Abstract Algebra — Chapter 5 Advanced ring theory

Chapter 5
Advanced ring theory

5.1. Unique factorization domains

In this section, we investigate the role that prime numbers play in the integers in greater generality. Recall that every nonzero integer can be written as plus or minus a product of distinct prime powers, and these prime powers are unique. Note that the units in $ℤ$ are $\pm 1$ , so we can say that every nonzero integer can be written as a product of prime powers times a unit. In this section, we investigate this property for a larger class of integral domains.

First, we introduce an analogue of prime numbers.

Definition 5.1.1. §

Let $R$ be an integral domain. A nonunit and nonzero element $p \in R$ is said to be an irreducible element if for every $a, b \in R$ with $p = 𝑎𝑏$ , either $a$ or $b$ is a unit.

Definition 5.1.2. §

Two elements $a$ and $b$ of a nonzero commutative ring $R$ are said to associates if $a = 𝑢𝑏$ with $u \in R^{\times}$ .

Of course, the property of being associate is an equivalence relation on an integral domain $R$ . The equivalence class of $0$ is ${0}$ and that of $1$ is $R^{\times}$ . We have the following simple lemma, which tells us that the equivalence class of an irreducible element consists of irreducible elements.

Lemma 5.1.3. §

If $R$ is an integral domain, and $p \in R$ is irreducible, then so is every associate of $p$ .

Proof.

That is, if $u \in R^{\times}$ and $𝑢𝑝 = 𝑎𝑏$ for $a, b \in R$ , then $p = (u^{- 1} a) b$ . As $p$ is irreducible, either $u^{- 1} a \in R^{\times}$ or $b \in R^{\times}$ . Finally, if $u^{- 1} a \in R^{\times}$ , then $a \in R^{\times}$ . □

Examples 5.1.4. §

a.

The irreducible elements of $ℤ$ are $\pm p$ for prime numbers $p$ . The elements $p$ and $- p$ are associates.

b.

The irreducible elements of $F [x]$ , for a field $F$ , are the irreducible polynomials of $F$ , since the units of $F [x]$ are the nonzero constant polynomials. Every nonzero polynomial has a unique associate with leading coefficient equal to $1$ .

c.

In the subring of $ℂ$ that is

ℤ [\sqrt{- 2}] = {a + b \sqrt{- 2} ∣ a, b \in ℤ},

a number of prime integers are no longer irreducible. For instance $2 = - {(\sqrt{- 2})}^{2}$ , and $\sqrt{- 2}$ is not a unit. Also, $3 = (1 + \sqrt{- 2}) (1 - \sqrt{- 2})$ , and neither $1 + \sqrt{- 2}$ nor $1 - \sqrt{- 2}$ is a unit, for if, e.g., $u \in ℤ [\sqrt{- 2}]$ with $u (1 + \sqrt{- 2}) = 1$ , then $3 u = 1 - \sqrt{- 2}$ , which is clearly impossible. On the other hand, it turns out that $5$ is irreducible, though we do not prove this now.

Definition 5.1.5. §

An integral domain $R$ is a unique factorization domain, or a UFD, if every nonzero, nonunit element $a \in R$ can be written as a product

a = p_{1} p_{2} \dots p_{r}

with $p_{1}, p_{2}, \dots, p_{r}$ irreducible elements of $R$ for some $r \geq 1$ , and moreover, this expression is unique in the sense that if

a = q_{1} q_{2} \dots q_{s}

with $q_{1}, q_{2}, \dots, q_{s}$ irreducible for some $s \geq 1$ , then $s = r$ and there exists a permutation $σ \in S_{r}$ such that $q_{σ (i)}$ and $p_{i}$ are associates for all $1 \leq i \leq r$ .

Remark 5.1.6. §

If one wants to allow units, one can rephrase Definition 5.1.5 to read that every nonzero element $a \in R$ can be written as $a = u p_{1} \dots p_{r}$ with $u \in R^{\times}$ and $p_{1}, p_{2}, \dots, p_{r}$ irreducible in $R$ for some $r = 0$ in a unique manner such that any such decomposition of $a = v q_{1} \dots q_{s}$ has $s = r$ and, after a reordering of the irreducibles, each $q_{i}$ is an associate of $p_{i}$ .

Example 5.1.7. §

The ring $ℤ$ is a unique factorization domain.

As we shall see later, $F [x]$ for a field $F$ is a unique factorization domain as well.

