and that $K$ is given by adjoining the elements of $A$ to $F$ .

Notation 6.1.8. §

Let $E ∕ F$ be a field extension and $α_{1}, α_{2}, \dots, α_{n} \in E$ for some $n \geq 0$ . We write $F (α_{1}, α_{2}, \dots, α_{n})$ for the subfield of $E$ generated by the set $α_{1}, α_{2}, \dots, α_{n}$ over $F$ .

Remark 6.1.9. §

One often says “ $F$ adjoin $α$ ” to refer to a field $F (α)$ .

Remark 6.1.10. §

Note that the field generated over $F$ by a set of elements $A$ of $E$ is well-defined, equal to the intersection of all subfields of $E$ containing both $F$ and $A$ .

Remark 6.1.11. §

Note that we distinguish between the field $F (x_{1}, x_{2}, \dots, x_{n})$ of rational functions, where $x_{1}, x_{2}, \dots, x_{n}$ are indeterminates, and $F (α_{1}, α_{2}, \dots, α_{n})$ , where $α_{1}, α_{2}, \dots, α_{n}$ are elements of an extension field of $F$ . These fields can be quite different. However, in that $F (α_{1}, α_{2}, \dots, α_{n})$ is the quotient field of $F [α_{1}, α_{2}, \dots, α_{n}]$ , every element of $F (α_{1}, α_{2}, \dots, α_{n})$ is a rational function in the elements $α_{i}$ with $1 \leq i \leq n$ .

Example 6.1.12. §

The fields $ℚ (\sqrt{2})$ and $ℚ (i)$ are extension fields of $ℚ$ inside $ℝ$ and $ℂ$ , respectively.

Proposition 6.1.13. §

Let $E ∕ F$ be a field extension, and let $α \in E$ . Then $F (α)$ is isomorphic to the quotient field of $F [α]$ .

Proof.

Since $F [α]$ is the smallest subring of $E$ containing $F$ and $α$ and $F (α)$ is the smallest subfield of $E$ containing $F$ and $α$ , inclusion provides an injective homomorphism

ι : F [α] \to F (α) .

Since $F (α)$ is a field, $ι$ induces an injective map $Q (ι) : Q (F [α]) \to F (α)$ . Since the image of $Q (ι)$ is a subfield of $F (α)$ containing $F$ and $α$ and $F (α)$ is the smallest such field in $E$ , we have that $Q (ι)$ is surjective as well. □

In many cases, an extension field generated by an element is actually equal to the ring generated by the element. We see this holds in a couple of simple examples.

Example 6.1.14. §

The fields $ℚ (\sqrt{2})$ and $ℚ (i)$ equal $ℚ [\sqrt{2}]$ and $ℚ [i]$ as subrings of $ℝ$ and $ℂ$ respectively. E.g., the elements of $ℚ (\sqrt{2})$ all may be written in the form $a + b \sqrt{2}$ for some $a, b \in ℚ$ .

The key in this example is that $\sqrt{2}$ and $i$ are roots of polynomials with coefficients in $ℚ$ . Let us examine this further.

Theorem 6.1.15. §

Let $E$ be an extension field of a field $F$ , and let $f \in F [x]$ be an irreducible polynomial that has a root $α \in E$ . Then the evaluation map $e_{α} : F [x] \to F (α)$ given by $e_{α} (g) = g (α)$ for $g \in F [x]$ induces an isomorphism

\bar{e_{α}} : F [x] ∕ (f) \to_{}^{\sim} F (α)

of fields such that $\bar{e_{α}} (a) = a$ for all $a \in F$ .

Proof.

First, note that we have the inclusion map $F \to F [x]$ and the quotient map $F [x] ∕ (f)$ , inducing a nonzero, and hence injective, map of fields $F \to F [x] ∕ (f)$ . This allows us to view $F$ as a subfield of $F [x] ∕ (f)$ . The map $e_{α}$ has kernel containing $f$ , so it induces a homomorphism $\bar{e_{α}}$ between fields as in the statement. The image of this map is then a field extension of $F$ containing $α$ , hence equals $F (α)$ by its definition as the field generated by $α$ over $F$ . □

Example 6.1.16. §

The field $ℚ (i)$ is isomorphic to the quotient ring $ℚ [x] ∕ (x^{2} + 1)$ .

We now obtain the following theorem as a corollary.

Theorem 6.1.17 (Kronecker). §

Let $F$ be a field, and let $f \in F [x]$ be a nonconstant polynomial. Then there exists a field extension $E$ of $F$ and an element $α \in E$ such that $f (α) = 0$ .

Proof.

First, we may assume that $f$ is irreducible by replacing $f$ by an irreducible polynomial dividing it which has $α$ as a root. We then set $E = F [x] ∕ (f)$ , which is a field. Let $α = x + (f)$ . Then $f (α)$ is the image of $f$ in $E$ , and so $f (α) = 0$ . □

Definition 6.1.18. §

Let $E ∕ F$ be a field extension. A nonconstant polynomial $f \in F [x]$ is said to split (or factor completely) in $E$ if it can be written as a product of linear polynomials in $E [x]$ .

Definition 6.1.19. §

Let $F$ be a field. A splitting field $E$ for $f \in F [x]$ over $F$ is an extension of $F$ such that $f$ splits in $E$ but not any proper subextension of $F$ in $E$ .

Examples 6.1.20. §

a.: The field $ℚ (\sqrt{2})$ is the splitting field of $x^{2} - 2$ , since it contains both of its roots. It is also the splitting field of ${(x - a)}^{2} - 2 b^{2}$ for any $a, b \in ℚ$ .
b.: The field $ℚ (\sqrt[3]{2})$ is not the splitting field of $x^{3} - 2$ , since it contains $\sqrt[3]{2}$ but not its other two roots. On the other hand, if $ω \in ℂ$ is a primitive cube root of unity, then $ℚ (\sqrt[3]{2}, ω \sqrt[3]{2}, ω^{2} \sqrt[3]{2})$ is a splitting field for $x^{3} - 2$ inside $ℂ$ . This field may be written more simply as $ℚ (ω, \sqrt[3]{2})$ .

As a corollary of Kronecker’s theorem, we have the following.

Corollary 6.1.21. §

Let $F$ be a field, and let $f \in F [x]$ . Then there exists a splitting field for $f$ over $F$ .

Proof.

Let $n = \deg f$ . The result is clearly true for $n = 1$ . Set $K = F [x] ∕ (f)$ . Then $f$ has a root $α$ in $K$ by Kronecker’s theorem. Set $g (x) = {(x - α)}^{- 1} f (x) \in E [x]$ . Then there exists a splitting field $E$ of $g$ over $K$ which is generated by the roots of $g$ over $F$ by induction. This $E$ is a splitting field of $f$ over $F$ , since it is generated by the roots of $f$ . □

We next distinguish two types of elements of extension fields of $F$ : those that are roots of polynomials and those that are not.

Definition 6.1.22. §

Let $E ∕ F$ be a field extension. An element $α \in E$ is called algebraic over $F$ if there exists a nonzero $f \in F [x]$ such that $f (α) = 0$ . Otherwise, $α$ is said to be transcendental over $F$ .

When speaking of elements of extensions of $ℚ$ , we speak simply of algebraic and transcendental numbers.

Definition 6.1.23. §

An element of $ℂ$ is said to be an algebraic number if it is algebraic over $ℚ$ and a transcendental number if it is transcendental over $ℚ$ .

Examples 6.1.24. §

The number $\sqrt{2}$ is an algebraic number, since it is a root of $x^{2} - 2$ . Similarly, $i$ is algebraic, being a root of $x^{2} + 1$ . However, the real number $π$ is transcendental, and the real number $e$ such that $\log e = 1$ is transcendental as well. We do not prove the latter two facts here.

Example 6.1.25. §

A real number given by repeated square roots

\sqrt{a_{1} + \sqrt{a_{2} + \dots + \sqrt{a_{n}}}}

with the $a_{i}$ positive rational numbers is algebraic: it is a root of

{(\dots {({(x^{2} - a_{1})}^{2} - a_{2})}^{2} \dots)}^{2} - a_{n} .

Example 6.1.26. §

If $F$ is a field and $α \in F$ , then $α$ is algebraic over $F$ , being a root of $x - α$ .

Note the following.

Proposition 6.1.27. §

Let $E ∕ F$ be a field extension, and let $α \in E$ . Then $α$ is transcendental over $F$ if and only if the evaluation homomorphism $e_{α} : F [x] \to E$ is injective.

Proof.

By definition, $α \in E$ is transcendental if and only if $g (α) \neq 0$ for every $g \in F [x]$ that is nonzero. But $g (α) = e_{α} (g)$ , so we are done. □

This allows us to give the prototypical example of a transcendental element.

Corollary 6.1.28. §

Let $F$ be a field. The element $x$ of the field $F (x)$ of rational functions of $F$ is transcendental over $F$ .

Proof.

Let $y$ be an indeterminate. Consider $e_{α} : F [y] \to F (x)$ given by $e_{α} (g) = g (x)$ . We have that the polynomial $g (x)$ is zero in $F (x)$ if and only if it is zero in $F [x]$ , and therefore if and only if $g = 0$ in $F [y]$ . □

Theorem 6.1.29. §

Let $E ∕ F$ be a field extension, and let $α \in E$ be algebraic over $F$ . Then there exists a unique monic, irreducible polynomial $f \in F [x]$ such that $f (α) = 0$ .

Proof.

Since $α$ is algebraic over $F$ , there exists a polynomial $g \in F [x]$ such that $g (α) = 0$ . Since $g$ factors as a product of irreducible polynomials, and $E$ is in particular an integral domain, one of the irreducible factors must have $α$ as a root, and by multiplying it by a constant, we may take it to be monic. So suppose that $f$ is a monic irreducible polynomial in $F [x]$ with $f (α) = 0$ . Without loss of generality, we may assume that $\deg f$ is minimal among all such polynomials. If $f^{'} \in F [x]$ satisfies $f^{'} (α) = 0$ , then the division algorithm provides $q, r \in F [x]$ with $r = f^{'} - 𝑞𝑓$ and either $\deg r < \deg f$ or $r = 0$ . Since $r (α) = 0$ , we must have $r = 0$ , but then $f^{'} = 𝑞𝑓$ , so $f$ divides $f^{'}$ . If $f^{'}$ were monic and irreducible, this would force $f^{'} = f$ , as desired. □

Definition 6.1.30. §

Let $E ∕ F$ be a field extension, and let $α \in E$ be algebraic over $F$ . The minimal polynomial of $α$ over $F$ is the unique monic irreducible polynomial in $F [x]$ which has $α$ as a root.

Examples 6.1.31. §

a.: If $F$ is a field and $α \in F$ , then $x - α$ is the minimal polynomial of $α$ over $F$ .
b.: The polynomial $x^{2} + 1$ is the minimal polynomial of $i$ over $ℚ$ .

6.2. Finite extensions

Remark 6.2.1. §

If $E ∕ F$ is an extension of fields, then $E$ is an $F$ -vector space via the restriction of the multiplication in $E$ to a map $F \times E \to E$ , which is then given by $a \cdot α = 𝑎𝛼$ for $a \in F$ and $α \in E$ . Moreover, $E$ actually contains $F$ , so $F$ is a $F$ -subspace of $E$ .

Definition 6.2.2. §

An extension $E ∕ F$ of fields is finite if $E$ is a finite-dimensional $F$ -vector space. Otherwise, $E ∕ F$ is said to be an infinite extension.

Example 6.2.3. §

The field $ℚ (\sqrt{2})$ is an extension of $ℚ$ with basis ${1, \sqrt{2}}$ and hence is a basis of $ℚ (\sqrt{2})$ over $ℚ$ .

Example 6.2.4. §

The field $F (x)$ of rational functions over a field $F$ is infinite. More generally, if $E ∕ F$ is an extension that contains an element that is transcendental over $F$ , then $E$ is an infinite extension of $F$ .

Definition 6.2.5. §

The degree $[E : F]$ of a finite extension $E$ of a field $F$ is defined to be the dimension $\dim_{F} E$ of $E$ as a vector space over $F$ . If $E ∕ F$ is an infinite extension, we say that the degree of $E$ over $F$ is infinite.

Example 6.2.6. §

The degree $[ℚ (\sqrt{2}) : ℚ]$ is $2$ , as $\sqrt{2}$ has minimal polynomial $x^{2} - 2$ . The set ${1, \sqrt{2}}$ forms a basis of $ℚ (\sqrt{2})$ as a $ℚ$ -vector space.

The following is essential to our studies.

Theorem 6.2.7. §

Let $F$ be a field, and let $f \in F [x]$ be an irreducible polynomial of degree $n$ . Then the field $F [x] ∕ (f)$ has degree $n$ over $F$ .

Proof.

Since $(f)$ contains only multiples of $f$ , it contains no nontrivial linear combinations of the monomials $x^{i}$ with $0 \leq i \leq n - 1$ . In other words, the $x^{i} + (f)$ with $0 \leq i \leq n - 1$ are linearly independent over $F$ . On the other hand, if $g \in F [x]$ , then $g = 𝑞𝑓 + r$ with $q, f \in F [x]$ and $\deg r < n$ , so $g + (f) = r + (f)$ , and therefore $g + (f)$ may be written as the image in the quotient of a linear combination of the monomials $x^{i}$ with $0 \leq i \leq n - 1$ . That is, the elements $x^{i} + (f)$ with $0 \leq i \leq n - 1$ form a basis of $F [x] ∕ (f)$ . □

The proof of Theorem 6.2.7, when taken together with Theorem 6.1.15, yields the following.

Corollary 6.2.8. §

Let $E ∕ F$ be a field extension, and let $α \in E$ be algebraic over $F$ . Let $n$ be the degree of the minimal polynomial of $F$ . Then $[F (α) : F] = n$ , and ${1, α, \dots, α^{n - 1}}$ is a basis of $F (α)$ over $F$ .

Proposition 6.2.9. §

If $E ∕ F$ is a finite extension and $α \in E$ , then $α$ is algebraic over $F$ .

Proof.

Since $[E : F]$ is finite, there is an $n \geq 1$ such that the set ${1, α, \dots, α^{n}}$ is $F$ -linearly dependent. We then have

\sum_{i = 0}^{n} c_{i} α^{i} = 0

for some $c_{i} \in F$ with $0 \leq i \leq n$ . Setting $f = \sum_{i = 0}^{n} c_{i} x^{i} \in F [x]$ , we see that $f (α) = 0$ , so $α$ is algebraic. □

Corollary 6.2.10. §

Every finite extension $E$ of a field $F$ has the form $E = F (α_{1}, α_{2}, \dots, α_{n})$ for some algebraic $α_{i} \in E$ for $1 \leq i \leq n$ .

Proof.

One may simply take ${α_{i} ∣ 1 \leq i \leq n}$ to be a basis of $E$ over $F$ . Since each $α_{i} \in E$ , we have $F (α_{1}, α_{2}, \dots, α_{n}) \subseteq E$ , and since every element in $E$ is a linear combination of the $α_{i}$ , we have the opposite containment. □

The following theorem, while stated for arbitrary field extensions, has a number of applications to finite extensions.

Theorem 6.2.11. §

Let $E$ be an extension of a field $F$ , and let $V$ be an $E$ -vector space. If $A$ is a basis of $E$ over $F$ and $B$ is an $E$ -basis of $V$ , then

𝐴𝐵 = {𝛼𝛽 ∣ α \in A, β \in B}

is a basis of $V$ over $F$ , and the map $A \times B \to 𝐴𝐵$ given by scalar multiplication in $V$ is a bijection.

Proof.

We first show that $𝐴𝐵$ spans $V$ . By definition of $B$ , any $γ \in V$ can be written as

γ = \sum_{j = 1}^{m} c_{j} β_{j}

with $c_{j} \in E$ and $β_{j} \in B$ for $1 \leq j \leq m$ and some $m \geq 1$ . Each $c_{j}$ is in the $F$ -span of some finite subset of $A$ . By taking the union of these subsets, we see that there is a single finite subset of $A$ such that every $c_{j}$ with $1 \leq j \leq m$ is in its span. That is, we may write

c_{j} = \sum_{i = 1}^{n} d_{𝑖𝑗} α_{i}

for some $d_{𝑖𝑗} \in F$ and $α_{i} \in E$ for $1 \leq i \leq n$ and some $n \geq 1$ . Plugging in, we obtain

γ = \sum_{i = 1}^{n} \sum_{j = 1}^{m} d_{𝑖𝑗} α_{i} β_{j},

so the set $𝐴𝐵$ spans $V$ over $F$ .

Now, if some $F$ -linear combination of the elements of $𝐴𝐵$ equals zero, then in particular (by throwing in terms with zero coefficients if needed) we may write

\sum_{i = 1}^{n} \sum_{j = 1}^{m} a_{𝑖𝑗} α_{i} β_{j} = 0

for some $α_{i} \in A$ , $β_{j} \in B$ , and $a_{𝑖𝑗} \in F$ for $1 \leq i \leq n$ and $1 \leq j \leq m$ , for some $m$ and $n$ . Since the $β_{j}$ are $E$ -linearly independent, this implies that

\sum_{i = 1}^{n} a_{𝑖𝑗} α_{i} = 0

for all $1 \leq j \leq m$ . Since the $α_{i}$ are $F$ -linearly independent, we then have that $a_{𝑖𝑗} = 0$ for all $i$ and $j$ . Therefore, the set $𝐴𝐵$ is a basis of $V$ over $F$ .

Note that we may also conclude that the surjection $A \times B \to 𝐴𝐵$ given by multiplication in $V$ is injective. If it were not, then we would have two distinct pairs $(α, β), (α^{'}, β^{'}) \in A \times B$ such that $𝛼𝛽 - α^{'} β^{'} = 0$ , contrary to what we have shown. □

Theorem 6.2.11 has the following almost immediate corollary.

Corollary 6.2.12. §

Let $E$ be a finite extension of a field $F$ , and let $K$ be a finite extension of $E$ . Then $K ∕ F$ is a finite extension, and we have

[K : F] = [K : E] [E : F] .

Proof.

Theorem 6.2.11 tells us that any basis of $K$ over $F$ has $[K : E] [E : F]$ elements, hence the result. □

This corollary has in turn the following two corollaries.

Corollary 6.2.13. §

Let $E$ be a finite extension of a field $F$ , and let $K$ be a finite extension of $E$ . Then $[K : E]$ and $[E : F]$ divide $[K : F]$ .

Corollary 6.2.14. §

Let $K ∕ F$ be a finite extension, and let $E$ be a subfield of $K$ containing $F$ . Then $K ∕ E$ and $E ∕ F$ are finite extensions.

Example 6.2.15. §

By Corollary 6.2.12, we have

[ℚ (i, \sqrt{2}) : ℚ] = [ℚ (i, \sqrt{2}) : ℚ (\sqrt{2})] [ℚ (\sqrt{2}) : ℚ],

and since $i \notin ℚ (\sqrt{2})$ , we have that $x^{2} + 1$ is irreducible in $ℚ (\sqrt{2}) [x]$ , so $[ℚ (i, \sqrt{2}) : ℚ (\sqrt{2})] = 2$ . Therefore, $[ℚ (i, \sqrt{2}) : ℚ] = 4$ .

We give another corollary of Corollary 6.2.12 that is a converse to Corollary 6.2.10.

Corollary 6.2.16. §

Let $K ∕ F$ be a field extension, and let $α_{1}, α_{2}, \dots, α_{n} \in K$ be algebraic. Then $F (α_{1}, α_{2}, \dots, α_{n})$ is a finite extension of $F$ .

Proof.

The corollary is true for $n = 1$ by definition of an algebraic element. Suppose by induction we know it for $n - 1$ , and let $E = F (α_{1}, α_{2}, \dots, α_{n - 1})$ , which is a finite extension of $F$ by induction. Note that $α_{n}$ is algebraic over $E$ in that it is algebraic over $F$ . Since $K = E (α_{n})$ , we therefore have that $K$ is a finite extension of $E$ . That $K ∕ F$ is a finite extension now follows from Corollary 6.2.12. □

Definition 6.2.17. §

A field extension $E ∕ F$ is said to be algebraic if every element of $E$ is algebraic over $F$ . Otherwise, $E ∕ F$ is said to be a transcendental extension.