Example 5.1.8. §

Consider the subring $F [x^{2}, 𝑥𝑦, y^{2}]$ of $F [x, y]$ . It consists exactly of the polynomials in $F [x, y]$ that can be written as polynomials in $x^{2}$ , $𝑥𝑦$ , and $y^{2}$ . These latter three elements are irreducible in $F [x^{2}, 𝑥𝑦, y^{2}]$ , but we have

x^{2} \cdot y^{2} = 𝑥𝑦 \cdot 𝑥𝑦,

so factorization is not unique.

A more standard example is the following.

Example 5.1.9. §

Consider the subring $ℤ [\sqrt{- 5}]$ of $ℂ$ . We have

6 = 2 \cdot 3 = (1 + \sqrt{- 5}) (1 - \sqrt{- 5}) .

The element $2$ divides only elements of the form $a + b \sqrt{- 5}$ with $a, b \in ℤ$ even, so it does not divide $1 + \sqrt{- 5}$ or $1 - \sqrt{- 5}$ . On the other hand, $2$ is irreducible since if $a + b \sqrt{- 5}$ divides $2$ , then so does its complex conjugate, and then

(a + b \sqrt{- 5}) (a - b \sqrt{- 5}) = a^{2} + 5 b^{2}

divides $2$ , which happens only if $a = \pm 1$ and $b = 0$ . Therefore, $ℤ [\sqrt{- 5}]$ is not a unique factorization domain.

One advantage of unique factorization domains is that they allow us to define a concept of greatest common divisor.

Definition 5.1.10. §

Let $R$ be a UFD. Let $a_{1}, a_{2}, \dots, a_{r} \in R$ be nonzero. A principal ideal $(d)$ for $d \in R$ is said to be the greatest common divisor, or GCD, of $a_{1}, a_{2}, \dots, a_{r}$ if $d$ divides $a_{i}$ for each $1 \leq i \leq r$ and if $d^{'}$ also divides each $a_{i}$ , then $d^{'}$ divides $d$ .

The element $d$ in the definition of GCD, if it exists, is only defined up to unit. On the other hand, $(d)$ is independent of this choice.

Lemma 5.1.11. §

Let $R$ be a UFD. Then every collection $a_{1}, a_{2}, \dots, a_{r}$ of nonzero elements of $R$ has a GCD.

Proof.

We sketch the proof. Factor each $a_{i}$ into a unit times a product of irreducibles. If there exists an irreducible element $p_{1}$ that divides each $a_{i}$ , an associate of it is one of the irreducibles appearing in the factorization of $a_{i}$ . We then have $b_{i} \in R$ with $a_{i} = p b_{i}$ for each $i$ , and the factorization of $b_{i}$ has one fewer irreducible element than that of $a_{i}$ . We repeat this process until the collection no longer has a common irreducible divisors, obtaining irreducibles $p_{1}, p_{2}, \dots, p_{k}$ such that $d = p_{1} p_{2} \dots p_{k}$ divides every $a_{i}$ .

We claim that $(d)$ is the GCD of $a_{1}, a_{2}, \dots, a_{k}$ . If not, then there exists $d^{'}$ that does not divide $d$ which divides every $a_{i}$ . This means that there exists an irreducible element $q \in R$ and some $n \geq 1$ such that $q^{n}$ divides $d^{'}$ but not $d$ . Then $q^{n}$ divides every $a_{i}$ , which means since $q^{n}$ does not divide $d$ that $q$ actually divides each $c_{i}$ such that $a_{i} = d c_{i}$ , in contradiction to the definition of $d$ . □

One advantage of having the notion of a GCD is that in quotient fields, it allows us to talk about fractions being in lowest terms.

Definition 5.1.12. §

Let $R$ be a UFD, and let $a, b \in R$ with $b \neq 0$ . We say that the fraction $\frac{a}{b}$ is reduced, or in lowest terms, if the GCD of $a$ and $b$ is $(1)$ .

Lemma 5.1.13. §

Let $R$ be a UFD. Every fraction in $Q (R)$ may be written in lowest terms.

Proof.

Let $a, b \in R$ with $b \neq 0$ . Let $(d)$ be the GCD of $a$ and $b$ . Then there exist $a^{'}, b^{'} \in R$ with $a = d a^{'}$ and $b = d b^{'}$ , and we have that the GCD of $a^{'}$ and $b^{'}$ is $(1)$ . We therefore have that $\frac{a^{'}}{b^{'}} = \frac{a}{b}$ , and the former form of the fraction is in lowest terms. □

Let us study factorization in principal ideal domains.