Proposition 6.2.18. §

Every finite extension is algebraic.

Proof.

If $α \in E$ , then $F (α) \subseteq E$ , so $F (α) ∕ F$ is finite. Hence, $α$ is algebraic over $F$ . □

In fact, we can do better.

Proposition 6.2.19. §

Let $E$ be an intermediate field in a field extension $K ∕ F$ . Then $K ∕ F$ is algebraic if and only if both $K ∕ E$ and $E ∕ F$ are algebraic.

Proof.

Suppose that $K ∕ E$ and $E ∕ F$ are algebraic. Let $α \in K$ , and let $f = \sum_{i = 0}^{n} a_{i} x^{i} \in E [x]$ be its minimal polynomial over $E$ . Since $E ∕ F$ is algebraic, the field $E_{f} = F (a_{1}, \dots, a_{n})$ is finite over of $F$ , and therefore so is $E_{f} (α)$ . In particular, $α$ is algebraic over $F$ , and therefore $K ∕ F$ is algebraic. The other direction is immediate. □

Remark 6.2.20. §

A transcendental field extension can never be finite.

Examples 6.2.21. §

a.: The field $ℝ$ is a transcendental extension of $ℚ$ .
b.: The field $K = ℚ (\sqrt{2}, \sqrt[3]{2}, \sqrt[4]{2}, \sqrt[5]{2}, \dots)$ is an algebraic extension of $ℚ$ , as the field generated by any finite list of these roots is equal to $ℚ (\sqrt[n]{2})$ for $n \geq 2$ , every element of $K$ is contained such a field, and each of these fields is algebraic over $ℚ$ .

6.3. Composite fields

Definition 6.3.1. §

Let $E_{1}$ and $E_{2}$ be subfields of a field $K$ . The compositum, or composite field, $E_{1} E_{2}$ of $E_{1}$ and $E_{2}$ is the smallest subfield of $K$ containing both $E_{1}$ and $E_{2}$ .

Remark 6.3.2. §

The compositum $E_{1} E_{2}$ of subfields $E_{1}$ and $E_{2}$ of a field $K$ is the intersection of all subfields of $K$ containing both $E_{1}$ and $E_{2}$ .

Example 6.3.3. §

Let $K ∕ F$ be a field extension, and let $α, β \in K$ . Then

F (α, β) = F (α) F (β) .

More generally, if $E$ is any subfield of $K$ containing $F$ and $α \in K$ , then

𝐸𝐹 (α) = E (α) .

We prove the following in the case of finite extensions. Note, though, that this finiteness is not needed, as seen through Corollary 6.3.11 below.

Proposition 6.3.4. §

Let $E_{1}$ and $E_{2}$ be finite extensions of a field $F$ contained in a field $K$ . Suppose that $A$ and $B$ are bases of $E_{1}$ and $E_{2}$ as $F$ -vector spaces, respectively. Then $E_{1} E_{2}$ is spanned by the set $𝐴𝐵$ .

Proof.

Set $m = [E_{1} : F]$ and $n = [E_{2} : F]$ , and let $A = {α_{1}, α_{2}, \dots, α_{m}}$ and $B = {β_{1}, β_{2}, \dots, β_{n}}$ . Clearly, we have

E_{1} E_{2} = F (α_{1}, α_{2}, \dots, α_{m}, β_{1}, β_{2}, \dots, β_{n}) .

As the elements of $A$ and $B$ are algebraic, we have $E (α_{i}) = E [α_{i}]$ and $E (β_{i}) = E [β_{i}]$ for any field $E$ containing $F$ , for all $i$ and $j$ . We then see by a simple recursion that every element of $E_{1} E_{2}$ may actually be expressed as a polynomial in the elements of $A$ and $B$ with coefficients in $F$ , not just a rational function. However, any monomial in the elements of $A$ lies in $E_{1}$ , hence may be written as a linear combination of the elements of $A$ . Similarly, any monomial in the elements of $B$ lies in $E_{2}$ , hence may be written as a linear combination of the elements of $B$ . Therefore, every monomial is the elements of $A$ and $B$ may be written as a product of a linear combination of elements of $A$ with a linear combination of elements of $B$ , which is the a linear combination of elements of $𝐴𝐵$ . Since every polynomial is a linear combination of monomials, we are done. □

Corollary 6.3.5. §

Let $E_{1}$ and $E_{2}$ be finite extensions of a field $F$ that are contained in a field $K$ . Then we have

[E_{1} E_{2} : F] \leq [E_{1} : F] [E_{2} : F] .

Proof.

Let $A$ (resp., $B$ ) be a basis of $E_{1}$ (resp., $E_{2}$ ) over $F$ . Then $𝐴𝐵$ has at most $[E_{1} : F] [E_{2} : F]$ elements and spans $E_{1} E_{2}$ over $F$ . □

Corollary 6.3.6. §

Let $E_{1}$ and $E_{2}$ be finite extensions of a field $F$ that are contained in a field $K$ , and suppose that $[E_{1} : F]$ and $[E_{2} : F]$ are relatively prime. Then we have

[E_{1} E_{2} : F] = [E_{1} : F] [E_{2} : F] .

Proof.

Both $[E_{1} : F]$ and $[E_{2} : F]$ divide $[E_{1} E_{2} : F]$ , so by their relative primality, their product $[E_{1} : F] [E_{2} : F]$ does as well. So we have $[E_{1} E_{2} : F] \geq [E_{1} : F] [E_{2} : F]$ , while Corollary 6.3.5 provides the opposite inequality. □

Definition 6.3.7. §

Let $n \geq 1$ . An $n$ th root of unity is an element of order dividing $n$ in the multiplicative group of a field.

That is, if $F$ is a field, $ζ \in F$ is an $n$ th root of unity if and only if $ζ^{n} = 1$ .

Example 6.3.8. §

Let $ω$ be a third root of unity in $ℂ$ that is not equal to $1$ . Note that $[ℚ (ω) : ℚ] = 2$ , since $ω^{2} + ω + 1 = 0$ . Then $\sqrt[3]{2}$ and $ω \sqrt[3]{2}$ are both cube roots of $2$ , and we have

ℚ (\sqrt[3]{2}, ω \sqrt[3]{2}) = ℚ (ω, \sqrt[3]{2})

We then see that

[ℚ (\sqrt[3]{2}) : ℚ] [ℚ (ω \sqrt[3]{2}) : ℚ] = 9,

while

[ℚ (ω, \sqrt[3]{2}) : ℚ] = [ℚ (\sqrt[3]{2}) : ℚ] [ℚ (ω) : ℚ] = 6

by Corollary 6.3.6.

More generally, we may define the compositum of a collection of fields.

Definition 6.3.9. §

Let ${E_{i} ∣ i \in I}$ be a collection of subfields of a field $K$ for some indexing set $I$ . Then the compositum of the fields $E_{i}$ for $i \in I$ is smallest subfield of $K$ containing all $E_{i}$ .

Let us give an alternate description of the compositum.

Lemma 6.3.10. §

Let ${E_{i} ∣ i \in I}$ be a collection of intermediate fields in an extension $K ∕ F$ for some indexing set $I$ . Then the compositum $E$ of the $E_{i}$ is equal to the union of its subfields $F (α_{1}, α_{2}, \dots, α_{n})$ , where $n \geq 0$ and each $α_{j}$ with $1 \leq j \leq n$ is an element of $E_{i}$ for some $i \in I$ .

Proof.

Clearly the above-described union $U$ is contained in $E$ and contains each $E_{i}$ . However, we must show that $U$ is a field, hence equal to $E$ . If $a, b \in U$ are nonzero, then $a \in F (α_{1}, α_{2}, \dots, α_{n})$ and $b \in F (β_{1}, β_{2}, \dots, β_{m})$ , where $n, m \geq 0$ and the $α_{j}$ and $β_{k}$ are elements of the $E_{i}$ . Then

a - b, a b^{- 1} \in F (α_{1}, α_{2}, \dots, α_{n}, β_{1}, β_{2}, \dots, β_{m}),

and the latter field is a subset of $U$ , so $U = E$ . □

We have the following corollary.

Corollary 6.3.11. §

Let ${E_{i} ∣ i \in I}$ be algebraic extensions of a field $F$ that are contained in a field $K$ , where $I$ is an indexing set. Then the compositum $E$ of the fields $E_{i}$ is an algebraic extension of $F$ .

Proof.

By Lemma 6.3.10, any $α \in E$ is an element of a subfield $F (α_{1}, α_{2}, \dots, α_{n})$ of $E$ , where each $α_{j} \in E_{i}$ for some $i \in I$ . Since $E_{i}$ is algebraic, $F (α_{j}) ∕ F$ is finite for all $1 \leq j \leq n$ , and therefore $F (α_{1}, α_{2}, \dots, α_{n}) ∕ F$ is finite by Corollary 6.3.5. □

6.4. Constructible numbers

In this section, we discuss a classical problem of the ancient Greeks, which we present as a game. The game begins with a line segment of length $1$ that has already been drawn on the plane. One is given two tools: a straightedge and a compass. At any step of the game, one can either use the straightedge to draw a line segment or the compass to draw a circle, in ways we will shortly make more specific. The goal of the game is to draw a line segment of a given desired length in a finite number of steps.

At any step, we consider a point to have been marked if it is either the endpoint of an already drawn line segment or the intersection of a drawn line segment or circle with another drawn line segment or circle. The straightedge allows us to draw a line segment between any two marked points and also to extend any previously drawn line segment until it meets any point that has already been drawn on the plane. The compass allows us to draw a circle that contains a given marked point and has as its center any other marked point.

Given these rules, we may now make the following definition.

Definition 6.4.1. §

A real number $α$ is said to be constructible if one can draw a line segment of length $| α |$ in the plane, starting from a line segment of length $1$ , using a straightedge and compass, in a finite number of steps.

We will denote a line segment between two distinct points $A$ and $B$ in $ℝ^{2}$ by $\bar{𝐴𝐵}$ . Its length will be denoted by $| \bar{𝐴𝐵} |$ . We prove a few preliminary results.

Lemma 6.4.2. §

Suppose that a line segment $\bar{𝐴𝐵}$ has been drawn in the plane.

a.: We may draw a line segment bisecting $\bar{𝐴𝐵}$ .
b.: We may draw a line segment $\bar{𝐴𝐶}$ perpendicular to $\bar{𝐴𝐵}$ .
c.: Given a point $D$ in the plane, we may draw a line segment $\bar{𝐷𝐸}$ parallel to $\bar{𝐴𝐵}$ .

Proof.

For part a, draw circles with center A and center B, both of radius $| \bar{𝐴𝐵} |$ . These intersect at two points, and the line segment between them is perpendicular to $\bar{𝐴𝐵}$ and passes through a midpoint $F$ of that segment.

For part b, by drawing the circle with center A and radius $| \bar{𝐴𝐹} |$ , we may mark a point $G$ on the line that contains $| \bar{𝐴𝐵} |$ that is on the opposite side of $A$ from $F$ and is such that $| \bar{𝐴𝐹} | = | \bar{𝐴𝐺} |$ . As we have already shown, we may then draw a line segment $\bar{𝐶𝐻}$ bisecting $\bar{𝐹𝐺}$ and passing through $A$ , which provides us with $\bar{𝐴𝐶}$ .

For part c, if $\bar{𝐵𝐷}$ is perpendicular to $\bar{𝐴𝐵}$ , we set $H = D$ . Otherwise, we draw a circle with center $D$ and passing through $B$ . It intersects the line containing $\bar{𝐴𝐵}$ in a second point $H$ . We draw a line segment through $D$ bisecting $\bar{𝐵𝐻}$ using part a. We then use part b to draw a perpendicular $\bar{𝐷𝐸}$ to $\bar{𝐵𝐻}$ , and it is by definition parallel to $\bar{𝐴𝐵}$ . □

We also have the following.

Lemma 6.4.3. §

Suppose we have drawn either a line segment of length $α$ or a circle of radius $α$ in the plane.

a.: We may draw a line segment of length $α$ with any marked point as an endpoint, along any line that contains at least one other marked point.
b.: We may draw a circle of radius $α$ with center any marked point.

Proof.

First we note that the two assumptions are equivalent. Given a line segment of length $α$ , we may use its endpoints to draw a circle of radius $α$ . Given a circle of radius $α$ and center $A$ , since we have at least one marked point other than its center in the plane, we can by drawing the line segment from the center to that point mark a point $B$ on the circle. The resulting line segment $B$ then has radius $α$ .

Suppose then that we are given a line segment $\bar{𝐴𝐵}$ of length $α$ and a marked point $C$ . We make two constructions using Lemma 6.4.2. We draw a line segment $\bar{𝐶𝐸}$ parallel to $\bar{𝐴𝐵}$ . We draw the line segment $\bar{𝐴𝐶}$ and then the parallel to $\bar{𝐴𝐶}$ passing through $B$ . It intersects the line through $C$ and $E$ at a point $D$ such that $| \bar{𝐶𝐷} | = α$ . The circle with center $C$ passing through $D$ then has radius $α$ . □

We prove the following.

Theorem 6.4.4. §

The set of constructible numbers is a subfield of $ℝ$ .

Proof.

Suppose that $α$ and $β$ are constructible and positive. Then we may draw a line segment $\bar{𝐴𝐵}$ of length $α$ in the plane, and we may then draw a line segment $\bar{𝐵𝐶}$ of length $β$ along the line defined by $\bar{𝐴𝐵}$ . If we do this so that it overlaps with $\bar{𝐴𝐵}$ , then we have constructed a line segment $\bar{𝐴𝐶}$ of length $| α - β |$ .

On the other hand, given $\bar{𝐴𝐵}$ of length $α$ , draw a line segment $\bar{𝐴𝐶}$ of length $β$ that is perpendicular to $\bar{𝐴𝐵}$ , and let $E$ be the point on the ray defined by $\bar{𝐴𝐶}$ such that $\bar{𝐴𝐸}$ has length $1$ . Draw the line segment $\bar{𝐶𝐵}$ , and use it to draw a parallel line segment from $E$ to a point $D$ on the ray defined by the segment $\bar{𝐴𝐵}$ . We then have that the triangle $𝐴𝐵𝐶$ is similar to the triangle $𝐴𝐷𝐸$ , so

| \bar{𝐴𝐷} | = \frac{| \bar{𝐴𝐷} |}{| \bar{𝐴𝐸} |} = \frac{| \bar{𝐴𝐵} |}{| \bar{𝐴𝐶} |} = \frac{α}{β} .

Therefore, $α β^{- 1}$ is constructible. □

Theorem 6.4.5. §

The field of constructible numbers consists exactly of the real numbers that can be obtained from $1$ by applying a finite sequence of the operations of addition, subtraction, multiplication, division (with nonzero denominators), and the taking of square roots (of positive numbers), using numbers already obtained from $1$ at an earlier point in the sequence.

Proof.

We first show that the square root of a constructible positive number $α$ is constructible. For this, draw a line segment $\bar{𝐴𝐷}$ of length $1 + α$ and mark a point $B$ at distance $1$ from $A$ and $α$ from $D$ along the segment. Find the midpoint $O$ of $\bar{𝐴𝐷}$ , and draw a circle with center $O$ and radius $| \bar{𝐴𝑂} | = (1 + α) ∕ 2$ . Draw a perpendicular to $\bar{𝐴𝐷}$ at the point $B$ , and let $C$ be a point where it intersects the drawn circle. Then the triangle $𝐴𝐵𝐶$ is similar to the triangle $𝐶𝐵𝐷$ , and therefore we have

| \bar{𝐵𝐶} | = \frac{| \bar{𝐵𝐶} |}{| \bar{𝐴𝐵} |} = \frac{| \bar{𝐵𝐷} |}{| \bar{𝐵𝐶} |} = \frac{α}{| \bar{𝐵𝐶} |},

and hence $| \bar{𝐵𝐶} | = \sqrt{α}$ .

Let $E$ be the set (or actually, field) of numbers that can be constructed from $1$ using field operations and square roots. Suppose that our initial line segment was between $(0, 0)$ and $(1, 0)$ on the plane. Suppose that all previously marked points have coordinates in $E$ . These points have been marked as the intersection points of lines and circles, where the lines are determined by previously marked points with $E$ -coordinates and the circles have centers previously marked points with $E$ -coordinates and are chosen to pass through marked points with $E$ -coordinates. Every drawn line thus has the form $𝑎𝑥 + 𝑏𝑦 + c = 0$ with $a, b, c \in E$ , and every drawn circle has the form $x^{2} + y^{2} + 𝑑𝑥 + 𝑒𝑦 + f = 0$ with $d, e, f \in E$ . The intersection of two such lines is easily seen to have coordinates obtained by field operations on the coefficients of the two lines in question. The coordinates of the intersection points of a line and a circle coming from the solution of a quadratic equation with coefficients are obtained by field operations on the coefficients of the line and the circle. Without loss of generality, we may suppose that our circle is $x^{2} + y^{2} = 1$ . The $x$ -coordinate of the intersection points satisfy $(a + 2 + b^{2}) x^{2} - 2 𝑎𝑐𝑥 + c^{2} - b^{2} = 0$ , which is a quadratic equation with cooefficients in $E$ . Finally, the intersection points of two circles, one of the above form and one of the form $x^{2} + y^{2} + d^{'} x + e^{'} y + f^{'} = 0$ are the intersection of the line $(d - d^{'}) x + (e - e^{'}) y + (f - f^{'}) = 0$ with either circle, which means said line is the common chord. These points have coordinates in $E$ by the previous case. □

Since the square root of a field element defines an extension of degree dividing $2$ of the field in which it lies, we have the following.

Corollary 6.4.6. §

Let $α$ be a constructible number. Then $α$ is an algebraic number, and $[ℚ (α) : ℚ]$ is a power of $2$ .

Corollary 6.4.7. §

The field of constructible numbers is an algebraic extension of $ℚ$ .

The ancient Greeks were in particular very concerned with three problems that they could not solve with a straightedge and compass. This was for good reason: they involved constructing line segments of unconstructible length. Yet, the Greeks never managed to prove this, and it was not until the 19th century that proofs were finally given. We list these three problems now.

Examples 6.4.8. §

a.

It is impossible to “double the cube.” That is, given a line segment, one cannot construct from it a new line segment such that a cube with the new line segment as one of its sides would have twice the volume of a cube with the original line segment as its side. Assuming the initial line segment had a constructible length $α$ , the new line segment would have to have length $\sqrt[3]{2} α$ , but then $\sqrt[3]{2}$ would be constructible, yet it defines an extension of degree $3$ over $ℚ$ , in contradiction to Corollary 6.4.6.

b.

It is impossible to “square the circle.” That is, given a drawn circle, it is impossible to construct a square with the same area. If the circle had radius $r$ , then the square would have side length $\sqrt{π} r$ , which would mean that $\sqrt{π}$ would be constructible, and hence $π$ would be as well, in contradiction to Corollary 6.4.6, since $π$ is transcendental.

c.

It is impossible to “trisect all angles.” That is, given an arbitrary angle between two drawn line segments with a common endpoint in a plane, it is not always possible to draw a line segment with the same endpoint having an angle with one of the line segments that is a third of the original angle. Note that an initial such angle $𝜃$ exists if and only if $\cos 𝜃$ is constructible, as seen by drawing a perpendicular from one line segment at point a distance one from the point of intersection until it intersects the line defined by the other. Therefore, the problem is, given a constructible number $α = \cos 𝜃$ , to show that $\cos (𝜃 ∕ 3)$ is constructible. However, we have a trigonometric identity

\cos 𝜃 = 4 \cos^{3} (𝜃 ∕ 3) - 3 \cos (𝜃 ∕ 3) .

Suppose that $𝜃 = π ∕ 3$ . Then $\cos (π ∕ 3) = \frac{1}{2}$ , and $\cos (π ∕ 9)$ would be a root of the polynomial $8 x^{3} - 6 x - 1$ , which is irreducible over $ℚ$ since it is irreducible in $ℤ [x]$ . (It has no roots, even modulo $2$ .) But then $\cos (π ∕ 9)$ would define a degree $3$ extension of $ℚ$ , contradicting Corollary 6.4.6 again.

6.5. Finite fields

In this section, we classify all finite fields, which is to say, fields of finite order.

Notation 6.5.1. §

We use $𝔽_{p}$ to denote $ℤ ∕ 𝑝ℤ$ when we consider it as a field.

Proposition 6.5.2. §

Every finite field contains $p^{n}$ elements for some $n \geq 1$ .

Proof.

Let $F$ be a finite field. Since it is finite, it has characteristic $p$ for some prime number $p$ , which means that it contains the field $𝔽_{p}$ , and moreover is a finite dimensional vector space over $𝔽_{p}$ . Therefore, $F$ has a finite $ℤ ∕ 𝑝ℤ$ -basis ${α_{1}, α_{2}, \dots, α_{n}}$ , so that the elements of $F$ are exactly the elements $c_{1} α_{1} + c_{2} α_{2} + \dots + c_{n} α_{n}$ with $c_{1}, c_{2}, \dots, c_{n} \in ℤ ∕ 𝑝ℤ$ . We therefore have $| F | = p^{n}$ . □

Definition 6.5.3. §

Let $F$ be a field and $n$ be a positive integer. The group $μ_{n} (F)$ of $n$ th roots of unity in $F$ is the subgroup of $F^{\times}$ with elements the $n$ th roots of $1$ in $F$ .

Lemma 6.5.4. §

Let $F$ be a field and $n$ be a positive integer. Then $μ_{n} (F)$ is a cyclic group of order dividing $n$ .

Proof.

Every element in $μ_{n} (F)$ has order dividing $n$ . Let $m$ be the exponent of $μ_{n} (F)$ . Then every element of $μ_{n} (F)$ is an $m$ th root of unity, so is a root of $x^{m} - 1$ , and hence the order of $μ_{n} (F)$ is at most $m$ . On the other hand, since $m$ is the exponent, the classification of finite abelian groups tells us that $μ_{n} (F)$ contains an element of order $m$ , so therefore $μ_{n} (F)$ is cyclic of order $m$ , which divides $n$ . □

Proposition 6.5.5. §

Let $F$ be a finite field of order $p^{n}$ for some prime $p$ and $n \geq 1$ . Then $F^{\times}$ is cyclic, and its multiplicative group is equal to $μ_{p^{n} - 1} (F)$ .

Proof.

Since $| F^{\times} | = p^{n} - 1$ , every element of $F^{\times}$ is a root of the polynomial $x^{p^{n} - 1} - 1$ , and conversely. Therefore, it follows from Lemma 6.5.4 that $F^{\times} = μ_{p^{n} - 1} (F)$ is cyclic. □

Corollary 6.5.6. §

The group ${(ℤ ∕ 𝑝ℤ)}^{\times}$ of units in $ℤ ∕ 𝑝ℤ$ is cyclic of order $p - 1$ .

Example 6.5.7. §

The cyclic group ${(ℤ ∕ 17 ℤ)}^{\times}$ of order $16$ is generated by $3$ .

Lemma 6.5.8. §

Let $F$ be a field of characteristic a prime $p$ , and let $α, β \in F$ . Then we have

{(α + β)}^{p^{n}} = α^{p^{n}} + β^{p^{n}}

for all $n \geq 0$ .

Proof.

It is easy to see that $(\binom{p}{i}) \equiv 0 𝑚𝑜𝑑 p$ for $1 \leq i \leq p - 1$ , and so we have the result for $n = 1$ by the binomial theorem. By induction, the result for general $n$ follows immediately. □

Theorem 6.5.9. §

Let $n$ be a positive integer. There exists a field $𝔽_{p^{n}}$ of order $p^{n}$ containing $𝔽_{p}$ , and it is unique up to isomorphism. Moreover, if $E$ is a finite field extension of $𝔽_{p}$ of degree a multiple of $n$ , then $E$ contains a unique subfield isomorphic to $𝔽_{p^{n}}$ .

Proof.

Let $F$ be the set of roots of $x^{p^{n}} - x$ in a splitting field $Ω$ of $x^{p^{n}} - x$ over $𝔽_{p}$ . If $α, β \in F$ are nonzero, then clearly ${(α β^{- 1})}^{p^{n}} = α β^{- 1}$ , so $α β^{- 1} \in F$ . Moreover, we have ${(α - β)}^{p^{n}} = α^{p^{n}} - β^{p^{n}}$ by Lemma 6.5.8, so ${(α - β)}^{p^{n}} = α - β$ . It follows that $F$ is a field in which $x^{p^{n}} - x$ splits, so it equals $Ω$ .

Now, $F$ has at most $p^{n}$ elements by definition. We must show that has exactly $p^{n}$ elements, so that its degree is $n$ over $𝔽_{p}$ . Clearly $x$ factors into $x^{p^{n}} - x$ exactly once. Let $a \in F^{\times}$ , and set

g (x) = \frac{x^{p^{n}} - x}{x - a} = \sum_{i = 1}^{p^{n} - 1} a^{i - 1} x^{p^{n} - i} .

Then we have

g (a) = \sum_{i = 1}^{p^{n} - 1} a^{p^{n} - 1} = (p^{n} - 1) a^{p^{n} - 1} = - 1 \neq 0,

so $x - a$ is not a factor of $g (x)$ , and therefore all roots of $x^{p^{n}} - x$ are distinct.

We prove the remaining claims. First, any finite field extension of $𝔽_{p}$ of degree a multiple $m$ of $n$ has $p^{m}$ elements, and Proposition 6.5.5 then implies that it consists of roots of $x^{p^{m}} - x$ . In particular, it contains a unique subfield of degree $n$ consisting of the roots of $x^{p^{n}} - x$ . Next, note that $𝐹 ≅ 𝐹 [x] ∕ (f)$ , where $f$ is the minimal polynomial of a generator of $μ_{p^{n} - 1} (F)$ . Given any other field $F^{'}$ of order $p^{n}$ , it also consists of the roots of $x^{p^{n}} - x$ , so contains a root of $f$ . This root then generates $F^{'}$ , being a primitive $(p^{n} - 1)$ th root of unity, so $F^{'} ≅ 𝐹 [x] ∕ (f)$ as well. □

Remark 6.5.10. §

Since $𝔽_{p^{n}}$ has order $p^{n}$ and is an $𝔽_{p}$ -vector space, we have $[𝔽_{p^{n}} : 𝔽_{p}] = n$ .

Corollary 6.5.11. §

The field $𝔽_{p^{n}}$ contains a subfield isomorphic to $𝔽_{p^{m}}$ if and only if $m$ divides $n$ .

From now on, for a prime $p$ and a positive integer $n$ , we will speak of $𝔽_{p^{n}}$ as being the unique (up to isomorphism) field of order $p^{n}$ .

Example 6.5.12. §

The field $𝔽_{9}$ consists of $0$ and $8$ th roots of unity. We have $𝔽_{9} = 𝔽_{3} (ζ)$ , where $ζ$ is a primitive $8$ th root of unity (or even a primitive fourth root of unity), so a root of $x^{4} + 1$ . Since $[𝔽_{9} : 𝔽_{3}] = 2$ , the minimal polynomial of $ζ$ must be of degree $2$ . Over $𝔽_{3}$ , we have only three irreducible polynomials of degree two: $x^{2} + 1$ , $x^{2} + x - 1$ and $x^{2} - x - 1$ . The product of the latter two is $x^{4} + 1$ , which is to say that the 2 of the primitive $8$ th roots of unity have minimal polynomial $x^{2} + x - 1$ and the other two $x^{2} - x - 1$ . On the other hand, we have $𝔽_{9} = 𝔽_{3} (ζ^{2})$ as well, and $ζ^{2}$ is a primitive $4$ th root of unity with minimal polynomial $x^{2} + 1$ .

The following result is rather useful.

Proposition 6.5.13. §

Let $q$ be a power of a prime $p$ . Let $m \geq 1$ , and let $ζ_{m}$ be a primitive $m$ th root of unity in an extension of $𝔽_{p}$ . Then $[𝔽_{q} (ζ_{m}) : 𝔽_{q}]$ is the order $k$ of $q$ in ${(ℤ ∕ 𝑚ℤ)}^{\times}$ . In other words, we have $𝔽_{q} (ζ_{m}) = 𝔽_{q^{k}}$ .

Proof.

Let $k = [𝔽_{q} (ζ_{m}) : 𝔽_{q}]$ . Then $𝔽_{q} (ζ_{m}) = 𝔽_{q^{k}}$ , and so $m$ divides $q^{k} - 1$ , and then $q$ has order dividing $k$ in ${(ℤ ∕ 𝑚ℤ)}^{\times}$ . On the other hand, since $𝔽_{q} (ζ_{m})$ is not contained in $𝔽_{q^{j}}$ for any $j < k$ , we have that $q^{j}$ is not $1$ in ${(ℤ ∕ 𝑚ℤ)}^{\times}$ . That is, $q$ has the desired order $k$ modulo $m$ . □

In order to apply the previous result, it is useful to understand the structure of the unit group of $ℤ ∕ 𝑚ℤ$ .

Proposition 6.5.14. §

Let $m \geq 2$ be an integer, and write $m = p_{1}^{r_{1}} p_{2}^{r_{2}} \dots p_{k}^{r_{k}}$ for distinct primes $p_{i}$ and positive integers $r_{i}$ for $1 \leq i \leq k$ , for some $k \geq 1$ . Then

{(ℤ ∕ 𝑚ℤ)}^{\times} ≅ \prod_{i = 1}^{k} {(ℤ ∕ p_{i}^{r_{i}} ℤ)}^{\times} .

Moreover, if $p$ is a prime number and $r$ is a positive integer, we have

{(ℤ ∕ p^{r} ℤ)}^{\times} ≅ {\begin{matrix} ℤ ∕ (p - 1) ℤ \times ℤ ∕ p^{r - 1} ℤ & if p is odd \\ ℤ ∕ 2 ℤ \times ℤ ∕ 2^{r - 2} ℤ & if p = 2 and r \geq 2 . \end{matrix}

Proof.

The first statement is a corollary of the Chinese remainder theorem for $ℤ$ . The reduction map ${(ℤ ∕ p^{r} ℤ)}^{\times} \to {(ℤ ∕ 𝑝ℤ)}^{\times} ≅ ℤ ∕ (p - 1) ℤ$ (noting Corollary 6.5.6) then has kernel the multiplicative group $(1 + 𝑝ℤ) ∕ (1 + p^{r} ℤ)$ of order $p^{r - 1}$ . If $p$ is odd, then ${(1 + p)}^{p^{i - 1}} - 1 \equiv p^{i} mod p^{i + 1}$ by the binomial theorem, so $1 + p$ has order $p^{r - 1}$ in the group. If $p = 2$ and $r \geq 2$ , then $5 = 1 + 4$ similarly generates the subgroup $(1 + 4 ℤ) ∕ (1 + 2^{r} ℤ)$ of order $2^{r - 2}$ . Clearly, this group does not contain $- 1$ , which has order $2$ . That is, $(1 + 2 ℤ) ∕ (1 + 2^{r} ℤ)$ is generated by the images of $- 1$ and $5$ and so is isomorphic to $ℤ ∕ 2 ℤ \times ℤ ∕ 2^{r - 2} ℤ$ . □

6.6. Cyclotomic fields

Let us explore the extensions of $ℚ$ generated by roots of unity, known as cyclotomic fields.

Notation 6.6.1. §

Let $n \geq 1$ . We will use $ζ_{n}$ to denote a primitive $n$ th root of unity in an extension of $ℚ$ . We can and therefore do choose these so that $ζ_{n}^{n ∕ m} = ζ_{m}$ if $m$ divides $n$ . For instance, one could take $ζ_{n} = e^{2 𝜋𝑖 ∕ n} \in ℂ$ .

Definition 6.6.2. §

Let $n \geq 1$ . Then $n$ th cyclotomic field is the extension of $ℚ$ generated by a primitive $n$ th root of unity $ζ_{n}$ .

Remark 6.6.3. §

The $n$ th cyclotomic field $ℚ (ζ_{n})$ is the splitting field of $x^{n} - 1$ in that all of the roots of $x^{n} - 1$ are powers of $ζ_{n}$ .

Definition 6.6.4. §

The $n$ th cyclotomic polynomial $Φ_{n}$ is the unique monic polynomial in $ℂ [x]$ with roots the primitive $n$ th roots of unity.

In Example 5.3.4, we saw that every

Φ_{p} = 1 + x + \dots + x^{p - 1},

where $p$ is prime, is irreducible using the Eisenstein criterion.

Remarks 6.6.5. §

Let $n$ be a positive integer.

a.

We have

x^{n} - 1 = \prod_{d ∣ n} Φ_{d},

with the sum taken over positive divisors of $n$ .

b.

Since

Φ_{n} = \frac{x^{n} - 1}{\prod_{\begin{array}{c} d ∣ n \\ d < n \end{array}} Φ_{d}},

we have by induction on $n$ that $Φ_{n} \in ℤ [x]$ .

c.

We have

Φ_{n} = \prod_{\begin{array}{c} i = 1 \\ \gcd (i, n) = 1 \end{array}}^{n} (x - ζ_{n}^{i}),

and therefore $Φ_{n}$ has degree $φ (n)$ , where $φ$ is the Euler-phi function. In particular, we have $\deg Φ_{n} = φ (n)$ .

Definition 6.6.6. §

The Möbius function $μ : ℤ_{> 0} \to {- 1, 0, 1}$ is defined by

μ (n) = {\begin{matrix} {(- 1)}^{k} & if n is a product of k distinct primes, where k \geq 0, \\ 0 & otherwise . \end{matrix}

We note the following.

Lemma 6.6.7. §

For any $n \geq 2$ , one has $\sum_{d ∣ n} μ (d) = 0$ .

Proof.

Since $μ (d)$ is zero if $d$ is divisible by a square of a prime, we have $\sum_{d ∣ n} μ (d) = \sum_{d ∣ m} μ (d)$ , where $m$ is the product of the primes dividing $n$ . If there are $k$ such primes, then there are $(\binom{k}{j})$ products of $j$ of them, each of which contributes ${(- 1)}^{j}$ to the sum. In other words,

\sum_{d ∣ n} μ (d) = \sum_{j = 0}^{k} (\binom{k}{j}) {(- 1)}^{j} = {(1 - 1)}^{k} = 0,

since $k \geq 1$ . □

Theorem 6.6.8 (Möbius inversion formula). §

Let $A$ be an abelian group and $f : ℤ_{> 0} \to A$ a function. Define $g : ℤ_{> 0} \to A$ by

g (n) = \sum_{d ∣ n} f (d)

for $n \geq 1$ . Then

f (n) = \sum_{d ∣ n} μ (d) g (\frac{n}{d}) .

Proof.

We calculate

\sum_{d ∣ n} μ (\frac{n}{d}) g (d) = \sum_{d ∣ n} \sum_{k ∣ d} μ (\frac{n}{d}) f (k) = \sum_{k ∣ n} \sum_{\begin{array}{c} d ∣ n \\ k ∣ d \end{array}} μ (\frac{n}{d}) f (k) = \sum_{k ∣ n} \sum_{c ∣ \frac{n}{k}} μ (\frac{n}{𝑘𝑐}) f (k) = f (n),

the last step by Lemma 6.6.7. □

Taking $A = ℚ {(x)}^{\times}$ , we have the following.

Lemma 6.6.9. §

Let $n \geq 1$ . Then

Φ_{n} = \prod_{\begin{array}{c} d ∣ n \\ d \geq 1 \end{array}} {(s^{n ∕ d} - 1)}^{μ (d)} .

The lemma can be used to calculate cyclotomic polynomials explicitly.

Examples 6.6.10. §

a.

We have $Φ_{1} = x - 1$ .

b.

For a prime $p$ and $k \geq 1$ , we have

Φ_{p^{k}} = \frac{x^{p^{k}} - 1}{x^{p^{k - 1}} - 1} = \sum_{i = 0}^{p - 1} x^{i p^{k - 1}} .

c.

For $p$ and $q$ distinct primes, we have

Φ_{𝑝𝑞} = \frac{(x^{𝑝𝑞} - 1) (x - 1)}{(x^{p} - 1) (x^{q} - 1)} = \frac{Φ_{q} (x^{p})}{Φ_{q} (x)}

For instance, taking $q = 2$ we obtain

Φ_{2 p} = \frac{x^{p} + 1}{x + 1} = Φ_{p} (- x),

and we have

Φ_{15} = x^{8} - x^{7} + x^{5} - x^{4} + x^{3} - x + 1 .

The $n$ th cyclotomic polynomial is in fact irreducible over $ℚ$ .

Theorem 6.6.11. §

Let $n \geq 1$ . Then the cyclotomic polynomial $Φ_{n}$ is irreducible in $ℚ [x]$ .

Proof.

Write $Φ_{n} = 𝑓𝑔$ with $f, g \in ℤ [x]$ and $f$ monic irreducible with $ζ$ as a root. Take any prime $p$ not dividing $n$ , and note that $ζ^{p}$ is also a root of $Φ_{n}$ .

If $ζ^{p}$ is a root of $g$ , then $g (x^{p})$ is divisible by the minimal polynomial $f (x)$ of $ζ$ . Let $\bar{f}$ and $\bar{g}$ denote the reductions modulo $p$ of $f$ and $g$ respectively. Then $\bar{g} (x^{p}) = \bar{g} {(x)}^{p} \in 𝔽_{p} [x]$ is divisible by $\bar{f} (x)$ , so $\bar{g}$ and $\bar{f}$ have a common factor. The reduction $ϕ_{n} = \bar{f} \bar{g}$ of $Φ_{n}$ modulo $p$ therefore has a multiple root in $\bar{𝔽_{p}}$ . In particular, $x^{n} - 1$ has a multiple root, but we know that it does not. That is, if we choose $k \geq 1$ so that $p^{k} \equiv 1 mod n$ , then the cyclic group $𝔽_{p^{k}}^{\times}$ of order $p^{k} - 1$ contains $n$ distinct $n$ th roots of unity.

Thus, $ζ^{p}$ is a root of $f$ for any prime $p$ and any root $ζ$ of $f$ . Since any integer $a$ prime to $n$ can be written as a product of primes not dividing $n$ , it follows that $ζ^{a}$ is a root of $f$ for all $a$ prime to $p$ . This forces $f = Φ_{n}$ , so $Φ_{n}$ is irreducible. □

6.7. Field embeddings

Definition 6.7.1. §

Let $E$ and $E^{'}$ be extensions of a field $F$ , and let $φ : E \to E^{'}$ be an isomorphism of fields. We say that $φ$ fixes $F$ if $φ (α) = α$ for all $α \in F$ .

Definition 6.7.2. §

Let $α$ and $β$ be elements of field extensions of a field $F$ . We say that $α$ and $β$ are conjugate over $F$ if there exists a field isomorphism $φ : F (α) \to F (β)$ fixing $F$ such that $φ (α) = β$ .

Proposition 6.7.3. §

Let $E$ and $E^{'}$ be extensions of a field $F$ , and let $α \in E$ , $β \in E^{'}$ be algebraic over $F$ . Then $α$ and $β$ are conjugate over $F$ if and only if the minimal polynomials of $α$ and $β$ in $F [x]$ are equal.

Proof.

Suppose that $α$ and $β$ are conjugate over $F$ , and let $φ : F (α) \to F (β)$ be a field isomorphism such that $φ (α) = β$ and $φ$ restricts to the identity map on $F$ . Then $φ (g (α)) = g (β)$ for all $g \in F [x]$ . Let $f \in F [x]$ be the minimal polynomial of $α$ . Then we have

0 = φ (0) = φ (f (α)) = f (β),

so $β$ is a root of $f$ . As $f$ is irreducible, it must be the minimal polynomial of $β$ .

Conversely, suppose that $α$ and $β$ have the same minimal polynomial $f \in F [x]$ . Then we have isomorphisms from $F [x] ∕ (f)$ to $F (α)$ and $F (β)$ as in Theorem 6.1.15, and composing the inverse of the first with the latter yields the desired isomorphism $F (α) \to F (β)$ . □

Example 6.7.4. §

Since $i$ and $- i$ are both roots of the irreducible polynomial $x^{2} + 1$ over $ℝ$ , they are conjugate elements of $ℂ$ . Therefore, there is a field isomorphism $ℂ \to ℂ$ that takes $i$ to $- i$ and fixes $ℝ$ . Such an isomorphism must take $a + 𝑏𝑖$ to its complex conjugate

\bar{a + 𝑏𝑖} = a - 𝑏𝑖

and is therefore the usual complex conjugation. In particular, complex conjugation is an isomorphism of fields, which is also easily verified directly. Moreover, we see that if $f \in ℝ [x]$ has a root $α$ , then $\bar{α}$ is a root as well, since $\bar{f (α)} = f (\bar{α})$ .