Definition 5.1.14. §

Let $X$ be a set, and let $\leq$ be a partial ordering on $X$ .

a.: An ascending chain in $X$ is a sequence ${(a_{i})}_{i \geq 1}$ of elements of $X$ such that $a_{i} \leq a_{i + 1}$ for all $i \geq 1$ .
b.: We say that $X$ satisfies the ascending chain condition, or ACC, if every ascending chain ${(a_{i})}_{i \geq 1}$ in $X$ is eventually constant: i.e., there exists $j \geq 1$ such that $a_{i} = a_{j}$ for all $i \geq j$ .

The following is an equivalent characterization of the ACC.

Proposition 5.1.15. §

A nonempty set $X$ with a partial ordering $\leq$ satisfies the ACC if and only if every nonempty subset of $X$ contains a maximal element.

Proof.

If every nonempty subset of $X$ contains a maximal element, then clearly ascending chains are eventually constant: i.e., their underlying sets are finite. For the other direction, it suffices to show that if $X$ satisfies the ACC, then it contains a maximal element. Let $C$ be a nonempty chain in $X$ , and suppose it does not have an upper bound. For each $x \in C$ , there exists $y \in C$ with $y > x$ , as otherwise $x$ would be an upper bound. We may therefore recursively pick $a_{i} \in X$ with $a_{i} < a_{i + 1}$ for each $i$ , but this is impossible. Thus $C$ has an upper bound, and therefore $X$ has a maximal element by Zorn’s lemma. □

Definition 5.1.16. §

We say that a commutative ring $R$ is noetherian if the set of its ideals satisfies the ascending chain condition with respect to containment of ideals.

Remark 5.1.17. §

We may rephrase the condition that $R$ be noetherian by saying that if ${(I_{n})}_{n \geq 1}$ is an ascending chain of ideals, then there exists $m \geq 1$ such that the union $I$ of the $I_{n}$ with $n \geq 1$ equals $I_{i}$ for all $i \geq m$ .

Remark 5.1.18. §

One may define a noncommutative ring to be left noetherian (resp., right noetherian rings) if it satisfies the ACC on left ideals (resp., right ideals). In general, a noetherian ring is taken to be one that is both left and right noetherian.

Theorem 5.1.19. §

A commutative ring $R$ is noetherian if and only if every ideal of $R$ is finitely generated.

Proof.

Suppose that every ideal of $R$ is finitely generated. Let ${(I_{n})}_{n \geq 1}$ be a chain of ideals of $R$ . Let $I$ be the union of the $I_{n}$ for $n \geq 1$ , which is an ideal by Lemma 3.10.10. Since $I$ is finitely generated, $I = (a_{1}, a_{2}, \dots, a_{r})$ , with $a_{k} \in I$ with $1 \leq k \leq r$ for some $r \geq 1$ . For each $k$ , there exists $m_{k} \geq 1$ with $a_{k} \in I_{m_{k}}$ , and if we let $m$ be the maximum of the $m_{k}$ , then $a_{k} \in I_{m}$ for every $a_{k}$ . Since $I$ is the smallest ideal of $R$ containing each $a_{k}$ , we have $I \subseteq I_{m}$ , which forces $I = I_{m}$ .

Conversely, suppose $R$ is noetherian, and let $I$ be an ideal of $R$ . Let $x_{1} \in I$ , and suppose inductively that we have constructed $x_{1}, x_{2}, \dots, x_{n} \in R$ with the property that if we set $I_{k} = (x_{1}, x_{2}, \dots, x_{k})$ for every $1 \leq k \leq n$ , then $I_{k} \subseteq I_{k + 1}$ for every $1 \leq k \leq n - 1$ . If $I_{n} \neq I$ , then let $x_{n + 1} \in I$ with $x_{n + 1} \notin I_{n}$ . Then $I_{n + 1} = (x_{1}, x_{2}, \dots, x_{n + 1})$ properly contains $I_{n}$ . If this process repeats indefinitely, then we have constructed an ascending chain ${(I_{n})}_{n \geq 1}$ that is not eventually constant, which would contradict the assumption that $R$ is noetherian. Therefore, there exists $m \geq 1$ such that $I_{m} = I$ , and so $I = (a_{1}, a_{2}, \dots, a_{m})$ is finitely generated. □

Corollary 5.1.20. §

Every principal ideal domain is noetherian.

Proposition 5.1.21. §

Let $R$ be a noetherian domain. Then every nonzero, nonunit $a \in R$ may be written as $a = p_{1} p_{2} \dots p_{r}$ with the $p_{i} \in R$ irreducible for all $1 \leq i \leq r$ and some $r \geq 1$ .

Proof.