Definition 6.7.5. §

An embedding of fields, or field embedding, is a ring homomorphism $φ : F \to F^{'}$ , where $F$ and $F^{'}$ are fields.

Remark 6.7.6. §

Any ring homomorphism between fields is injective, so field embeddings are injective.

Definition 6.7.7. §

Let $φ : F \to M$ be a field embedding, and let $E ∕ F$ be an extension field. We say that a field embedding $Φ : E \to M$ extends $φ$ , and is an extension of $φ$ , if $Φ |_{F} = φ$ .

Example 6.7.8. §

We have a field embedding $ι : ℚ \to ℝ$ . There are two field embeddings $ι^{'} : ℚ (\sqrt{2}) \to ℝ$ extending $ι$ . Either we take $ι^{'} (a + b \sqrt{2}) = a + b \sqrt{2}$ for $a, b \in ℚ$ , or we set $ι^{'} (a + b \sqrt{2}) = a - b \sqrt{2}$ . On the other hand, there is no field embedding $κ : ℚ (i) \to ℝ$ extending $ι$ , since there is no element of $i$ that would satisfy $κ {(i)}^{2} + 1 = 0$ , but there is no element of $ℝ$ with this property.

Let us give a slight extension of one direction of Proposition 6.7.3.

Theorem 6.7.9. §

Let $E ∕ F$ be a field extension, and let $α \in E$ be algebraic over $F$ . Let $φ : F \to M$ be a field embedding, and consider the induced map $\tilde{φ} : F [x] \to M [x]$ . Let $f \in F [x]$ be the minimal polynomial of $α$ . Then there is a bijection between the set of field embeddings $F (α) \to M$ extending $φ$ and the set of roots of $\tilde{φ} (f)$ in $M$ taking an extension $Φ$ of $φ$ to $Φ (α)$ .

Proof.

Suppose that $β$ is a root of $\tilde{φ} (f)$ . Let $e_{β} : M [x] \to M$ denote the evaluation map at $β$ . The composition $e_{β} \circ \tilde{φ}$ has kernel containing $(f)$ , and the kernel then equals $(f)$ by the maximality of $(f)$ and the fact that the composition is nonzero. The first isomorphism theorem yields a field embedding $F [x] ∕ (f) \to M$ sending the coset of $x$ to $β$ . The map $Φ$ is then obtained by composing with the isomorphism $F (α) \to F [x] ∕ (f)$ of Theorem 6.1.15, and it sends $α$ to $β$ . Moreover, if $κ$ is any other lift of $φ$ such that $κ (α) = β$ , we have

κ (\sum_{i = 0}^{\deg f - 1} c_{i} α^{i}) = \sum_{i = 0}^{\deg f - 1} φ (c_{i}) β^{i} = Φ (\sum_{i = 0}^{\deg f - 1} c_{i} α^{i})

for all $c_{i} \in F$ for $1 \leq i < \deg f$ , so $κ = Φ$ .

Conversely, suppose $Φ : F (α) \to M$ is an extension of $φ$ . Then we have $\tilde{φ} (f) (Φ (α)) = Φ (f (α)) = 0$ . □

Corollary 6.7.10. §

Let $E ∕ F$ be a field extension, let $α \in E$ be algebraic over $F$ , and let $φ : F \to M$ be a field embedding. Let $\tilde{φ} : F [x] \to M [x]$ denote the induced map on polynomial rings. The number of extensions of $φ$ to an embedding $Φ : F (α) \to M$ is the number of distinct roots of $\tilde{φ} (f)$ in $M$ , where $f \in F [x]$ is the minimal polynomial of $α$ .

Remark 6.7.11. §

In the setting of Corollary 6.7.10, we may identify $F$ with its isomorphic image $φ (F)$ . This allows us to think of $F$ as a subfield of $M$ . In this case, $f \in F [x]$ may be thought of as itself having roots in $M$ , and the number of embeddings of $F (α)$ in $M$ is the number of distinct roots of $f$ in $M$ .

In general, for finite extensions, we have the following.

Corollary 6.7.12. §

Let $E ∕ F$ be a finite extension of fields. Let $φ : F \to M$ be a field embedding. Then the number of extensions $Φ : E \to M$ of $F$ is finite, less than or equal to $[E : F]$ .

Proof.

Since any finite extension $E ∕ F$ is finitely generated, it suffices by the multiplicativity of degrees of field extensions in Corollary 6.2.12 and recursion to prove the result in the case that $E = F (α)$ for some $α \in E$ . In this case, the degree of the minimal polynomial of $α$ is equal to $[E : F]$ and is greater than or equal to the number of distinct roots in $M$ of the image of the minimal polynomial of $α$ . The result is therefore a consequence of Corollary 6.7.10. □

Example 6.7.13. §

As seen in Example 6.7.8, there are exactly two embeddings of $ℚ (\sqrt{2})$ in $ℝ$ , but no embeddings of $ℚ (i)$ in $ℝ$ .

Example 6.7.14. §

There are four embeddings of $ℚ (\sqrt{2}, \sqrt{3})$ in $ℝ$ . If $φ$ is such an embedding, then we have $φ (\sqrt{2}) = \pm \sqrt{2}$ and $φ (\sqrt{3}) = \pm \sqrt{3}$ , and the signs determine the embedding uniquely.

Finally, we note the following.

Proposition 6.7.15. §

Let $E ∕ F$ be an algebraic field extension, and let $σ : E \to E$ be a field embedding fixing $F$ . Then $σ$ is an isomorphism.

Proof.

Let $β \in E$ , and let $f \in F [x]$ be its minimal polynomial. By Proposition 6.7.3, every root of $f$ in $E$ is sent by $σ$ to another root of $f$ in $E$ . As $σ$ is injective and the set of roots of $f$ in $E$ is finite, $σ$ permutes these roots. In particular, there exists a root $α$ of $f$ in $E$ such that $σ (α) = β$ . Therefore, we have $σ (E) = E$ , as desired. □

6.8. Algebraically closed fields

We begin with the notion of an algebraically closed field.

Definition 6.8.1. §

A field $L$ is algebraically closed if contains a root of every nonconstant polynomial $f \in L [x]$ .

The following theorem has analytic, topological, geometric, and algebraic proofs (though all in a sense require some very basic analysis).

Theorem 6.8.2 (Fundamental theorem of algebra). §

The field $ℂ$ of complex numbers is algebraically closed.

We defer an algebraic proof of this theorem until after our treatment of Galois theory. For the reader’s enjoyment, here are sketches of three proofs which require some knowledge of subjects outside of this course. The first two use complex analysis:

Remark 6.8.3. §

if $p \in ℂ [x]$ has no roots in $ℂ$ , then $p^{- 1}$ is entire. It is also bounded as a function on $ℂ$ , being the inverse of a polynomial. Hence, it is constant by Liouville’s theorem.

Remark 6.8.4. §

A nonconstant polynomial $p \in ℂ [x]$ defines a nonconstant continuous map from the Riemann sphere $ℙ^{1} (ℂ)$ to itself. Its image is closed as $ℙ^{1} (ℂ)$ is compact Hausdorff, while its image is open by the holomorphicity of $p$ and the open mapping theorem, so the image is $ℙ^{1} (ℂ)$ .

Next, algebraic topology:

Remark 6.8.5. §

Suppose that $p \in ℂ [x]$ is monic of degree $n$ and has no roots in $ℂ$ , and choose $r > 0$ such that $| p (z) - z^{n} | < r^{n}$ for all $z \in ℂ$ with $| z | = r$ . The map $F : S^{1} \to S^{1}$ with

F (z) = \frac{p (𝑟𝑧)}{| p (𝑟𝑧) |}

is homotopic to $z \mapsto z^{n}$ by a homotopy $H : [0, 1] \times S^{1} \to S^{1}$ given by

H (t, z) = \frac{(1 - t) (p (𝑟𝑧) - {(𝑟𝑧)}^{n}) + {(𝑟𝑧)}^{n}}{| (1 - t) p (𝑟𝑧) + t {(𝑟𝑧)}^{n} |} .

Now $F$ extends to a map $ℂ \to S^{1}$ on the simply connected space $ℂ$ by the same formula, so induces the zero map on $π_{1} (S^{1}) ≅ ℤ$ . But $z \mapsto z^{n}$ induces multiplication by $n$ on $π_{1} (S^{1})$ , so $n = 0$ .

Proposition 6.8.6. §

Let $L$ be an algebraically closed field, and let $f \in L [x]$ be nonconstant. Then $f$ splits in $L$ .

Proof.

We prove this by induction, as it is clear for $\deg f = 1$ . Suppose we know the result for all polynomials of degree less than $n = \deg f$ . Since $f$ has a root $α$ in $L$ , we have $f = (x - α) g$ for some $g \in L [x]$ of degree $n - 1$ . By induction, $g$ factors into linear terms. □

Corollary 6.8.7. §

Let $M$ be an algebraic extension of an algebraically closed field $L$ . Then $M = L$ .

Proof.

Let $α \in M$ . As $M$ is algebraic over $L$ , there exists a nonconstant $f \in L [x]$ with $f (α) = 0$ , and by Proposition 6.8.6, the polynomial $f$ is divisible by $x - α$ in $L [x]$ (recalling that $M [x]$ is a UFD). Therefore, we have $α \in L$ . □

We next show that extensions of field embeddings into algebraically closed fields always exist, when the extension is algebraic.

Theorem 6.8.8. §

Let $E ∕ F$ be an algebraic extension of fields. Let $φ : F \to M$ be a field embedding, where $M$ is an algebraically closed field. Then there exists a field embedding $Φ : E \to M$ extending $φ$ .

Proof.

Let $X$ denote the nonempty set of all pairs $(K, σ)$ , where $K$ is an intermediate subfield of $E ∕ F$ and $σ : K \to M$ is an extension of $φ$ . We say that $(K, σ) \leq (K^{'}, σ^{'})$ for $(K, σ)$ and $(K^{'}, σ^{'}) \in X$ if $K^{'}$ contains $K$ and $σ^{'} |_{K} = σ$ . Let $C$ be a chain in $X$ , set

L = ⋃_{(K, σ) \in C} K

and define $τ : L \to M$ by $τ |_{K} = σ$ for all $(K, σ) \in C$ . It is easy to see that $τ$ is a well-defined field embedding, since $C$ is a chain, and therefore, $(L, τ) \in X$ is an upper bound for $C$ .

By Zorn’s lemma, we therefore have that $X$ contains a maximal element, which we call $(Ω, λ)$ . We claim that $E = Ω$ . To see this, let $α \in E$ , and let $f \in Ω [x]$ be the minimal polynomial of $α$ over $Ω$ . If $f = \sum_{i = 0}^{n} a_{i} x^{i}$ with $a_{i} \in Ω$ for $0 \leq i \leq n$ and $n = \deg f$ , then we set

g = \sum_{i = 0}^{n} λ (a_{i}) x^{i} .

Since $M$ is algebraically closed, $g$ has a root $β$ in $M$ . By Proposition 6.7.9, we may then extend $λ$ to an embedding $λ^{'} : Ω (α) \to M$ . We then have $(Ω, λ) \leq (Ω (α), λ^{'})$ , and the maximality of $Ω$ forces $E = Ω$ . Setting $Φ = λ$ , we are done. □

Proposition 6.8.9. §

The set of all algebraic elements over a field $F$ in an extension $E$ is a subfield of $E$ , and it is equal to the the largest intermediate extension of $E ∕ F$ that is algebraic over $F$ .

Proof.

Let $M$ denote the set of all algebraic elements over $F$ in $E$ , and let $α, β \in M$ . Then $F (α, β) ∕ F$ is a finite extension, so every element of it is algebraic. In particular, $α - β$ and, if $β \neq 0$ , the element $α β^{- 1}$ are elements of $F (α, β)$ , so they are algebraic elements over $F$ , hence contained in $M$ . Therefore, $M$ is a field. The second statement is then an immediate consequence of the definition of $M$ . □

Corollary 6.8.10. §

The set $\bar{ℚ}$ of algebraic numbers in $ℂ$ forms a field.

Definition 6.8.11. §

An algebraic closure of a field $F$ is an algebraically closed, algebraic extension of $F$ .

Remark 6.8.12. §

Since an algebraic closure is algebraic, every element of the the algebraic closure of a field $F$ has to be the root of a polynomial with $F$ -coefficients. On the other hand, since $\bar{F}$ is algebraically closed, it contains all roots of every polynomial with coefficients in $\bar{F}$ (i.e., every polynomial in $\bar{F} [x]$ factors completely). Thus, if an algebraic closure exists, and we shall see that it does, it consists exactly of all roots of polynomials in $F$ , and every root of a polynomial with coefficients in $\bar{F}$ is actually the root of a polynomial with coefficients in $F$ .

In fact, if a field is contained in an algebraically closed field, then we can see that it does in fact have an algebraic closure quite directly.

Proposition 6.8.13. §

Let $F$ be a field, and suppose that $M$ is an algebraically closed extension field of $F$ . Then $M$ contains a unique algebraic closure of $F$ , equal to the field of elements of $M$ that are algebraic over $F$ .

Proof.

Let $\bar{F}$ denote the field consisting of all elements of $M$ that are algebraic over $F$ . We claim that $\bar{F}$ is algebraically closed. For this, suppose that $f \in \bar{F} [x]$ , and $α \in M$ be a root. As $α$ is algebraic over $\bar{F}$ , we have by Proposition 6.2.19 that $α$ is also algebraic over $F$ . That is, $α$ is an element of $\bar{F}$ . □

Corollary 6.8.14. §

The field $\bar{ℚ}$ of algebraic numbers in $ℂ$ is an algebraic closure of $ℚ$ .

Using Zorn’s lemma, we may prove that every field has an algebraic closure. This is the first result on extension fields in which we do not have a previously given field that contains the field of interest, which makes the proof rather more tricky.

Theorem 6.8.15. §

Every field $F$ has an algebraic closure.

Proof.

Let $F$ be a field. Let $Ω$ be a set that is the disjoint union of finite sets $R_{f}$ for each monic irreducible $f \in F [x]$ , where the number of elements in $R_{f}$ is the number of distinct roots of $f$ in a splitting field. (We will end up identifying the elements of $R_{f}$ with roots of $f$ , but they do not start as such.) We may view $F$ as a subset of $Ω$ by identifying $a \in F$ with the unique element of $R_{x - a}$ . Let $X$ be the nonempty set of all algebraic extensions of $F$ , the underlying sets of which are contained in $Ω$ in the sense if $E \in X$ , then every $α \in E$ lies in $R_{f}$ for $f \in F [x]$ the minimal polynomial of $α$ . We put a partial ordering on $X$ by $E \leq E^{'}$ for $E, E^{'} \in X$ if and only if $E \subseteq E^{'}$ and $E^{'} ∕ E$ is a field extension.

Let $𝒞$ be chain in $X$ , and let $K$ be the union of the fields in $𝒞$ . Then $K$ is a field, as any two elements $α, β \in K$ satisfy $α, β \in E$ for some $E \in 𝒞$ (taking the larger of the two fields in which $α$ and $β$ are contained by definition), and then $α - β \in E$ and $α β^{- 1} \in E$ if $β \neq 0$ . Since $K \in X$ , the chain $𝒞$ has an upper bound, and we may apply Zorn’s lemma to the set $X$ to find a maximal element $\bar{F} \in X$ .

Let $f \in F [x]$ , and let $g \in \bar{F} [x]$ be a monic, nonconstant irreducible polynomial dividing $f$ . We then have that $E = \bar{F} [x] ∕ (g)$ is an extension of $\bar{F}$ that is algebraic over $F$ . We may view the underlying set of $E$ as being contained in $Ω$ as follows. If $h \in F [x]$ is a monic irreducible polynomial with a root in $E$ , we may identify those of its distinct roots in $E$ that are not contained in $\bar{F}$ with distinct elements of $R_{h}$ that are not in $\bar{F}$ . Since $\bar{F} \in X$ is maximal, we must have $E = \bar{F}$ . In particular, $f$ must factor completely in $\bar{F} [x]$ .

It remains only to show that $\bar{F}$ is algebraically closed. Any root $β$ of an irreducible polynomial $g \in \bar{F} [x]$ in an extension of $\bar{F}$ is algebraic over $F$ by Proposition 6.2.19. Therefore, $g$ divides the minimal polynomial $f \in F [x]$ of $β$ , which by the argument we have just given splits over $\bar{F}$ . In particular, we have $β \in \bar{F}$ . □

We next remark that the algebraic closure of any field is in fact unique up to isomorphism.

Proposition 6.8.16. §

Let $M$ and $M^{'}$ be algebraic closures of a field $F$ . Then there exists an isomorphism $Φ : M \to M^{'}$ fixing $F$ .

Proof.

Theorem 6.8.8 applied to the case that $φ = {id}_{F}$ , $E = M$ , and $M = M^{'}$ implies that there exists a field embedding $Φ : M \to M^{'}$ extending $F$ . To see that it is an isomorphism, note that the image of $Φ$ is algebraic over $F$ , being contained in $M^{'}$ , and algebraically closed over $F$ since a root of a polynomial in $F [x]$ in $M$ maps under $Φ$ to a root of the same polynomial. Therefore, $Φ (M)$ is an algebraic closure of $F$ contained in $M^{'}$ , and hence must be $M^{'}$ itself. □

Remark 6.8.17. §

As any two algebraic closures of a field $F$ are isomorphic via an isomorphism that fixes $F$ , we usually refer to “the” algebraic closure of $F$ , denoting it by $\bar{F}$ .

Remark 6.8.18. §

If $E$ is an algebraic extension of $F$ and $\bar{E}$ is the algebraic closure of $E$ , then it is also an algebraic closure of $F$ . In particular, there exists an algebraic closure of $F$ containing $E$ .

Corollary 6.8.19. §

Let $Φ : M \to M^{'}$ be field embedding of algebraic closures of a field $F$ fixing $F$ . Then $Φ$ is an isomorphism.

Proof.

Let $Ψ : M^{'} \to M$ be an isomorphism fixing $F$ , which exists by Proposition 6.8.16. Then $Ψ \circ Φ : M \to M$ is a field embedding fixing $F$ , which is then an isomorphism by Proposition 6.7.15. □

6.9. Transcendental extensions

Definition 6.9.1. §

A field extension $E ∕ F$ is purely transcendental, or totally transcendental, if every element of $E - F$ is transcendental over $E$ .

Terminology 6.9.2. §

For a ring $R$ and an indexing set $I$ , we may speak of the polynomial ring $R [{(x_{i})}_{i \in I}]$ in the variables $x_{i}$ for $i \in I$ . It is simply the union over all finite lists $i_{1}, \dots, i_{n}$ of distinct elements of $I$ of the polynomial rings $R [x_{i_{1}}, \dots, x_{i_{n}}]$ , with the operations being induced by the operations on these rings. If $F$ is a field, then the rational function field $F ({(x_{i})}_{i \in I})$ is the fraction field of $F [{(x_{i})}_{i \in I}]$ . This field is itself the union of the rational function fields $F (x_{i_{1}}, \dots, x_{i_{n}})$ .

Proposition 6.9.3. §

For any indexing set $I$ , the field $F ({(t_{i})}_{i \in I})$ of rational functions in the variables $t_{i}$ for $i \in I$ is purely transcendental over $F$ .

Proof.

Consider first the extension $F (t) ∕ F$ given by the $F$ -rational function field in a single variable $t$ . Let $α = \frac{f}{g} \in F (t) - F$ , where $f, g \in F [t]$ and $g \neq 0$ . We may view $f (x)$ and $g (x)$ as elements of $F [x]$ . Then $α \cdot g (x) \in F (α) [x]$ is not an element of $F [x]$ , so the polynomial $f (x) - 𝛼𝑔 (x)$ is nonzero but does have a root $t$ , which is then algebraic over $F (α)$ . Since $t$ is transcendental over $F$ , this forces $α$ to be as well. This gives the result for a single variable, and the case of finitely many variables follows immediately by induction. Since $F ({(t_{i})}_{i \in I})$ is the union of the rational function fields $F (t_{i_{1}}, \dots, t_{i_{n}})$ with $i_{1}, \dots, i_{n} \in I$ , the case of finitely many variables yields the general case. □

Proposition 6.9.4. §

Every extension of fields is a purely transcendental extension of an algebraic extension.