Let

X = {(a) ∣ a \in R - (R^{\times} \cup {0}) is not a product of irreducible elements} .

which we order by inclusion of ideals. If $X \neq \emptyset$ , then $X$ contains a maximal element $(a)$ by Proposition 5.1.15. Since $a$ is not irreducible, we may write $a = 𝑏𝑐$ for some $b, c \notin R^{\times}$ . Then $(b)$ and $(c)$ properly contain $(a)$ as $R$ is a domain. By maximality of $(a)$ , we may write both $b$ and $c$ as products of irreducible elements. Since $a = 𝑏𝑐$ , we have that $a$ is product of irreducible elements as well, which is a contradiction. Thus $X$ is empty, as required. □

Lemma 5.1.22. §

Let $R$ be a PID, and let $a \in R$ be nonzero. Then $(a)$ is maximal if and only if $a$ is an irreducible element.

Proof.

Clearly, $a$ cannot be a unit for either condition to hold. If $a = 𝑏𝑐$ with $b$ and $c$ non-units, then $(a) ⊊ (b) ⊊ R$ , so $(a)$ is not maximal. And if $(a)$ is not maximal, then there exists an proper ideal $I = (c)$ of $R$ properly containing $(a)$ , so we may write $a = 𝑏𝑐$ with $b \in R$ . Since the containment is proper, $b$ is not a unit, and $c$ is not a unit by definition. Therefore, $a$ is reducible. □

Definition 5.1.23. §

An element $p$ of a domain $R$ is prime if for every $a, b \in R$ such that $p ∣ 𝑎𝑏$ , either $p ∣ a$ or $p ∣ b$ .

Lemma 5.1.24. §

In a domain, prime elements are irreducible.

Proof.

Let $R$ be a domain. Let $p \in R$ be prime and suppose $p = 𝑎𝑏$ for $a, b \in R$ . Then $p ∣ a$ or $p ∣ b$ as $p$ is prime. Suppose then without loss of generality that $a = 𝑝𝑐$ for some $c \in R$ . We then have $p = 𝑝𝑏𝑐$ , and by cancellation this forces $𝑏𝑐 = 1$ , so $b \in R^{\times}$ . Thus, $p$ is irreducible. □

Lemma 5.1.25. §

In a UFD, irreducible elements are prime.

Proof.

Let $R$ be a UFD, and let $π \in R$ be irreducible. Suppose that $a, b \in R$ with $π ∣ 𝑎𝑏$ . Write $𝑎𝑏 = 𝜋𝑐$ for $c \in R$ . Write $a$ , $b$ , and $c$ as products of irreducible elements times a unit. Since $R$ is a UFD, we must then have that $π$ is associate to one of the irreducible elements of which $a$ and $b$ are products. That is, $π$ divides $a$ or $b$ . □

Example 5.1.26. §

In Example 5.1.9, we showed that $2$ is irreducible in $ℤ [\sqrt{- 5}]$ , and $6 = (1 + \sqrt{- 5}) (1 - \sqrt{- 5})$ , but $2 ∤ (1 \pm \sqrt{- 5})$ . Thus, $2$ is not prime.

Proposition 5.1.27. §

Let $R$ be a domain in which every nonzero, non-unit element $a$ has a factorization $a = p_{1} p_{2} \dots p_{r}$ with $p_{i}$ prime for $1 \leq i \leq r$ and $r \geq 1$ . Then $R$ is a UFD.

Proof.

Let $a \in R$ be a nonzero, nonunit element, and write $a = p_{1} p_{2} \dots p_{r}$ with $p_{1}, p_{2}, \dots, p_{r}$ prime. By Lemma 5.1.24, this is also a factorization of $a$ into irreducible elements. Suppose that

a = q_{1} q_{2} \dots q_{s}

with $q_{1}, q_{2}, \dots, q_{s}$ irreducible. If $s = 1$ , then $a$ is irreducible, so $r = 1$ and $p_{1} = q_{1}$ . Suppose by induction we have proven uniqueness whenever there is a decomposition of $a$ with fewer than $s \geq 2$ irreducibles.

As $p_{r}$ is prime, we have that $p_{r}$ divides some $q_{i}$ for some $1 \leq i \leq s$ . Since $q_{i}$ is irreducible, this means that $q_{i} = w p_{r}$ with $w \in R^{\times}$ . Since $R$ is an integral domain, we then have

p_{1} p_{2} \dots p_{r - 1} = w q_{1} q_{2} \dots q_{i - 1} q_{i + 1} \dots q_{s} .

As $s \geq 2$ by assumption, note that $w q_{1}$ is an associate to $q_{1}$ and the expression on the right is a product of $s - 1$ irreducible elements. By induction, we have $r = s$ , and there exists a bijective function

σ : {1, 2, \dots, r - 1} \to {1, 2, \dots, i - 1, i + 1, \dots, r}

with $q_{σ (i)}$ and $p_{i}$ associates for each $1 \leq i \leq r - 1$ . We may extend $σ$ to an element of $S_{r}$ by setting $σ (r) = i$ , and then $q_{σ (r)} = q_{i}$ is an associate of $p_{r}$ as well, proving uniqueness. □

In a principal ideal domain, irreducible elements are in fact prime.

Lemma 5.1.28. §

Let $R$ be a PID. If $p \in R$ is irreducible, then $p$ is prime.

Proof.

Let $a, b \in R$ with $p ∣ 𝑎𝑏$ . Then $𝑎𝑏 \in (p)$ , and $(p)$ is maximal by Lemma 5.1.22. Since every maximal ideal of $R$ is prime, we have that $(p)$ is prime, and therefore either $a \in (p)$ or $b \in (p)$ . □

The following key theorem is now a consequence of results we have already proven.

Theorem 5.1.29. §

Every principal ideal domain is a unique factorization domain.

Proof.

Suppose that $R$ is a PID. As $R$ is noetherian, Proposition 5.1.21 states that we may factor every nonzero, nonunit element in $R$ as a product of irreducible elements. As irreducible elements of a PID are prime, Proposition 5.1.27 tells us that $R$ is a UFD. □

Given that every polynomial ring over a field is a PID, we have the following corollary. It is an interesting exercise to prove it directly.

Corollary 5.1.30. §

For any field $F$ , the ring $F [x]$ is a unique factorization domain.

Corollary 3.9.2 tells us what we may already have known from experience, that we can factor one-variable polynomials into irreducible factors over a field, and there is only one way to do this.

5.2. Polynomial rings over commutative rings

Now that we know that every PID is a UFD, the question arises: is every UFD also a PID? The answer, in fact, is no. For this, let us examine polynomial rings over integral domains in a bit more detail.

Definition 5.2.1. §

Let $R$ be an integral domain. A polynomial $f \in R [x]$ is said to be primitive if the only elements of $R$ that divide all of the coefficients of $f$ are units.

In a UFD, we can actually talk about the GCD of the coefficients of a polynomial.

Definition 5.2.2. §

Let $R$ be a UFD. The content of the a polynomial in $R [x]$ is the GCD of its coefficients.

Remark 5.2.3. §

If $R$ is a UFD, then a polynomial in $R [x]$ is primitive if and only if the GCD of its coefficients is $(1)$ .

Definition 5.2.4. §

A polynomial in $R [x]$ for a nonzero ring $R$ is said to be monic is its leading coefficient is $1$ .

Remark 5.2.5. §

Monic polynomials in $R [x]$ , where $R$ is a UFD, are primitive.

Lemma 5.2.6. §

Let $R$ be a UFD. If $(c)$ is the content of $f \in R [x]$ for, then there exists a primitive polynomial $g \in R [x]$ with $f = 𝑐𝑔$ .

Proof.

By definition, $c$ divides each coefficient of $f$ , so $f = 𝑐𝑔$ for some $g \in R [x]$ . Let $d \in R$ be such that $(d)$ is the content of $g$ . Then $g = 𝑑h$ for some $h \in R [x]$ , so we have $f = 𝑐𝑑h$ . But this implies that $𝑐𝑑$ divides every coefficient of $f$ , so $𝑐𝑑$ divides the content $c$ , forcing $d$ to be a unit as $R$ is a domain. Therefore, $g$ is primitive. □

Example 5.2.7. §

The polynomial $f = 25 x^{2} + 10 x - 15$ in $ℤ [x]$ has content $5$ , and so it is not primitive. In fact, $f = 5 g$ , where $g = 5 x^{2} + 2 x - 3$ , and $g$ is primitive.

Lemma 5.2.8 (Gauss’s Lemma). §

Let $R$ be a UFD. Then the product of any two primitive polynomials in $R [x]$ is primitive.

Proof.