Proof.

Given a field extension $K ∕ F$ , we may consider its subfield $E$ of elements algebraic over $F$ . If $α \in K - E$ , then $α$ cannot be algebraic over $E$ . That is, if it were, then it would also be algebraic over $F$ in that $E ∕ F$ is algebraic. □

Definition 6.9.5. §

Let $K ∕ F$ be a field extension. We say that a subset $S$ of $K$ is algebraically independent over $F$ , or $F$ -algebraically independent, if $f (s_{1}, \dots, s_{n}) \neq 0$ for all nonzero polynomials $f$ in $n$ variables over $F$ and distinct $s_{1}, \dots, s_{n} \in S$ for some $n \geq 1$ .

Here are a couple of straightfoward lemmas.

Lemma 6.9.6. §

Let $K ∕ F$ be a field extension, and let $S \subset K$ be algebraically independent over $F$ . Then $t \in K$ is transcendental over the field $F (S)$ generated by $S$ over $F$ if and only if $S \cup {t}$ is algebraically independent over $F$ .

Lemma 6.9.7. §

A subset $S$ of a field extension $K$ of $F$ is algebraically independent over $F$ if and only if each $s \in S$ is transcendental over $K (S - {s})$ .

Definition 6.9.8. §

A subset $S$ of an extension $K$ of a field $F$ is a transcendence basis of $K ∕ F$ if and only if $S$ is algebraically independent over $F$ and $K$ is algebraic over $F (S)$ .

The following equivalent conditions for being a transcendence basis nearly mimic the usual equivalent conditions for a subset of a vector space to be a basis. (That is, a subset is a basis if and only if it is a maximal linearly independent subset and if and only if it is a minimal spanning set.)

Proposition 6.9.9. §

Let $S$ be a subset of an extension $K$ of a field $F$ . The following are equivalent:

i.: $S$ is a transcendence basis of $K ∕ F$ ,
ii.: $S$ is a maximal $F$ -algebraically independent subset of $K$ ,
iii.: $S$ is a minimal subset of $K$ such that $K$ is algebraic over $F (S)$ .

Proof.

The equivalence of (i) and (ii) is a direct consequence of Lemma 6.9.6, and the equivalence of (i) and (iii) is a direct consquence of Lemma 6.9.7. □

Theorem 6.9.10. §

Every $F$ -algebraically independent subset of an extension $K ∕ F$ is contained in a transcendence basis, and every subset of $K$ that generates an extension over which $K$ is algebraic contains a transcendence basis.

Proof.

Let $A$ be an $F$ -algebraically independent subset of $K$ . Let $X$ be the set of $F$ -algebraically independent subsets of $K$ containing $A$ , ordered by inclusion. We may take the union of any chain $𝒞$ in $X$ , and it is $F$ -algebraically independent in that any finitely many elements of the union on which we would test algebraic independence is contained in some element of the chain. This union is an upper bound, and thus by Zorn’s lemma, $X$ contains a maximal element $B$ . To finish the proof, we need only see that $K$ is algebraic over $F (B)$ . But this is clear, since if $t \in K - B$ is transcendental over $F (B)$ , then $B \cup {t}$ is $F$ -algebraically independent, contradicting the maximality of $B$ .

Now, let $S \subseteq K$ be such that $K ∕ F (S)$ is algebraic. Consider the set $Y$ of $F$ -algebraically independent subsets of $K$ contained in $S$ , again ordered by inclusion. Every chain has an upper bound as before, so $Y$ contains a maximal element $T$ . We need only see that $K$ is algebraic over $F (T)$ . If not, then since $K$ is algebraic over $F (S)$ and $T \subset S$ , we must have that there exists $s \in S - T$ that is transcendental over $F (T)$ , and then $T \cap {s} \in Y$ , contradicting the maximality of $T$ . □

Corollary 6.9.11. §

Every extension of fields $K ∕ F$ has an intermediate field $E$ such that $K ∕ E$ is algebraic and $E ∕ F$ is purely transcendental.

We omit a proof of the following.

Theorem 6.9.12. §

If $S$ and $T$ are transcendence bases of an extension $K ∕ F$ , then $S$ and $T$ have the same cardinality.

In particular, we may make the following definition.

Definition 6.9.13. §

We say that a field extension $K ∕ F$ has finite transcendence degree if it has a finite transcendence basis, in which case the number of elements in a transcendence basis is called the transcendence degree. Otherwise, we say that $K ∕ F$ has infinite transcendence degree.

6.10. Separable extensions

We shall use $\bar{F}$ to denote an algebraic closure of a field $F$ .

Definition 6.10.1. §

Let $F$ be a field. Let $f \in F [x]$ be nonzero, and let $α \in \bar{F}$ be a root of $f$ . The multiplicity of $α$ as a root of $f$ is the largest positive integer $m$ such that ${(x - α)}^{m}$ divides $f$ in $\bar{F} [x]$ .

Example 6.10.2. §

Let $f = x^{p} - t \in 𝔽_{p} (t) [x]$ , which is irreducible. In $\bar{𝔽_{p} (t)} [x]$ , we have

f = x^{p} - t = {(x - t^{1 ∕ p})}^{p},

so $t^{1 ∕ p}$ has multiplicity $p$ as a root of $f$ .

Lemma 6.10.3. §

Let $F$ be a field, and let $f \in F [x]$ be irreducible. Then every root of $f$ in an algebraic closure $\bar{F}$ of $F$ has the same multiplicity.

Proof.

Let $α, β \in \bar{F}$ be roots of $f$ . Fix an field isomorphism $σ : F (α) \to F (β)$ taking $α$ to $β$ , and extend it to an embedding $τ : \bar{F} \to \bar{F}$ . Let $\tilde{τ} : \bar{F} [x] \to \bar{F} [x]$ map induced by $τ$ . If $m$ denotes the multiplicity of $α$ , then

\tilde{τ} ({(x - α)}^{m}) = {(x - β)}^{m} .

Since ${(x - α)}^{m}$ divides $f$ in $\bar{F} [x]$ and $\tilde{τ} (f) = f$ , the multiplicity of $β$ is then at least $m$ , but this was independent of the choice of $α$ and $β$ , so $α$ and $β$ have the same multiplicity. □

Corollary 6.10.4. §

Let $F$ be a field. The number of distinct roots of an irreducible polynomial $f \in F [x]$ in an algebraic closure $\bar{F}$ of $F$ divides the degree of $f$ .

Definition 6.10.5. §

Let $F$ be a field. We say that a nonconstant polynomial $f \in F [x]$ is separable if every root of $f$ has multiplicity $1$ .

Definition 6.10.6. §

Let $F$ be a field and $\bar{F}$ be an algebraic closure of $F$ . An element $α \in \bar{F}$ is separable over $F$ if and only if its minimal polynomial over $F$ is separable.

Definition 6.10.7. §

We say that an algebraic extension $E ∕ F$ is separable if every $α \in E$ is separable over $F$ .

Lemma 6.10.8. §

Let $E$ be an algebraic extension of a field $F$ , and let $K$ be an algebraic extension of $E$ . If $K ∕ F$ is separable, then so are $K ∕ E$ and $E ∕ F$ .

Proof.

Suppose that $K ∕ F$ is separable. By definition, if $α \in E$ , then $α \in K$ , so its minimal polynomial over $F$ is separable. Thus, $E ∕ F$ is separable. Moreover, the minimal polynomial of any $β \in K$ over $E$ divides the minimal polynomial of $β$ over $F$ , so $β$ is separable over $E$ . In other words, $K ∕ E$ is separable. □

Notation 6.10.9. §

Let $K$ and $L$ be extensions of a field $F$ . We will denote the set of field embeddings of $K$ into $L$ that fix $F$ by ${Emb}_{F} (K, L)$ . If $K$ is algebraic over $F$ and $L$ is taken to be a fixed algebraic closure of $F$ , we will simply write ${Emb}_{F} (K)$ (despite the dependence on the algebraic closure).

Lemma 6.10.10. §

Let $K ∕ F$ be a finite extension of fields, and let $E$ be an intermediate field in $K ∕ F$ . Then the number of extensions of $φ \in {Emb}_{F} (E)$ to $K$ equals the order of ${Emb}_{E} (K)$ .

Proof.

Fix an extension of $φ$ to an embedding of an algebraic closure of $K$ into an algebraic closure $\bar{F}$ of $F$ , which exists by Theorem 6.8.8. It is then an isomorphism by Corollary 6.8.19, and we use it to identify $\bar{F}$ as an algebraic closure of $K$ .

We define a bijection from the set of extensions $Φ : K \to \bar{F}$ of $φ$ to the set ${Emb}_{E} (K)$ as follows. Fix such an extension $K \to \bar{F}$ of $φ$ , and let $Ψ$ be an extension of it to an isomorphism $\bar{F} \to \bar{F}$ . Now, for any extension $Φ : K \to \bar{F}$ of $φ$ , we associate $Ψ^{- 1} \circ Φ : K \to \bar{F}$ . Since $Ψ$ and $Φ$ are extensions of $φ : E \to \bar{F}$ , the composition $Ψ^{- 1} \circ Φ$ fixes $E$ . The association $Φ \mapsto Ψ^{- 1} \circ Φ$ is then the desired binjection, with inverse given by composition with $Ψ$ . □

Lemma 6.10.11. §

Let $E ∕ F$ be a field extension, and let $α \in E$ be algebraic over $F$ . Then $α$ is separable over $F$ if and only if $F (α) ∕ F$ is separable.

Proof.

We prove the nontrivial direction, which results from several applications of Theorem 6.7.9. Fix an algebraic closure $\bar{F}$ of $F$ . For a given $β \in F (α)$ , the number $e = | {Emb}_{F} (F (β)) |$ is at most the degree $[F (β) : F]$ , with equality if and only if $β$ is separable. Since $α$ is separable over $F$ , we have

| {Emb}_{F} (F (α)) | = [F (α) : F] .

Moreover, $α$ is separable over $F (β)$ as well, since its minimal polynomial over $F (β)$ divides its minimal polynomial over $F$ . Thus, the number of embeddings of $F (α)$ in $\bar{F}$ extending a given embedding of $F (β)$ into $\bar{F}$ is exactly $[F (α) : F (β)]$ . Therefore, we have that

[F (α) : F] = [F (α) : F (β)] e,

which means that $e = [F (β) : F]$ , so $β$ is separable. □

We also have the following.

Proposition 6.10.12. §

Let $E ∕ F$ be a finite extension. Fix an algebraic closure $\bar{F}$ of $F$ .

a.: The number of embeddings of $E$ into $\bar{F}$ that fix $F$ divides $[E : F]$ .
b.: The number of embeddings of $E$ into $\bar{F}$ that fix $F$ is equal to $[E : F]$ if and only if $E ∕ F$ is separable.

Proof.

Let $e = | {Emb}_{F} (E) |$ . Write $E = F (α_{1}, α_{2}, \dots, α_{n})$ , and let $E_{i} = F (α_{1}, α_{2}, \dots, α_{i - 1})$ for $1 \leq i \leq n + 1$ . Then $E_{i + 1} = E_{i} (α_{i})$ for $i \leq n$ , and by Theorem 6.7.9, the number $e_{i}$ of embeddings of $E_{i + 1}$ into $\bar{F}$ extending an embedding $φ_{i}$ of $E_{i}$ into $\bar{F}$ is the number of distinct roots of the minimal polynomial of $α_{i}$ over $E_{i}$ . This number, in turn, is a divisor of $[E_{i + 1} : E_{i}]$ , with equality if and only if $α_{i}$ is separable over $E_{i}$ . Since

e = \prod_{i = 1}^{n} e_{i}

and

[E : F] = \prod_{i = 1}^{n} [E_{i + 1} : E_{i}],

we therefore have that $e$ divides $[E : F]$ , with equality if and only if $e_{i} = [E_{i + 1} : E_{i}]$ for each $i$ , and in particular, noting Lemma 6.10.8, if $E ∕ F$ is separable.

Conversely, suppose that $e = [E : F]$ . For $β \in E$ , the number of distinct roots $c$ of its minimal polynomial is the number of embeddings of $F (β)$ into $\bar{F}$ fixing $F$ . The number of embeddings $d$ of $E$ into $\bar{F}$ extending one of those embeddings is equal to the number of embeddings of $d$ into $\bar{F}$ fixing $F (β)$ , which divides $[E : F (β)]$ by what we have shown. We then have $[E : F] = 𝑐𝑑$ , forcing $c = [F (β) : F]$ . That is, $E ∕ F$ is separable. □

Proposition 6.10.13. §

Let $K$ be an algebraic extension of a field $F$ , and let $E$ be an intermediate field in $K ∕ F$ . Then $K ∕ F$ is separable if and only if $K ∕ E$ and $E ∕ F$ are.

Proof.

By Lemma 6.10.8, we are reduced to showing that if $K ∕ E$ and $E ∕ F$ are separable, then $K ∕ F$ is separable. Proposition 6.10.12 implies this immediately if $K ∕ F$ is finite. In general, take $α \in K$ , and note that any minimal polynomial $g$ of $α$ over $E$ actually has coefficients in some finite subextension $E^{'}$ of $E$ , in that $E ∕ F$ is algebraic. Then $E^{'} (α) ∕ E^{'}$ is separable since $g$ is, and $E^{'} ∕ F$ is separable by Lemma 6.10.8. As $E^{'} (α) ∕ F$ is finite, we have the result. □

Definition 6.10.14. §

We say an algebraic extension $E ∕ F$ is purely inseparable if $E$ contains no nontrivial separable subextensions of $F$ .

Proposition 6.10.13 tells us that it suffices to check the separability of an extension on a generating set. It also implies the following.

Corollary 6.10.15. §

Let $K ∕ F$ be an algebraic extension. The set $E$ of all separable elements in $K ∕ F$ is a subfield of $K$ . Moreover, the extension $K ∕ E$ is purely inseparable.

Definition 6.10.16. §

Let $K ∕ F$ be a finite extension, and let $E$ be the maximal separable subextension of $F$ in $K$ .

i.: The degree of separability ${[K : F]}_{s}$ of $K ∕ F$ is $[E : F]$ .
ii.: The degree of inseparability ${[K : F]}_{i}$ of $K ∕ F$ is $[K : E]$ .

We have the following multiplicativity of separable and inseparable degrees.

Lemma 6.10.17. §

Let $K ∕ F$ be a finite extension and $E$ an intermediate field in $K ∕ F$ . Then

\begin{array}{rcl} {[K : F]}_{s} = {[K : E]}_{s} {[E : F]}_{s} & and & {[K : F]}_{i} = {[K : E]}_{i} {[E : F]}_{i} . \end{array}

Proof.

By the multiplicativity of degrees of field extensions, it suffices to consider separable degrees. Fix an algebraic closure $\bar{F}$ of $F$ . Given a field embedding of $E$ into $\bar{F}$ fixing $F$ , the number of extensions of it to $K$ is ${[K : E]}_{s}$ by Corollary 6.7.10 and Lemma 6.10.10. The number of such embeddings being ${[E : F]}_{s}$ , we have the result. □

Let us investigate the circumstances under which all finite extensions of a given field are separable.

Definition 6.10.18. §

A field $F$ is perfect is every finite extension of it is separable.

Example 6.10.19. §

The field $𝔽_{p}$ is perfect. To see this, recall the field $𝔽_{p^{n}}$ for $n \geq 1$ is equal to the set of roots of the polynomial $x^{p^{n}} - x$ , which are all distinct (since there need to be $p^{n}$ of them). Since the minimal polynomial of any $α \in 𝔽_{p^{n}}$ , divides $x^{p^{n}} - x$ , that polynomial is separable, and therefore $𝔽_{p^{n}} ∕ 𝔽_{p}$ is separable.

Lemma 6.10.20. §

Let $E ∕ F$ be an algebraic field extension. Let $f \in E [x]$ be monic, and let $m \geq 1$ be such that $f^{m} \in F [x]$ . Then, either $m = 0$ in $F$ or $f \in F [x]$ .

Proof.

Suppose that $f \notin F [x]$ . Write $f = \sum_{i = 0}^{n} a_{i} x^{i}$ with $n = \deg f$ and $a_{n} = 1$ . Let $i \leq n - 1$ be maximal such that $a_{i} \notin F$ . The coefficient $c$ of $x^{(m - 1) n + i}$ in $f^{m}$ is a polynomial in the coefficients $a_{i}, a_{i + 1}, \dots, a_{n - 1}$ such that $c - m a_{i}$ is a polynomial in $a_{i + 1}, \dots, a_{n - 1}$ , which are elements of $F$ . Since $c \in F$ , we have $m a_{i} \in F$ , which forces either $m = 0$ in $F$ or $a_{i} \in F$ . □

Theorem 6.10.21. §

Let $F$ be a field of characteristic $0$ . Then $F$ is perfect.

Proof.

If $f \in F [x]$ is monic and irreducible, then every root of $f$ in an algebraic closure $\bar{F}$ occurs with some multiplicity $m \geq 1$ . It follows that

f = \prod_{i = 1}^{d} {(x - α_{i})}^{m}

for some $d \geq 1$ and distinct $α_{1}, α_{2}, \dots, α_{m} \in \bar{F}$ , so $f = g^{m}$ for some $g \in \bar{F} [x]$ . Since the characteristic of $F$ is zero, Lemma 6.10.20 tells us that $m = 1$ . □

The following tells us that the degree of inseparability of a finite field extension is the power of the characteristic of the fields.

Proposition 6.10.22. §

Let $F$ be a field of characteristic $p$ . If $E ∕ F$ is purely inseparable and $α \in E$ , then $α^{p^{k}} \in F$ for some minimal $k \geq 0$ , and the minimal polynomial of $α$ over $F$ is $x^{p^{k}} - α^{p^{k}} = {(x - α)}^{p^{k}}$ .

Proof.

Fix an algebraic closure $\bar{F}$ of $F$ containing $E$ . Let $f \in F [x]$ be the minimal polynomial of some element $α$ of $E$ not in $F$ . Again we have

f = \prod_{i = 1}^{d} {(x - α_{i})}^{m}

for some $d \geq 1$ and distinct $α_{1}, α_{2}, \dots, α_{m} \in \bar{F}$ , so $f = g^{m}$ for some $g \in \bar{F} [x]$ . We must show that $m$ is a $p$ -power and $d = 1$ .

Write $m = p^{k} t$ with $p ∤ t$ and $k \geq 1$ . The fact that $f = {(g^{p^{k}})}^{t} \in F [x]$ forces $g^{p^{k}} \in F [x]$ by Lemma 6.10.20. Since $f$ is irreducible, we have $t = 1$ .

Now set $a_{i} = α_{i}^{p^{k}}$ and write

f = \prod_{i = 1}^{d} (x^{p^{k}} - a_{i}) .

Then $f (x) = h (x^{p^{k}})$ for $h = \prod_{i = 1}^{d} (x - a_{i})$ . The polynomial $h$ lies in $F [x]$ since it has the same set of coefficients as $f$ , it is irreducible as any factorization of $h$ would give rise to a factorization of $f$ , and it has $α^{p^{k}}$ as a root. Also, the $a_{i}$ are distinct elements, since there are no nontrivial $p^{k}$ th roots of unity in a field of characteristic $p$ , which tells us that raising to the $p^{k}$ th power is injective. As $E ∕ F$ is purely inseparable and any root of $h$ generates a separable extension of $F$ , we must have $d = 1$ . □

Corollary 6.10.23. §

Let $F$ be a field of characteristic $p$ , and let $E ∕ F$ be a finite extension. Then ${[E : F]}_{i}$ is a power of $p$ .

We then have the following.

Proposition 6.10.24. §

The degree of separability ${[K : F]}_{s}$ of a finite extension $K ∕ F$ is equal to the number of embeddings of $K$ fixing $F$ into a given algebraic closure of $F$ .

Proof.