Let

\begin{array}{rcl} f = \sum_{i = 0}^{n} a_{i} x^{i} & and & g = \sum_{j = 0}^{m} b_{j} x^{j} \end{array}

be primitive polynomials in $R [x]$ . The $k$ th coefficient of $𝑓𝑔$ is $c_{k} = \sum_{i = 0}^{k} a_{i} b_{k - i}$ . If $p$ is an irreducible element of $R$ , then since $f$ and $g$ are primitive, there exist minimal nonnegative integers $r$ and $s$ such that $p ∤ a_{r}$ and $p ∤ b_{s}$ . Since $p ∣ a_{i}$ for $i < r$ and $p ∣ b_{j}$ for $j < s$ , which is to say that $p ∣ b_{r + s - i}$ for $r < i \leq r + s$ , we have that $p$ divides every term of $c_{r + s}$ except $a_{r} b_{s}$ , which it does not divide. Therefore, $p$ does not divides $c_{r + s}$ . Since $p$ was arbitrary, $𝑓𝑔$ is primitive. □

Note that we can speak about polynomials being irreducible in $R [x]$ for any integral domain $R$ , since we have a notion of irreducible element in such a ring. For a field $F$ , this coincides with the usual notion of an irreducible polynomial.

Proposition 5.2.9. §

Let $R$ be an integral domain, and let $F = Q (R)$ .

a.: If $f \in R [x]$ is a primitive polynomial that is irreducible as an element of $F [x]$ , then $f$ is irreducible in $R [x]$ . In particular, if $f$ is primitive and cannot be written as a product of two nonconstant polynomials in $R [x]$ , then it is irreducible in $R [x]$ .
b.: Suppose that $R$ is a UFD. If $f \in R [x]$ is irreducible, then it is irreducible as an element of $F [x]$ as well. In fact, if $f \in R [x]$ and $f = 𝑔h$ for nonconstant $g, h \in F [x]$ , then there exists $α \in F^{\times}$ such that $g^{'} = 𝛼𝑔$ and $h^{'} = α^{- 1} h$ are in $R [x]$ and therefore $f = g^{'} h^{'}$ in $R [x]$ .

Proof.

First, we treat part a. If $f \in R [x]$ is primitive and reducible (which is to say, not irreducible and not a unit or zero), then we can write $f = 𝑔h$ for nonunits $g, h \in R [x]$ . If $g$ or $h$ is constant, then $f$ is not primitive, so neither is constant, and therefore $f$ is reducible in $F [x]$ .

Next, we turn to part b. Suppose that $f \in R [x]$ can be written as $f = 𝑔h$ with $g, h \in F [x]$ nonconstant. Let $(d)$ (resp., $(e)$ ) be a multiple of all of the denominators of the coefficients of $g$ (resp., $h$ ), written in lowest terms. Then $𝑑𝑒𝑓 = g^{'} h^{'}$ , where $g^{'}, h^{'} \in R [x]$ are nonconstant. The content of $𝑑𝑒𝑓$ is contained in $(𝑑𝑒)$ , so the content of $g^{'} h^{'}$ is as well. By unique factorization in $R$ , we may write $𝑑𝑒 = d^{'} e^{'}$ , where $d^{'} \in R$ divides the content of $g^{'}$ and $e^{'}$ divides the content of $h^{'}$ , and we may then divide $g^{'}$ by $d^{'}$ and $h^{'}$ by $e^{'}$ to obtain $g^{″}$ and $h^{″}$ in $R [x]$ such that $f = g^{″} h^{″}$ . Therefore, $f$ is reducible in $R [x]$ , and the remaining statement of the lemma holds as well. □

We are now ready to prove the following.

Theorem 5.2.10. §

If $R$ is a UFD, then $R [x]$ is a UFD as well.

Proof.

Let $f \in R [x]$ be a nonzero element that is not a unit. Write

f = f_{1} f_{2} \dots f_{r}

with $f_{i} \in R [x]$ nonconstant, where $r$ is maximal such that this can be done. Note that such a maximal $r$ exists as the degree of $f$ is finite. For $1 \leq i \leq r$ , let $(c_{i})$ be the content of $f_{i}$ , and define $g_{i} \in R [x]$ by $f_{i} = c_{i} g_{i}$ . Set $c = c_{1} c_{2} \dots c_{r}$ , and set $g = g_{1} g_{2} \dots g_{r}$ . Now, if any $g_{i}$ were not irreducible in $F [x]$ for $F = Q (R)$ , then it would not be irreducible in $R [x]$ by Proposition 5.2.9b. Moreover, since $g_{i}$ is primitive, it would then be written as a product of two nonconstant polynomials in $R [x]$ , which would contradict the maximality of $r$ . Therefore, each $g_{i}$ is irreducible. Since $R$ is a UFD, we may also write $c = p_{1} p_{2} \dots p_{k}$ with $p_{i} \in R$ irreducible for $1 \leq i \leq k$ and some $k \geq 0$ , and so

f = p_{1} p_{2} \dots p_{k} g_{1} g_{2} \dots g_{r}

is a factorization of $f$ into irreducibles in $R [x]$ .