Let $E$ be the maximal separable subextension of $F$ in $K$ . We know that there are ${[K : F]}_{s}$ elements of ${Emb}_{F} (E)$ . Any $α \in K - E$ has minimal polynomial ${(x - α)}^{p^{n}}$ over $E$ for some $n \geq 1$ , so $α$ has only one conjugate over $E$ in $K$ . Thus, any $φ \in {Emb}_{F} (E)$ extends uniquely to an embedding of $E (α)$ in $\bar{F}$ . Replacing $E$ by $E (α)$ and repeating this last argument, we obtain recursively that $φ$ has a unique extension to all of $K$ . Since every element of ${Emb}_{F} (K)$ is an extension of its restriction to $E$ , the number of such elements is ${[K : F]}_{s}$ . □

Finally, we show that finite separable extensions can be generated by a single element.

Definition 6.10.25. §

We say that a finite field extension $E ∕ F$ is simple if there exists $α \in E$ such that $E = F (α)$ . In that case, $α$ is said to be a primitive element for $E ∕ F$ .

Theorem 6.10.26 (Primitive element theorem). §

Every finite, separable field extension is simple.

Proof.

Note that if $F$ is finite, then it is isomorphic to $𝔽_{p^{n}}$ for some prime $p$ and $n \geq 1$ , and by Proposition 6.5.5, it equals $𝔽_{p} (ξ)$ for some primitive $(p^{n} - 1)$ th root of unity in $F$ . So we may assume that $F$ is infinite.

Since every finite extension is finitely generated by Corollary 6.2.10, it suffices by recursion to show that if $E ∕ F$ is a finite field extension with $E = F (α, β)$ for some $α, β \in E$ , then there exists $c \in F$ such that $E = F (α + 𝑐𝛽)$ .

Since $F$ is infinite, we can and do choose $c \in F^{\times}$ such that

c \neq - \frac{α^{'} - α}{β^{'} - β}

for all conjugates $α^{'}$ of $α$ over $F$ with $α^{'} \neq α$ and all conjugates $β^{'}$ of $β$ over $F$ with $β^{'} \neq β$ . Set $γ = α + 𝑐𝛽$ . Then $γ \neq α^{'} + c β^{'}$ for all $α^{'}$ and $β^{'}$ as above. Let $f$ be the minimal polynomial of $α$ , and let $h (x) = f (γ - 𝑐𝑥) \in F (γ) [x]$ . Then $h (β) = f (α) = 0$ and $h (β^{'}) \neq 0$ for $β^{'}$ . Since $h$ shares the root $β$ with the minimal polynomial $g$ of $β$ over $F$ , but not any other root, and the minimal polynomial $q$ of $β$ over $F (γ)$ divides both of the latter polynomials, we must have $q = x - β$ , which is to say that $β \in F (γ)$ , which then implies that $α \in F (γ)$ as well. We therefore have $F (γ) = F (α, β)$ , as desired. □

Remark 6.10.27. §

Much as with algebraic closure, we have the notion of a separable closure of a field. A field $L$ is separably closed if it contains a root of every monic, separable polynomial with coefficients in $L$ . Algebraically closed fields are therefore separably closed. A separable closure of a field $F$ is a separable extension $F^{sep}$ of $F$ that is separably closed. If $F$ is a subfield of any separably closed field $L$ , the set of all roots in $L$ of all monic, separable polynomials in $F [x]$ is a subfield that is a separable closure of $F$ . Separable closures exist: in fact, given a field $F$ , take an algebraic closure $\bar{F}$ of $F$ , and it then contains a separable closure $F^{sep}$ , which is the union of all finite separable subextensions of $F$ in $\bar{F}$ . Of course, if $F$ is perfect, then the notions of separable closure and algebraic closure of $F$ coincide.

6.11. Normal extensions

We extend the definition of a splitting field to include sets of polynomials.

Definition 6.11.1. §

Let $F$ be a field, and let $S$ be a subset of $F [x]$ consisting of nonconstant polynomials. A splitting field $E$ for $S$ over $F$ is an extension of $F$ such that every polynomial in $S$ splits in $E$ and which contains no proper subextension of $F$ in which this occurs.

Example 6.11.2. §

The field $ℚ (\sqrt{2}, \sqrt{3})$ is the splitting field of ${x^{2} - 2, x^{2} - 3}$ . It is then also the splitting field of $(x^{2} - 2) (x^{2} - 3)$ .

Example 6.11.3. §

An algebraic closure $\bar{F}$ of a field $F$ is a splitting field of the set of all nonconstant polynomials in $F [x]$ .

Remark 6.11.4. §

An algebraic closure $\bar{F}$ of a field $F$ will always contain a unique splitting field for any subset $S$ of $F [x]$ . This field is equal to the intersection of all subfields of $\bar{F}$ in which every polynomial in $S$ splits.

Definition 6.11.5. §

We say that an algebraic field extension $E ∕ F$ is normal if $E$ is the splitting field of some set of polynomials in $F [x]$ .

Lemma 6.11.6. §

If $E ∕ F$ is normal, then so is $E ∕ F^{'}$ , where $F^{'}$ is any intermediate field in $E ∕ F$ .

Proof.

If $E$ is the splitting field of a set $S$ of nonconstant polynomials in $F [x]$ , then $E$ is generated over $F$ by the roots of the polynomials in $S$ . Then $E$ is also generated over $F^{'}$ by these roots, so $E$ is also a splitting field over $F^{'}$ . □

Theorem 6.11.7. §

An algebraic field extension $E ∕ F$ is normal if and only if every field embedding $Φ$ of $E$ that fixes $F$ into an algebraic closure $\bar{F}$ of $F$ containing $E$ satisfies $Φ (E) = E$ . Moreover, under these conditions, $E$ is equal to the splitting field over $F$ of the set of minimal polynomials over $F$ of every element of $E$ .

Proof.

Suppose first that $E ∕ F$ is normal, and let $S \subseteq F [x]$ be a set of polynomials of which $E$ is a splitting field. Let $Φ \in {Emb}_{F} (E)$ . By definition, $E$ is generated over $F$ by the roots of all polynomials in $S$ . Let $f \in S$ , and let $α \in E$ be a root. By Theorem 6.7.9, we must have that $Φ (α)$ is a root of $f$ in $\bar{F}$ . But every root of $f$ in $\bar{F}$ lies in the subfield $E$ , since $f$ splits in $E$ , so $Φ (α) \in E$ . As every element of $E$ may be written as a rational function in the roots of polynomials in $S$ with coefficients in $F$ , we therefore have $Φ (E) \subseteq E$ . Noting Proposition 6.7.15, we then have that $Φ (E) = E$ .

Conversely, suppose that $Φ (E) = E$ for every $Φ \in {Emb}_{F} (E)$ . Let $α \in E$ , and let $f$ be its minimal polynomial over $F$ . Then for any root $β \in \bar{F}$ of $f$ , we have an isomorphism $φ : F (α) \to F (β)$ sending $α$ to $β$ . We may then extend the resulting embedding $F (α) \to \bar{F}$ to an embedding $Φ : E \to \bar{F}$ . Since $Φ (E) = E$ , we therefore have $β \in E$ . So $E$ contains the splitting field of every polynomial of $F$ that has a root in $E$ . Since $E$ is algebraic and therefore consists entirely of roots of polynomials in $F$ , it is therefore equal to said splitting field. □

Corollary 6.11.8. §

Let $E ∕ F$ be a normal field extension, and let $f \in F [x]$ be an irreducible polynomial that has a root in $E$ . Then $f$ splits in $E$ .

Proof.

This follows directly from the final statement of Theorem 6.11.7. □

For composite extensions, we have the following.

Proposition 6.11.9. §

Let $F$ be a field and $\bar{F}$ an algebraic closure of $F$ . Suppose that $E$ and $K$ are subfields of $\bar{F}$ that are normal over $F$ . Then $𝐸𝐾 ∕ F$ is normal as well.

Proof.

We note that any $φ \in {Emb}_{F} (𝐸𝐾)$ restricts to embeddings of $E$ and of $K$ into $\bar{F}$ . Since $E ∕ F$ and $K ∕ F$ are normal, we have $φ (E) = E$ and $φ (K) = K$ . Every element in $𝐸𝐾$ is a rational function in the elements of $E \cup K$ , so $φ (𝐸𝐾)$ is contained in $𝐸𝐾$ (and thus equal to $𝐸𝐾$ ) as well. By Theorem 6.11.7, $𝐸𝐾 ∕ F$ is normal. □

Definition 6.11.10. §

Let $E$ be a field. An automorphism of $E$ is an isomorphism of rings from $E$ to itself.

Examples 6.11.11. §

a.: The identity map ${id}_{F}$ is an automorphism of any field $F$ , known as the trivial automorphism. It is the identity element in $Aut (F)$ , and it is often denoted by $1$ .
b.: Complex conjugation is an automorphism of $ℂ$ fixing $ℝ$ .
c.: The map $ϕ : ℚ (\sqrt{2}) \to ℚ (\sqrt{2})$ sending $a + b \sqrt{2}$ to $a - b \sqrt{2}$ for all $a, b \in ℚ$ is an automorphism of $ℚ (\sqrt{2})$ .
d.: The only automorphism of $ℚ$ is the trivial automorphism, as the fact that $ϕ (1) = 1$ forces $ϕ (a) = a$ for all $a \in ℚ$ using the properties of a ring homomorphism.
e.: The Frobenius map $φ_{p} : {\bar{𝔽}}_{p} \to {\bar{𝔽}}_{p}$ defined by $φ_{p} (x) = x^{p}$ is an automorphism of ${\bar{𝔽}}_{p}$ fixing $𝔽_{p}$ .

Remark 6.11.12. §

The set of automorphisms of a field form a group under composition. That is, the composition of two automorphisms is also an automorphism, as is the inverse of one.

Definition 6.11.13. §

The automorphism group $Aut (E)$ of a field $E$ is the group of automorphisms of $E$ with the operation of composition.

Often, we are interested in automorphisms fixing a subfield $F$ of $E$ . It is easy to see that these form a subgroup of $Aut (E)$ .

Notation 6.11.14. §

We let ${Aut}_{F} (E)$ denote the subgroup of $Aut (E)$ for a field $E$ consisting of automorphisms that fix a subfield $F$ .

Remark 6.11.15. §

If $E$ is of characteristic $0$ , then ${Aut}_{ℚ} (E) = Aut (E)$ .

Example 6.11.16. §

Note that $ℂ = ℝ (i)$ , and $i$ has minimal polynomial $x^{2} + 1$ . Any automorphism of $ℂ$ fixing $ℝ$ must take $i$ to $i$ or $- i$ , which then determines the automorphism uniquely. That is, the group ${Aut}_{ℝ} (ℂ)$ consists of exactly two elements, the trivial automorphism and complex conjugation.

The following is an immediate corollary of Theorem 6.11.7, Proposition 6.10.12a, and Proposition 6.10.24.

Corollary 6.11.17. §

Let $E$ be a finite normal extension of a field $F$ . Then ${Emb}_{F} (E) = {Aut}_{F} (E)$ , which has order ${[E : F]}_{s}$ .

Example 6.11.18. §

Consider the splitting field $E = ℚ (ω, \sqrt[3]{2})$ of $x^{3} - 2$ , where $ω$ is a primitive cube root of unity. Since $E$ is normal, any embedding of $E$ in an algebraic closure of $ℚ$ containing $E$ has image $E$ , so gives rise to an automorphism of $E$ . Theorem 6.7.9 then tells us that we may choose such an automorphism uniquely as follows. First, we choose another root of the minimal polynomial $x^{2} + x + 1$ of $ω$ and send $ω$ to it, i.e., to $ω$ or $ω^{2}$ . This yields an automorphism of $ℚ (ω)$ . Then, we extend this automorphism to an automorphism of $E$ by sending $\sqrt[3]{2}$ to a root of its minimal polynomial over $ℚ (ω)$ . Since the degree of $ℚ (ω)$ , i.e. 2, is prime to the degree of $ℚ (\sqrt[3]{2})$ , i.e. 3, over $ℚ$ , we have $[ℚ (ω, \sqrt[3]{2}) : ℚ (ω)] = 3$ , so $x^{3} - 2$ is still irreducible over $ℚ (ω)$ . Therefore, we can send $\sqrt[3]{2}$ to any of $\sqrt[3]{2}$ , $ω \sqrt[3]{2}$ , and $ω^{2} \sqrt[3]{2}$ . That is, there are exactly $6$ , or $[ℚ (ω, \sqrt[3]{2}) : ℚ]$ , elements of ${Aut}_{ℚ} (ℚ (ω, \sqrt[3]{2}))$ .

6.12. Galois extensions

Definition 6.12.1. §

An algebraic field extension is said to be Galois if it is both normal and separable.

Remark 6.12.2. §

By Theorem 6.10.21, an algebraic extension of a field of characteristic $0$ is Galois if and only if it is normal.

Examples 6.12.3. §

a.: The extensions $ℚ (\sqrt{2})$ and $ℚ (i)$ of $ℚ$ are Galois.
b.: The extension $ℚ (\sqrt[3]{2}) ∕ ℚ$ is not Galois. It is separable but not normal.
c.: The extension $𝔽_{p} (t^{1 ∕ p}) ∕ 𝔽_{p} (t)$ is not Galois. It is normal but not separable.
d.: The field $\bar{ℚ}$ is a Galois extension of $ℚ$ .
e.: For any $n \geq 1$ , the field $𝔽_{p^{n}}$ is a Galois extension of $𝔽_{p}$ .

Definition 6.12.4. §

Let $E ∕ F$ be a Galois extension. The Galois group $Gal (E ∕ F)$ of $E ∕ F$ is the group of automorphisms of $E$ that fix $F$ .

Remark 6.12.5. §

The group $Gal (E ∕ F)$ is just ${Aut}_{F} (E)$ in our earlier notation. The notation $Gal (E ∕ F)$ is used only for Galois extensions, whereas ${Aut}_{F} (E)$ can be used for arbitrary extensions.

Notation 6.12.6. §

We often write

A field extension. A full diagram description follows.

Diagram description: A field extension

The upper field E is an extension of the lower field F. The connecting line denotes the field extension, not a directed field homomorphism.

Objects, listed by row and column:

Row 1, from left to right: column 1: E.
Row 2, from left to right: column 1: F.

Arrows and lines:

An undirected line joins E and F, without a label.

to indicate that $E$ is a field extension of $F$ , and if $E ∕ F$ is Galois with Galois group $G$ , we indicate this by the diagram

A Galois extension and its Galois group. A full diagram description follows.

Diagram description: A Galois extension and its Galois group

The upper field E is a Galois extension of F, with Galois group G. The line denotes the field extension.

Objects, listed by row and column:

Row 1, from left to right: column 1: E.
Row 2, from left to right: column 1: F.

Arrows and lines:

An undirected line joins E and F, labelled G.

Drawings such as these are known as field diagrams and are useful in illustrating examples.

We will be concerned here only with finite Galois extensions. The following is immediate from Corollary 6.11.17 and Proposition 6.10.12b.

Proposition 6.12.7. §

Let $E ∕ F$ be a finite Galois extension of fields. Then $Gal (E ∕ F)$ is a finite group of order $[E : F]$ .

Lemma 6.12.8. §

Let $E$ be a field, and let $G$ be a subgroup of $Aut (E)$ . Then the set of elements of $E$ that are fixed by every element of $G$ is a subfield of $E$ .

Proof.

Let $a, b \in E$ with $b \neq 0$ . Let $σ \in G$ . Then we have

\begin{array}{rcl} σ (a - b) = σ (a) - σ (b) = a - b & and & σ (a b^{- 1}) = σ (a) σ {(b)}^{- 1} = a b^{- 1}, \end{array}

so $a - b$ and $a b^{- 1}$ are elements of $E$ fixed by $G$ . □

With Lemma 6.12.8 in hand, we may make the following definition.

Definition 6.12.9. §

Let $G$ be a subgroup of $Aut (E)$ . The fixed field $E^{G}$ of $E$ under $G$ is the largest subfield of $E$ fixed by $G$ .

Note the following.

Lemma 6.12.10. §

Let $K ∕ F$ be a Galois extension, and let $E$ be an intermediate field in $K ∕ F$ . Then $K$ is a Galois extension of $E$ . Moreover, $E ∕ F$ is Galois if and only if it is normal.

Proof.

The extension $K ∕ E$ is normal by Lemma 6.11.6 and separable by Lemma 6.10.8. The extension $E ∕ F$ is also separable by Lemma 6.10.8, hence the second claim. □

Proposition 6.12.11. §

Let $K ∕ F$ be a finite Galois extension. Then the fixed field of $K$ under $Gal (K ∕ F)$ is $F$ .

Proof.

Let $E = K^{Gal (K ∕ F)}$ . Clearly $F \subseteq E$ , and we must show the other containment. By Lemma 6.12.10, the extension $K ∕ E$ is Galois. On the other hand, every element of $Gal (K ∕ F)$ fixes $E$ , so $Gal (K ∕ F)$ is equal to its subgroup $Gal (K ∕ E)$ of automorphisms fixing $E$ . By Proposition 6.12.7, we have that

[K : F] = | Gal (K ∕ F) | = | Gal (K ∕ E) | = [K : E],

which means that $[E : F] = 1$ , and therefore $E = F$ . □

Notation 6.12.12. §

If $K ∕ F$ is a finite Galois extension and $E$ is an intermediate field, then the restriction of $σ \in K$ to an embedding of $E$ into $K$ is denoted $σ |_{E}$ .

Remark 6.12.13. §

If $K ∕ F$ is a finite Galois extension and $E$ is an intermediate field in $K ∕ F$ such that $E ∕ F$ is Galois, then $σ |_{E}$ is an automorphism of $E$ , so $σ |_{E} \in Gal (E ∕ F)$ .

Definition 6.12.14. §

Let $K ∕ F$ be a finite Galois extension, and let $E$ be an intermediate field in $K ∕ F$ such that $E ∕ F$ is Galois. Then the restriction map from $K$ to $E$ (over $F$ ) is the homomorphism of groups $Gal (K ∕ F) \to Gal (E ∕ F)$ takes $σ \in Gal (K ∕ F)$ to $σ |_{E}$ .

Lemma 6.12.15. §

Let $K ∕ F$ be a Galois extension, and let $E$ be an intermediate field in $K ∕ F$ . Then there exists a bijection of sets

{res}_{E} : Gal (K ∕ F) ∕ Gal (K ∕ E) \to {Emb}_{F} (E), {res}_{E} (σ Gal (K ∕ E)) = σ |_{E}

for $σ \in Gal (K ∕ F)$ , where $\bar{F}$ is an algebraic closure of $F$ containing $K$ .

Proof.

Let $σ, τ \in Gal (K ∕ F)$ . We have that $σ |_{E} = τ |_{E}$ if and only if $σ^{- 1} τ$ fixes $E$ , or equivalently, is an element of $Gal (K ∕ E)$ . In other words, $σ |_{E} = τ |_{E}$ if and only if $σ Gal (K ∕ E) = τ Gal (K ∕ E)$ . Therefore, ${res}_{E}$ is both well-defined and one-to-one.

Given an embedding $τ$ of $E$ into $\bar{F}$ fixing $F$ , we may extend it to an embedding $σ$ of $K$ into $\bar{F}$ . Since $K ∕ F$ is normal, $σ$ is an automorphism of $K$ . That is, $σ$ is an element $Gal (K ∕ F)$ with $σ |_{E} = τ$ , so ${res}_{E}$ is surjective. □

Proposition 6.12.16. §

Let $K ∕ F$ be a Galois extension, and let $E$ be an intermediate field in $K ∕ F$ . Then $E ∕ F$ is Galois if and only if $Gal (K ∕ E)$ is a normal subgroup of $Gal (K ∕ F)$ . If $E ∕ F$ is Galois, then restriction induces an isomorphism

{res}_{E} : Gal (K ∕ F) ∕ Gal (K ∕ E) \to_{}^{\sim} Gal (E ∕ F) .

Proof.

If $E ∕ F$ is Galois, then the restriction map from $Gal (K ∕ E)$ to $Gal (K ∕ F)$ is a surjective homomorphism with kernel exactly $Gal (K ∕ E)$ by Lemma 6.12.15. So, $Gal (K ∕ E)$ is normal in $Gal (K ∕ F)$ , and we have the stated isomorphism.