Now, if

f = q_{1} q_{2} \dots q_{l} h_{1} h_{2} \dots h_{s}

with $q_{i} \in R$ irreducible and $h_{i} \in R [x]$ irreducible and nonconstant, then $(q_{1} q_{2} \dots q_{l})$ is the content of $f$ by Gauss’s lemma, and so $q_{1} q_{2} \dots q_{l}$ agrees with $c$ up to unit in $R$ . Since $R$ is a UFD, it follows that $l = k$ and there exists $σ \in S_{k}$ such that each $q_{σ (i)}$ is an associate of $p_{i}$ . Next, we have

g_{1} g_{2} \dots g_{r} = u h_{1} h_{2} \dots h_{s}

for some unit $u \in R^{\times}$ , and by uniqueness of factorization in $F [x]$ , we have that $s = r$ , and there exists $τ \in S_{r}$ such that $h_{τ (i)} = v_{i} g_{i}$ for some $v_{i} \in F^{\times}$ for each $1 \leq i \leq r$ . But the content of each $g_{i}$ and each $h_{j}$ is $(1)$ , since these elements are irreducible in $R [x]$ , and therefore writing $v_{i} = \frac{a_{i}}{b_{i}}$ with $a_{i}, b_{i} \in R$ , the fact that $b_{i} h_{τ (i)} = a_{i} g_{i}$ implies that $(a_{i}) = (b_{i})$ , since both sides must have the same content. In other words, $v_{i} \in R^{\times}$ , and so $h_{τ (i)}$ and $g_{i}$ are associates in $R [x]$ , finishing the proof of uniqueness. □

Examples 5.2.11. §

a.: Since $ℤ$ is a UFD, so is $ℤ [x]$ . However, $ℤ [x]$ is not a PID, since $(p, x)$ is not principal.
b.: Since $ℚ [x]$ is a UFD, so is $ℚ [x, y]$ . Again, $ℚ [x, y]$ is not a PID, since $(x, y)$ is not principal.
c.: If $R$ is any UFD, then $R [x_{1}, x_{2}, \dots x_{n}]$ is a UFD for any $n \geq 1$ .

We next prove that a polynomial ring over a noetherian commutative ring is noetherian.

Theorem 5.2.12 (Hilbert’s basis theorem). §

The polynomial ring $R [x]$ over a commutative noetherian ring $R$ is noetherian.

Proof.

Let $I$ be an ideal of $R [x]$ . We must show that $I$ is finitely generated. Let $L$ be the set the leading coefficients of the elements of $I$ . Then $L$ is clearly an ideal: if $a \in L$ is the leading coefficient of $f \in I$ and $r \in R$ , then $𝑟𝑎$ is the leading coefficient of $𝑟𝑓 \in I$ , and if $a, b \in L$ are the leading coefficients of $f$ and $g$ , respectively, then $x^{\deg g} f + x^{\deg f} g \in I$ has leading coefficient $a + b$ . Since $R$ is noetherian, there exist $a_{1}, a_{2}, \dots, a_{k} \in L$ such that $L = (a_{1}, a_{2}, \dots, a_{k})$ . Let $f_{i} \in I$ of degree $n_{i} \geq 0$ have leading coefficient $a_{i}$ for $1 \leq i \leq k$ . Let $n = \max {n_{i} ∣ 1 \leq i \leq k}$ .

Next, for $m \geq 0$ , let $L_{m}$ be the set of all leading coefficients of polynomials in $I$ of degree $m$ . This, again, is clearly an ideal of $R$ , so we have $L_{m} = (b_{m, 1}, b_{m, 2}, \dots, b_{m, l_{m}})$ for some $l_{m} \geq 1$ and $b_{m, i} \in L_{m}$ for $1 \leq i \leq l_{m}$ . For each such $i$ , let $g_{m, i} \in I$ be a polynomial of degree $m$ with leading coefficient $b_{m, i}$ . We claim that $I$ is generated by

X = {f_{i} ∣ 1 \leq i \leq k} \cup ⋃_{m = 0}^{n} {g_{m, i} ∣ 1 \leq i \leq l_{m}} .