Conversely, suppose that $Gal (K ∕ E)$ is a normal subgroup of $Gal (K ∕ F)$ . We already know that $E ∕ F$ is separable by Lemma 6.10.8. To show that $E ∕ F$ is normal, it suffices by Theorem 6.11.7 to show that $φ (α) \in E$ for all $α \in E$ and field embeddings $φ : E \to \bar{F}$ fixing $F$ , where $\bar{F}$ is an algebraic closure of $F$ containing $E$ . Since $K ∕ E$ is Galois, and since $E$ is the fixed field of $Gal (K ∕ E)$ , we have $φ (α) \in K$ , and we will have $φ (α) \in E$ if we can show that $σ (φ (α)) = φ (α)$ for all $σ \in Gal (K ∕ E)$ . Since $K ∕ F$ is Galois, we may lift $φ$ to $τ \in Gal (K ∕ F)$ . The desired equality then amounts to $𝜎𝜏 (α) = τ (α)$ , or $τ^{- 1} 𝜎𝜏 (α) = α$ . Since $Gal (K ∕ E)$ is normal in $Gal (K ∕ F)$ , we have that $τ^{- 1} 𝜎𝜏$ fixes $E$ , and in particular $α$ . □

The final ingredient we need is as follows.

Proposition 6.12.17. §

Let $K ∕ F$ be a finite Galois extension, and let $H$ be a subgroup of $Gal (K ∕ F)$ . Then we have $Gal (K ∕ K^{H}) = H$ .

Proof.

By definition, $H$ fixes $K^{H}$ , so we have $H \leq Gal (K ∕ K^{H})$ . Since $K ∕ F$ is separable, so is $K ∕ K^{H}$ , and the primitive element theorem tells us that $K = K^{H} (α)$ for some $α \in K$ . Define

f = \prod_{σ \in H} (x - σ (α)) \in K [x] .

For $σ \in H$ , let $\tilde{σ} : K [x] \to K [x]$ denote the induced homomorphism. We then have $\tilde{σ} (f) = f$ for all $σ \in H$ , which means that $f \in K^{H} [x]$ . In particular, the minimal polynomial of $α$ over $K^{H}$ divides $f$ , and the degree of that polynomial is $[K : K^{H}]$ , while the degree of $f$ is $| H |$ . This implies that $[K : K^{H}] \leq | H |$ , which since $H \leq Gal (K ∕ K^{H})$ , forces equality on both counts. □

Definition 6.12.18. §

Let $P$ and $Q$ be sets of subsets of a set $X$ and a set $Y$ , respectively, and suppose that $ϕ : P \to Q$ is a function. We say that $ϕ$ is inclusion-reversing if whenever $A, B \in P$ with $A \subseteq B$ , one has $ϕ (B) \subseteq ϕ (A)$ .

We may now state the fundamental theorem of Galois theory, which is essentially just a combination of results we have proven above.

Theorem 6.12.19 (Fundamental theorem of Galois theory). §

Let $K ∕ F$ be a finite Galois extension. Then there are inverse inclusion-reversing bijections

Finite Galois correspondence. A full diagram description follows.

Diagram description: Finite Galois correspondence

The two displayed maps are mutually inverse, inclusion-reversing bijections between the specified collections.

Objects, listed by row and column:

Row 1, from left to right: column 1: left brace intermediate fields inK / F right brace; column 2: left brace subgroups of Gal (K / F) right brace.

Arrows and lines:

An arrow from left brace intermediate fields inK / F right brace to left brace subgroups of Gal (K / F) right brace, labelled psi.
An arrow from left brace subgroups of Gal (K / F) right brace to left brace intermediate fields inK / F right brace, labelled theta.

defined on intermediate fields $E$ in $K ∕ F$ and subgroups $H$ of $Gal (K ∕ F)$ by

\begin{array}{rcl} ψ (E) = Gal (K ∕ E) & and & 𝜃 (H) = K^{H} . \end{array}

Moreover, for such $E$ and $H$ , we have

\begin{array}{rcl} [K : E] = | Gal (K ∕ E) | & and & | H | = [K : K^{H}] . \end{array}

These correspondences restrict to bijections

Normal extensions and normal subgroups. A full diagram description follows.

Diagram description: Normal extensions and normal subgroups

The two displayed maps are mutually inverse, inclusion-reversing bijections between the specified collections.

Objects, listed by row and column:

Row 1, from left to right: column 1: left brace normal extensions of F in K right brace; column 2: left brace normal subgroups of Gal (K / F) right brace.

Arrows and lines:

An arrow from left brace normal extensions of F in K right brace to left brace normal subgroups of Gal (K / F) right brace, labelled psi.
An arrow from left brace normal subgroups of Gal (K / F) right brace to left brace normal extensions of F in K right brace, labelled theta.

Moreover, if $E$ is normal over $F$ $($ resp., $H ⊴ Gal (K ∕ F))$ , then restriction induces an isomorphism

Gal (K ∕ F) ∕ Gal (K ∕ E) \to_{}^{\sim} Gal (E ∕ F) (resp., Gal (K ∕ F) ∕ H \to_{}^{\sim} Gal (K^{H} ∕ F)) .

Proof.

Let $E$ and $H$ be as in the statement of the theorem. We have that

𝜃 (ψ (E)) = 𝜃 (Gal (K ∕ E)) = K^{Gal (K ∕ E)} = E

by Proposition 6.12.11 and

ψ (𝜃 (H)) = ψ (K^{H}) = Gal (K ∕ K^{H}) = H

by Proposition 6.12.17, so $𝜃$ and $ψ$ are inverse bijections. The inclusion-reversing properties of $𝜃$ and $ψ$ are immediate from the definitions of Galois groups and fixed fields. The statements on orders and indices then follow immediately from Proposition 6.12.7, and the statements on normal extensions and subgroups then become simply Proposition 6.12.16. □

Example 6.12.20. §

The extension $ℚ (\sqrt{2}, i) ∕ ℚ$ is Galois with Galois group isomorphic to the Klein four group. We have the complete field diagram

Field lattice for the biquadratic extension generated by square root of two and i. A full diagram description follows.

Diagram description: Field lattice for the biquadratic extension generated by square root of two and i

This is a field-inclusion lattice. Each undirected line joins a larger field above to a subfield below; a group written beside a line is the Galois group of that extension. Unlabelled lines do not assert that the corresponding extension is Galois.

Objects, listed by row and column:

Row 1, from left to right: column 3: blackboard Q (square root of (2), i).
Row 2, from left to right: column 1: blackboard Q (square root of (2)); column 4: blackboard Q (square root of (minus 2)); column 5: blackboard Q (i).
Row 3, from left to right: column 3: blackboard Q.

Arrows and lines:

An undirected line joins blackboard Q (square root of (2), i) and blackboard Q (square root of (2)), labelled blackboard Z / 2 blackboard Z.
An undirected line joins blackboard Q (square root of (2), i) and blackboard Q, labelled ( blackboard Z / 2 blackboard Z ) superscript (2).
An undirected line joins blackboard Q (square root of (2), i) and blackboard Q (square root of (minus 2)), labelled blackboard Z / 2 blackboard Z.
An undirected line joins blackboard Q (square root of (2), i) and blackboard Q (i), labelled blackboard Z / 2 blackboard Z.
An undirected line joins blackboard Q (square root of (2)) and blackboard Q, labelled blackboard Z / 2 blackboard Z.
An undirected line joins blackboard Q (square root of (minus 2)) and blackboard Q, labelled blackboard Z / 2 blackboard Z.
An undirected line joins blackboard Q (i) and blackboard Q, labelled blackboard Z / 2 blackboard Z.

That is, $Gal (ℚ (\sqrt{2}, i) ∕ ℚ)$ is abelian with two generators $σ$ and $τ$ such that $σ (\sqrt{2}) = - \sqrt{2}$ , $σ (i) = i$ , $τ (\sqrt{2}) = \sqrt{2}$ , and $τ (i) = - i$ .

Example 6.12.21. §

Let $G = Gal (ℚ (ω, \sqrt[3]{2}) ∕ ℚ)$ , where $ω$ is a primitive $3$ rd root of unity. We have the field diagram

Field lattice for the splitting field of the cube root of two. A full diagram description follows.

Diagram description: Field lattice for the splitting field of the cube root of two

Objects, listed by row and column:

Row 1, from left to right: column 2: blackboard Q ( omega ,root of degree 3 of (2)).
Row 2, from left to right: column 1: blackboard Q ( omega ); column 3: blackboard Q (root of degree 3 of (2)); column 4: blackboard Q ( omega root of degree 3 of (2)); column 5: blackboard Q ( omega superscript (2)root of degree 3 of (2)).
Row 3, from left to right: column 2: blackboard Q.

Arrows and lines:

An undirected line joins blackboard Q ( omega ,root of degree 3 of (2)) and blackboard Q ( omega ), labelled blackboard Z / 3 blackboard Z.
An undirected line joins blackboard Q ( omega ,root of degree 3 of (2)) and blackboard Q, labelled G.
An undirected line joins blackboard Q ( omega ,root of degree 3 of (2)) and blackboard Q (root of degree 3 of (2)), labelled blackboard Z / 2 blackboard Z.
An undirected line joins blackboard Q ( omega ,root of degree 3 of (2)) and blackboard Q ( omega root of degree 3 of (2)), labelled blackboard Z / 2 blackboard Z.
An undirected line joins blackboard Q ( omega ,root of degree 3 of (2)) and blackboard Q ( omega superscript (2)root of degree 3 of (2)), labelled blackboard Z / 2 blackboard Z.
An undirected line joins blackboard Q ( omega ) and blackboard Q, labelled blackboard Z / 2 blackboard Z.
An undirected line joins blackboard Q (root of degree 3 of (2)) and blackboard Q, without a label.
An undirected line joins blackboard Q ( omega root of degree 3 of (2)) and blackboard Q, without a label.
An undirected line joins blackboard Q ( omega superscript (2)root of degree 3 of (2)) and blackboard Q, without a label.

As a consequence of the fundamental theorem of Galois theory, we have $𝐺 ≅ S_{3}$ , since there are only two groups of order $6$ up to isomorphism and the cyclic one has a unique subgroup of order $3$ . lt follows that our field diagram contains all of the intermediate fields in $ℚ (ω, \sqrt[3]{2}) ∕ ℚ$ . The fact that the extension $ℚ (ω^{i} \sqrt[3]{2})$ is not Galois for $0 \leq i \leq 2$ corresponds to the fact $Gal (ℚ (ω, \sqrt[3]{2}) ∕ ℚ (\sqrt[3]{2}))$ is not a normal subgroup of $G$ , and the other two non-normal intermediate fields correspond to conjugate subgroups.

One can also see this explicitly: note that $Gal (ℚ (ω, \sqrt[3]{2}) ∕ ℚ (ω))$ is generated by an element $τ$ such that $τ (\sqrt[3]{2}) = ω \sqrt[3]{2}$ , and $Gal (ℚ (ω, \sqrt[3]{2}) ∕ ℚ (\sqrt[3]{2}))$ is generated by an element $σ$ such that $σ (ω) = ω^{2}$ . Then $τ^{3} = 1$ , $σ^{2} = 1$ , and

\begin{array}{rcl} 𝜎𝜏 σ^{- 1} (ω) = 𝜎𝜏 (ω^{2}) = σ (ω^{2}) = ω = τ^{- 1} (ω) \\ 𝜎𝜏 σ^{- 1} (\sqrt[3]{2}) = 𝜎𝜏 (\sqrt[3]{2}) = σ (ω \sqrt[3]{2}) = ω^{2} \sqrt[3]{2} = τ^{- 1} (\sqrt[3]{2}), \end{array}

so $𝜎𝜏 σ^{- 1} = τ^{- 1}$ , and $G = ⟨ σ, τ ⟩$ is a nonabelian group of order $6$ , isomorphic to $D_{3} ≅ S_{3}$ .

More generally, we have the following results on Galois groups of composite fields.

Proposition 6.12.22. §

Let $K$ and $E$ be extensions of $F$ in an algebraic closure $\bar{F}$ of $F$ such that $K ∕ F$ is finite Galois. Then $𝐸𝐾 ∕ E$ and $K ∕ (E \cap K)$ are finite Galois, and the restriction map

{res}_{K} : Gal (𝐸𝐾 ∕ E) \to Gal (K ∕ (E \cap K)), {res}_{K} (σ) = σ |_{K} for σ \in Gal (𝐸𝐾 ∕ E)

is an isomorphism.

Proof.

First, note that $𝐸𝐾 ∕ E$ is normal as it is the splitting field of the same set of polynomials in $F [x]$ that $K$ is over $F$ . Since $K ∕ F$ is finite and separable, we have $K = F (β)$ for some $β \in K$ , so $𝐸𝐾 = E (β)$ , and the fact that the minimal polynomial of $β$ is separable over $F$ tells us that it is over $E$ as well, and therefore $𝐸𝐾 ∕ E$ is separable as well. Thus, $𝐸𝐾 ∕ E$ is Galois, and $K ∕ (E \cap K)$ is Galois by Lemma 6.12.10.

Now, suppose that $σ \in Gal (𝐸𝐾 ∕ E)$ and ${res}_{K} (σ) = σ |_{K} = 1$ . By definition, we have $σ |_{E} = 1$ as well, so $σ$ fixes every rational function over $E$ in $β$ , and therefore $σ$ fixes $𝐸𝐾$ , which is to say that $σ = 1$ , or ${res}_{K}$ is injective. Now, let $H$ be the image of ${res}_{K}$ . The elements of $K$ fixed by $H$ are exactly the elements of $K$ fixed by $Gal (𝐸𝐾 ∕ E)$ , so we have

K^{H} = K^{Gal (𝐸𝐾 ∕ E)} = {(𝐸𝐾)}^{Gal (𝐸𝐾 ∕ E)} \cap K = E \cap K,

and therefore $H = Gal (K ∕ K^{H}) = Gal (K ∕ (E \cap K))$ , so ${res}_{K}$ is surjective as well. □

Proposition 6.12.23. §

Let $L ∕ F$ be an algebraic extension, and let $K$ and $E$ be finite Galois extensions of $F$ in $L$ . Then $𝐸𝐾 ∕ F$ and $E \cap K ∕ F$ are Galois, and the product of restriction maps

π : Gal (𝐸𝐾 ∕ F) \to Gal (K ∕ F) \times Gal (E ∕ F), π (σ) = (σ |_{K}, σ |_{E}) for σ \in Gal (𝐸𝐾 ∕ F)

is an injective homomorphism that is an isomorphism if and only if $E \cap K = F$ .

Proof.

That $𝐸𝐾 ∕ F$ is separable is Corollary 6.10.13 applied to $𝐸𝐾 ∕ E$ and $E ∕ F$ , and that it is normal is Proposition 6.11.9. If $β \in E \cap K$ , then both $E$ and $K$ contain all roots in $\bar{F}$ of its minimal polynomial, so $E \cap K$ also contains these roots, hence is normal over $K$ . That $E \cap K$ is separable over $F$ follows from the fact that $E$ is.

The kernel of $π$ is exactly those elements of $Gal (𝐸𝐾 ∕ F)$ that fix both $K$ and $E$ , and hence fix all of $𝐸𝐾$ , since every element of $𝐸𝐾$ is a rational function in the elements of $E$ and $K$ . Thus $π$ is injective. Since $π$ is injective, it is surjective if and only if the orders of its domain and codomain are the same, which is to say if and only if

[𝐸𝐾 : F] = [E : F] [K : F] .

By Proposition 6.12.22, we have

[𝐸𝐾 : F] = [𝐸𝐾 : K] [K : F] = [E : E \cap K] [K : F],

so $π$ is surjective if and only if $E \cap K = F$ . □

Definition 6.12.24. §

Let $K ∕ F$ be a Galois extension.

a.: We say that $K ∕ F$ is abelian if $Gal (K ∕ F)$ is abelian.
b.: We say that $K ∕ F$ is cyclic if $Gal (K ∕ F)$ is cyclic.

Examples 6.12.25. §

We revisit Examples 6.12.20 and 6.12.21.

a.: The field $ℚ (i, \sqrt{2})$ is the compositum of the normal extensions $ℚ (i)$ and $ℚ (\sqrt{2})$ , which both have Galois group $ℤ ∕ 2 ℤ$ and satisfy $ℚ (i) \cap ℚ (\sqrt{2}) = ℚ$ . By Proposition 6.12.23, we have $Gal (ℚ (i, \sqrt{2}) ∕ ℚ) ≅ {(ℤ ∕ 2 ℤ)}^{2}$ . The extension $ℚ (i, \sqrt{2}) ∕ ℚ$ is abelian.
b.: Take $G = Gal (ℚ (ω, \sqrt[3]{2}) ∕ ℚ)$ . Take $K = ℚ (ω)$ and $E = ℚ (\sqrt[3]{2})$ . Then $G = Gal (𝐸𝐾 ∕ ℚ)$ , and we set $N = Gal (𝐸𝐾 ∕ K)$ and $H = Gal (𝐸𝐾 ∕ E)$ . The map ${res}_{K} : G \to Gal (K ∕ ℚ)$ is a surjection with kernel $N$ that restricts to an isomorphism on $H$ by Proposition 6.12.22. In particular, $H$ is a complement to $N$ , and $G$ is a semidirect product $N ⋊ H$ , nontrivial as $E ∕ ℚ$ is not normal. In our case, $𝑁 ≅ ℤ ∕ 3 ℤ$ and $𝐻 ≅ ℤ ∕ 2 ℤ$ , so $G$ is nonabelian of order $6$ , isomorphic to $S_{3}$ .

The following example is worth being stated as a proposition, as it tells us that all Galois groups of all extensions of finite fields are cyclic.

Proposition 6.12.26. §

Let $q$ be a prime power and $n \geq 1$ . Then $𝔽_{q^{n}} ∕ 𝔽_{q}$ is cyclic of degree $n$ .

Proof.

The group $Gal (𝔽_{q^{n}} ∕ 𝔽_{q})$ contains the Frobenius element $φ_{q}$ with $φ_{q} (α) = α^{q}$ for all $α \in 𝔽_{q^{n}}$ . For $φ_{q}^{r} (α) = α^{q^{r}}$ to equal $α$ would mean that that $α$ is a $(q^{r} - 1)$ th root of unity, which in turn could only happen for all $α \in 𝔽_{q^{n}}$ if and only if $r$ is a multiple of $n$ . That is, the order of $φ_{q}$ is $n$ . Therefore, $G_{n}$ must be cyclic of order $n$ , generated by $φ_{q}$ . We have that $𝔽_{q^{m}}$ is a subfield of $𝔽_{q^{n}}$ if and only if $m$ divides $n$ , in which case $Gal (𝔽_{q^{n}} ∕ 𝔽_{q^{m}}) = ⟨ φ_{q}^{m} ⟩$ is a cyclic group of order $n ∕ m$ . In particular, every finite Galois extension of finite fields is cyclic. □

We can also determine the structure of the Galois groups of cyclotomic extensions of $ℚ$ . We note that the extension $ℚ (ζ_{n}) ∕ ℚ$ is Galois in that $ℚ (ζ_{n})$ is the splitting field of $x^{n} - 1$ .

Terminology 6.12.27. §

For $n \geq 1$ and $a \in {(ℤ ∕ 𝑛ℤ)}^{\times}$ , we will take $ζ_{n}^{a}$ to be $ζ_{n}^{ã}$ for any $ã \in ℤ$ with $a = ã + 𝑛ℤ$ .

Definition 6.12.28. §

For every $n \geq 1$ , the $n$ th cyclotomic character is the unique map

χ_{n} : Gal (ℚ (ζ_{n}) ∕ ℚ) \to {(ℤ ∕ 𝑛ℤ)}^{\times}

such that $σ (ζ_{n}) = ζ_{n}^{χ_{n} (σ)}$ for all $σ \in Gal (ℚ (ζ_{n}) ∕ ℚ)$ .

Proposition 6.12.29. §

The $n$ th cyclotomic character is an isomorphism for every $n \geq 1$ .

Proof.

We note first that $χ_{n}$ is a homomorphism. That is, for $σ, τ \in Gal (ℚ (ζ_{n}) ∕ ℚ)$ , we have

ζ_{n}^{χ_{n} (𝜎𝜏)} = 𝜎𝜏 (ζ_{n}) = σ (ζ_{n}^{χ_{n} (τ)}) = σ {(ζ_{n})}^{χ_{n} (τ)} = ζ_{n}^{χ_{n} (σ) χ_{n} (τ)} .

Next, note that $χ_{n}$ is injective since an element of $Gal (ℚ (ζ_{n}) ∕ ℚ)$ is determined by its value on the generator $ζ_{n}$ of the extension. Finally, Theorem 6.6.11 implies that $[ℚ (ζ_{n}) : ℚ] = φ (n)$ , so the orders of the two groups are the same. □

Corollary 6.12.30. §

The $n$ th cyclotomic field is a finite abelian extension of $ℚ$ .

Remark 6.12.31. §

The Kronecker-Weber theorem, a proof of which is beyond the scope of these notes, states that every finite abelian extension of $ℚ$ is contained inside some cyclotomic field.

6.13. Permutations of roots

We first recall that every finite Galois extension is the splitting field of some polynomial (and in fact we may take that polynomial to be irreducible by the primitive element theorem).

Theorem 6.13.1. §

Let $K ∕ F$ be the splitting field of a separable degree $n$ polynomial in $F [x]$ . Then $Gal (K ∕ F)$ is isomorphic to a subgroup of $S_{n}$ .

Proof.

Let $K$ be the splitting field of $f \in F [x]$ , and let $X$ be the set of $n$ roots of $f$ . For $α \in X$ and $σ \in Gal (K ∕ F)$ , we have $f (σ (α)) = σ (f (α)) = 0$ , so $σ (α) \in X$ . In other words, $Gal (K ∕ F)$ acts on $X$ , and thus we have an induced permutation representation $ρ : Gal (K ∕ F) \to S_{X}$ . Note that $K$ is given by adjoining the elements of $X$ to $F$ , so if $σ \in Gal (K ∕ F)$ fixes every element of $X$ , it fixes every element of $K$ and is therefore tirival. Thus, the action of $Gal (K ∕ F)$ on $X$ is faithful, so $ρ$ is injective. □

Corollary 6.13.2. §

Let $K ∕ F$ be the splitting field of a separable degree $n$ polynomial in $F [x]$ . Then $[K : F]$ divides $n!$ .

Examples 6.13.3. §

Again, we revisit Examples 6.12.20 and 6.12.21.

a.

The field $ℚ (\sqrt{2}, i)$ is the splitting field of $(x^{2} - 2) (x^{2} + 1)$ over $ℚ$ , which has $4$ roots. The image of $Gal (ℚ (\sqrt{2}, i) ∕ ℚ)$ under any permutation representation on these roots is conjugate to $⟨ (1 2), (3 4) ⟩$ .

b.

If we label the roots of $x^{3} - 2$ in the order $\sqrt[3]{2}$ , $ω \sqrt[3]{2}$ , $ω^{2} \sqrt[3]{2}$ , then we have a permutation representation

ρ : G = Gal (ℚ (ω, \sqrt[3]{2}) ∕ ℚ (ω)) \to_{}^{\sim} S_{3} .

We have $G = ⟨ σ, τ ⟩$ as in Example 6.12.21 with $ρ (σ) = (2 3)$ and $ρ (τ) = (1 2 3)$ .

One might ask if every subgroup of $S_{n}$ , and therefore every finite group, occurs as the Galois group of some extension of fields. As we shall see, the answer is yes.

Definition 6.13.4. §

Let $F$ be a field, and let $x_{1}, x_{2}, \dots, x_{n}$ be indeterminates. For $1 \leq k \leq n$ , the $k$ th elementary symmetric polynomial $s_{n, k}$ in $F [x_{1}, x_{2}, \dots, x_{n}]$ is

s_{n, k} (x_{1}, \dots, x_{n}) = \sum_{1 \leq i_{1} < i_{2} < \dots < i_{k} \leq n} x_{i_{1}} x_{i_{2}} \dots x_{i_{k}} .

Remark 6.13.5. §

Put differently, $s_{k}$ is the sum over the subsets of $X_{n} = {1, 2, \dots, n}$ of order $k$ of the products of variables with indices in the sets. That is,

s_{n, k} (x_{1}, \dots, x_{n}) = \sum_{\begin{array}{c} P \subseteq X_{n} \\ | P | = k \end{array}} \prod_{i \in P} x_{i} .

As a consequence, $s_{k, n}$ is a sum of $(\binom{n}{k})$ monomials.

Examples 6.13.6. §

We have $s_{n, 1} = x_{1} + x_{2} + \dots + x_{n}$ and $s_{n, n} = x_{1} x_{2} \dots x_{n}$ . For $n = 3$ , we also have $s_{3, 2} = x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3}$ , and for $n = 4$ , we have

\begin{array}{rcl} s_{4, 2} = x_{1} x_{2} + x_{1} x_{3} + x_{1} x_{4} + x_{2} x_{3} + x_{2} x_{4} + x_{3} x_{5} & and & s_{4, 3} = x_{1} x_{2} x_{3} + x_{1} x_{2} x_{4} + x_{1} x_{3} x_{4} + x_{2} x_{3} x_{4} . \end{array}

Proposition 6.13.7. §

The function field $F (x_{1}, x_{2}, \dots, x_{n})$ is a finite Galois extension of its subfield $F (s_{n, 1}, s_{n, 2}, \dots, s_{n, n})$ , with Galois group isomorphic to $S_{n}$ .

Proof.

Let $E = F (s_{n, 1}, s_{n, 2}, \dots, s_{n, n})$ and $K = ℚ (x_{1}, x_{2}, \dots, x_{n})$ . The polynomial

f (y) = \prod_{i = 1}^{n} (y - x_{i}) = \sum_{i = 0}^{n} {(- 1)}^{n - i} s_{n, n - i} y^{i} \in E [y]

has roots $x_{i}$ with $1 \leq i \leq n$ . Thus $K$ is the splitting field of $f$ over $E$ . To $ρ \in S_{n}$ , we can associate a unique $ϕ (ρ) \in {Aut}_{F} (K)$ by

ϕ (ρ) (h (x_{1}, x_{2}, \dots, x_{n})) = h (x_{ρ (1)}, x_{ρ (2)}, \dots, x_{ρ (n)})

for $h \in K$ . As $S_{n}$ acts on the set of subsets of $X_{n}$ of order $k$ , Remark 6.13.5 implies that $ϕ (ρ) (s_{n, k}) = s_{n, k}$ for all $k$ , so $ϕ (ρ) \in Gal (K ∕ E)$ . The map $ϕ : S_{n} \to Gal (K ∕ E)$ is a homomorphism that is injective by definition and surjective by Theorem 6.13.1. □

We have the following consequence.

Corollary 6.13.8. §

Every finite group is isomorphic to the Galois group of some field extension.

Proof.

Let $G$ be a finite group, and choose $n$ such that $H$ is isomorphic to a subgroup of $S_{n}$ , which exists by Cayley’s theorem. Proposition 6.13.7 yields an extension $K ∕ E$ of fields with $Gal (K ∕ E) ≅ S_{n}$ . Then $G$ is isomorphic to some subgroup $H$ of $Gal (K ∕ E)$ , and we have $𝐻 ≅ Gal (K ∕ K^{H})$ . □

Definition 6.13.9. §

Let $F$ be a field. The discriminant of a monic, degree $n$ polynomial $f \in F [x]$ is

D (f) = \prod_{1 \leq i < j \leq n} {(α_{i} - α_{j})}^{2},

where $f = \prod_{i = 1}^{n} (x - α_{i})$ in a splitting field of $F$ .

The following lemma is obvious from the definition of the discriminant.

Lemma 6.13.10. §

The discriminant of a monic polynomial $f$ is $0$ if and only if $f$ is inseparable.

In fact, the discriminant of a monic polynomial lies in the ground field of the extension, from which it easily follows that it is well-defined independently of the choice of splitting field in its definition.

Proposition 6.13.11. §

The discriminant of a monic polynomial $f \in F [x]$ lies in $F$ .

Proof.

By Lemma 6.13.10, we may suppose that $f$ is separable. Let $K$ be a splitting field of $F$ , and let $σ \in Gal (K ∕ F)$ . As $σ$ permutes the roots $α_{i}$ of $f$ , it induces an element $ρ \in S_{n}$ such that $σ (α_{i}) = α_{ρ (i)}$ . Taking $Δ = \prod_{1 \leq i < j \leq n} (x_{i} - x_{j})$ , we know by Proposition 4.12.1 that $ρ (Δ) = sign (ρ) Δ$ for the standard action of $S_{n}$ on polynomials in variables $x_{1}, x_{2}, \dots, x_{n}$ . But then $ρ (Δ^{2}) = Δ^{2}$ , so plugging in $α_{i}$ for $x_{i}$ , we obtain $σ (D (f)) = D (f)$ . Since $D (f)$ is fixed by $Gal (K ∕ F)$ , it lies in $F$ . □

Remark 6.13.12. §

Supposing that $char F \neq 2$ , the proof of Proposition 6.13.11 shows that an element of $Gal (K ∕ F)$ for the splitting field $K$ of a separable polynomial $f$ of degree $n$ induces an even permutation of the roots of $f$ if and only if it fixes $\prod_{1 \leq i < j \leq n} (α_{i} - α_{j})$ .

As a direct consequence of Remark 6.13.12, we have the following.

Proposition 6.13.13. §

The discriminant $D (f)$ of a monic, separable polynomial $f \in F [x]$ is a square in $F^{\times}$ if and only if the Galois group of its splitting field has image a subgroup of $A_{n}$ via its permutation representation on the roots of $f$ .

We explore the consequences of Proposition 6.13.13 for polynomials of low degree.

Example 6.13.14. §

Suppose $char F \neq 2$ . Let $f = x^{2} + 𝑎𝑥 + b \in F [x]$ . Let $α$ , $β$ be the roots of $F$ in an algebraic closure of $F$ . The extension $F (α) ∕ F$ is normal, being that it is of degree $1$ or $2$ , so $F (α) = F (β)$ . Note that $- a = α + β$ and $b = 𝛼𝛽$ , so

D (f) = α^{2} + β^{2} - 2 𝛼𝛽 = a^{2} - 4 b .

Proposition 6.13.13 tells us that $a^{2} - 4 b$ is a square if and only if $α \in F$ . This can also be seen by the quadratic formula, which tells us in particular that $F (α) = F (\sqrt{D})$ if $char F \neq 2$ .

The case of degree $3$ polynomials is rather more involved.

Example 6.13.15. §

Let $f = x^{3} + a x^{2} + 𝑏𝑥 + c \in F [x]$ . If $char F \neq 3$ , then setting $y = x + \frac{a}{3}$ , we obtain

\begin{array}{l} f & = {(y - \frac{a}{3})}^{3} + a {(y - \frac{a}{3})}^{2} + b (y - \frac{a}{3}) + c \\ = (y^{3} - a y^{2} + \frac{a^{2}}{3} y - \frac{a^{3}}{27}) + (a y^{2} - \frac{2 a^{2}}{3} y + \frac{a^{3}}{9}) + (𝑏𝑦 - \frac{𝑎𝑏}{3}) + c \\ = y^{3} + (- \frac{a^{2}}{3} + b) y + (\frac{2 a^{3}}{27} - \frac{𝑎𝑏}{3} + c) . \end{array}

Set $p = \frac{1}{3} (- a^{2} + 3 b)$ and $q = \frac{1}{27} (2 a^{3} - 9 𝑎𝑏 + 27 c)$ , and let $g = x^{3} + 𝑝𝑥 + q \in F [x]$ .

Let $K$ be a splitting field of $f$ over $F$ , and let $α, β, γ \in K$ be the roots of $g$ . Then $α + β + γ = 0$ , $s_{3, 2} (α, β, γ) = p$ , and $- 𝛼𝛽𝛾 = q$ . Note that this implies that

0 = {(α + β + γ)}^{2} = α^{2} + β^{2} + γ^{2} + 2 p

(6.13.1)

and

p^{2} = {(𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾)}^{2} = 2 𝛼𝛽𝛾 (α + β + γ) + α^{2} β^{2} + α^{2} γ^{2} + β^{2} γ^{2} = α^{2} β^{2} + α^{2} γ^{2} + β^{2} γ^{2} .

(6.13.2)

Note that the formal derivative of $g$ is

3 x^{2} + p = s_{3, 2} (x - α, x - β, x - γ),

and we can plug $α$ into this, for instance, to obtain

3 α^{2} + p = (α - β) (α - γ) .

Doing this also for $β$ and $γ$ and taking the ordering of the differences into account, we obtain by (6.13.1) and (6.13.2) that

\begin{array}{l} - D (g) & = (3 α^{2} + p) (3 β^{2} + p) (3 γ^{2} + p) \\ = 27 α^{2} β^{2} γ^{2} + 9 p (α^{2} β^{2} + α^{2} γ^{2} + β^{2} γ^{2}) + 3 p^{2} (α^{2} + β^{2} + γ^{2}) + p^{3} \\ = 27 q^{2} + 9 p^{3} - 6 p^{3} + p^{3} = 27 q^{2} + 4 p^{3} . \end{array}

That is, $D (g) = - 4 p^{3} - 27 q^{2}$ . Since the roots of $f$ and $g$ differ by $\frac{a}{3}$ , the differences of the roots of the two are the same, so $D (f) = D (g)$ , and one may then compute that

D (f) = a^{2} b^{2} - 4 a^{3} c + 18 𝑎𝑏𝑐 - 4 b^{3} - 27 c^{2} .

In fact, the latter formula holds even if $char F = 3$ .

Now, suppose that $f$ is irreducible and $char F \neq 2$ . Then $Gal (K ∕ F)$ is isomorphic to a subgroup of $S_{3}$ of order divisible by $3$ , so it is either isomorphic to $A_{3} ≅ ℤ ∕ 3 ℤ$ or $S_{3}$ , depending on whether $D (f)$ is a square or not, respectively. If $D (f) \in F^{\times 2}$ , then $K$ is given by adjoining any single root of $f$ . If $D (f) \notin F^{\times 2}$ , then $K$ has a unique intermediate extension $F (D {(f)}^{1 ∕ 2})$ of degree $2$ , and $K$ is given by adjoining to this any root of $f$ .

We go into a bit less detail for polynomials of degree $4$ .

Example 6.13.16. §

Suppose that $char F \neq 2$ . Let $K$ be the splitting field of a monic, irreducible polynomial $f$ of degree $4$ in $F [x]$ . If we suppose that $f$ has the form $f = x^{4} + p x^{2} + 𝑞𝑥 + r$ , which may be accomplished by a simple change of variables that preserves the discriminant, then

D (f) = 16 p^{4} r - 4 p^{3} q^{2} - 128 p^{2} r^{2} + 144 p q^{2} r - 27 q^{4} + 256 r^{3} .

If $D (f) \in F^{\times 2}$ , then $Gal (K ∕ F)$ is isomorphic to a subgroup of $A_{4}$ of degree divisible by $4$ , so $A_{4}$ or the Klein $4$ -group $V_{4}$ . If $D (f) \notin F^{\times 2}$ , then $Gal (K ∕ F)$ is isomorphic to $ℤ ∕ 4 ℤ$ , $D_{4}$ , or $S_{4}$ .

Let $α_{1}, α_{2}, α_{3}, α_{4}$ be the roots of $f$ , and set $β_{3} = (α_{1} + α_{2}) (α_{3} + α_{4})$ , $β_{2} = (α_{1} + α_{3}) (α_{2} + α_{4})$ , and $β_{1} = (α_{1} + α_{4}) (α_{2} + α_{3})$ . The set ${β_{1}, β_{2}, β_{3}}$ is a union of orbits under $Gal (K ∕ F)$ , so we can set

g = (x - β_{1}) (x - β_{2}) (x - β_{3}) \in F [x] .

From $β_{2} - β_{3} = (α_{1} - α_{4}) (α_{2} - α_{3})$ and the corresponding equalities for the other differences, one sees that $D (g) = D (f)$ . One may also compute that $g = x^{3} - 2 p x^{2} + (p^{2} - 4 r) x + q^{2}$ for $f$ of the above form. Let $E$ be the splitting field of $g$ over $F$ , and note that $K = E (α_{i})$ for each $i \in {1, 2, 3, 4}$ .

Let $ρ_{K} : Gal (K ∕ F) \to S_{4}$ (resp., $ρ_{E} : Gal (E ∕ F) \to S_{3}$ ) be the permutation map for the given ordering of the $α_{i}$ (resp., $β_{i}$ ). Then we have $π : S_{4} \to S_{3}$ with kernel $⟨ (1 2) (3 4), (1 3) (2 4) ⟩$ and restricting to the identity on $⟨ (1 2), (1 2 3) ⟩$ such that $π (ρ_{K} (σ)) = ρ_{E} (σ |_{E})$ for all $σ \in Gal (K ∕ F)$ .

If $g$ splits, which is to say $E = F$ , then $Gal (K ∕ F) ≅ V_{4}$ . If $g$ factors as a linear polynomial times an irreducible quadratic so that $E = F (D {(f)}^{1 ∕ 2})$ , then $Gal (K ∕ F) ≅ D_{4}$ if $f$ is irreducible over $E$ and $Gal (K ∕ F) ≅ ℤ ∕ 4 ℤ$ otherwise. If $g$ is irreducible and $D (f) \in F^{\times 2}$ , then $Gal (E ∕ F) ≅ ℤ ∕ 3 ℤ$ , which forces $Gal (K ∕ F) ≅ A_{4}$ since $4$ divides $[K : F]$ . If $g$ is irreducible and $D (f)$ is not a square in $F$ , then $Gal (E ∕ F) ≅ S_{3}$ , which forces $Gal (K ∕ F) ≅ S_{4}$ .

We next present a proof of the fundamental theorem of algebra that uses Galois theory. We will use the fact that every polynomial of odd degree has a real root (by the intermediate value theorem). We also recall that quadratic polynomials in $ℂ [x]$ split completely, as is seen via the quadratic formula and the fact that complex numbers have square roots in $ℂ$ .

Proof of the fundamental theorem of algebra.

First, let $f \in ℂ [x]$ be monic and irreducible, and let $\bar{f} \in ℂ [x]$ given by applying complex conjugation to its coefficients. The polynomial $g = f \bar{f}$ lies in $ℝ [x]$ since complex conjugation permutes $f$ and $\bar{f}$ , and it suffices to show that $g$ has a root in $ℂ$ . So, we can and do assume that $f \in ℝ [x]$ .

Let $n = \deg f$ , and write $n = 2^{k} m$ for some odd $m$ and $k \geq 0$ . If $k = 0$ , then $f$ has odd degree and hence a real root, so we suppose $k \geq 1$ . By induction, suppose we know that all polynomials in $ℝ [x]$ of degree $2^{k - 1}$ times an odd number have a root in $ℂ$ . Let $α_{1}, α_{2}, \dots, α_{n}$ be the roots of $f$ in a splitting field $Ω$ of $f$ over $ℂ$ .

For $t \in ℝ$ , define

h_{t} (x) = \prod_{1 \leq i < j \leq n} (x - (α_{i} + α_{j} + t α_{i} α_{j})) \in Ω [x] .

Any permutation of the $α_{i}$ ’s preserves $h_{t}$ , so $Gal (Ω ∕ ℝ)$ fixes $h_{t}$ , and thus $h_{t} \in ℝ [x]$ . Note that $\deg h_{t} = (\binom{n}{2}) = 2^{k - 1} m^{'}$ for some odd $m^{'}$ , and thus by induction $h_{t}$ has a root in $ℂ$ , which necessarily has the form $α_{i} + α_{j} + t α_{i} α_{j}$ for some $i < j$ . In fact, we have such a root for every $t \in ℝ$ , and since that is an infinite set of $t$ , there exist $i < j$ and $s, t \in ℝ$ such that $α_{i} + α_{j} + s α_{i} α_{j}$ and $α_{i} + α_{j} + t α_{i} α_{j}$ are both in $ℂ$ from which it follows that $α_{i} + α_{j} \in ℂ$ and $α_{i} α_{j} \in ℂ$ . But then $(x - α_{i}) (x - α_{j}) \in ℂ [x]$ , which being quadratic, has a root in $ℂ$ . □