Let $J$ be the ideal of $R [x]$ generated by $X$ , which is contained in $I$ . Let $h \in I$ , and let $c \in L$ be its leading coefficient. We want to show that $h \in J$ . Write $c = \sum_{i = 1}^{k} r_{i} a_{i}$ with $r_{i} \in R$ . If $d \geq n$ , then $c$ is the leading coefficient of

h^{'} = \sum_{i = 1}^{k} r_{i} x^{d - n_{i}} f_{i} \in J,

so $h - h^{'}$ has degree less than $d$ . We can then replace $h$ by $h - h^{'}$ and repeat the process until $d < n$ .

We are reduced to showing that if $h \in I$ has degree $d < n$ and leading coefficient $c$ , then $h \in J$ . In this case, we have $c = \sum_{i = 1}^{l_{d}} s_{i} b_{d, i}$ with $s_{i} \in S$ , and $c$ is the leading coefficient of

h^{'} = \sum_{i = 1}^{l_{d}} s_{i} g_{d, i} \in J

Then $h - h^{'} \in J$ has degree less than $d$ . Replacing $h - h^{'}$ by $h$ and repeating the process, we see that $h \in J$ . □

Corollary 5.2.13. §

Let $R$ be a noetherian ring. Then $R [x_{1}, x_{2}, \dots, x_{n}]$ is noetherian for every $n \geq 1$ .

5.3. Irreducibility of polynomials

In this section, we investigate criteria for determining if a polynomial is irreducible or not.

Definition 5.3.1. §

Let $R$ be an integral domain, and let $𝔭$ be a prime ideal of $R$ . We say that a nonconstant polynomial $f = \sum_{i = 0}^{n} a_{i} x^{i}$ in $R [x]$ is an Eisenstein polynomial (with respect to $𝔭$ ) if $a_{n} \notin 𝔭$ , $a_{i} \in 𝔭$ for all $0 \leq i \leq n - 1$ , and $a_{0} \notin 𝔭^{2}$ .

Theorem 5.3.2 (Eistenstein criterion). §

Let $R$ be an integral domain, and let $f \in R [x]$ be an Eiseinstein polynomial.

a.: If $R$ is a UFD, then $f$ is irreducible in $Q (R) [x]$ .
b.: If $f$ is primitive, then it is irreducible in $R [x]$ .

Proof.

Suppose $f = \sum_{i = 0}^{n} a_{i} x^{i}$ is of degree $n$ and Eisenstein with respect to a prime ideal $𝔭$ of $R$ . By Proposition 5.2.9, it suffices for each part to show that $f$ is not a product of two nonconstant polynomials in $R [x]$ . So, let $g = \sum_{i = 0}^{s} b_{i} x^{i}$ and $h = \sum_{j = 0}^{t} c_{j} x^{j}$ be polynomials in $R [x]$ with $f = 𝑔h$ , where $s + t = n$ . We then have

a_{k} = \sum_{i = 0}^{k} b_{i} c_{k - i}

for all $0 \leq k \leq n$ . In particular, $a_{0} = b_{0} c_{0}$ is an element of $𝔭$ but not $𝔭^{2}$ . Since $𝔭$ is prime, at least one of $b_{0}$ and $c_{0}$ lies in $𝔭$ , but as $a_{0} \notin 𝔭^{2}$ , at least one does not lie in $𝔭$ as well.

Without loss of generality, suppose that $b_{0} \in 𝔭$ and $c_{0} \notin 𝔭$ . As $a_{n} = b_{s} c_{t} \notin 𝔭$ , we have $b_{s} \notin 𝔭$ . Let $k \geq 1$ be minimal such that $b_{k} \notin 𝔭$ . If $k < n$ , then $a_{k} \in 𝔭$ and $b_{i} \in 𝔭$ for $i < k$ , so we have $b_{k} c_{0} \in 𝔭$ , which therefore forces $c_{0} \in 𝔭$ by the primality of $𝔭$ . Therefore, $k = n$ , which means that $h$ is constant, proving the result. □

We will most commonly be concerned with the Eisenstein criterion in the case that $R = ℤ$ .

Example 5.3.3. §

For any prime number $p$ and integer $n \geq 1$ , the polynomial $x^{n} - p$ is irreducible by the Eisenstein criterion. That is, we take our prime ideal to be $(p)$ in the ring $ℤ$ .

Example 5.3.4. §

For a prime number $p$ , set

Φ_{p} = \frac{x^{p} - 1}{x - 1} = x^{p - 1} + x^{p - 2} + \dots + 1 .

This polynomial has as its roots in $ℂ$ the distinct $p$ th roots of unity that are not equal to $1$ . Over $ℚ$ , we claim it is irreducible. For this, consider the polynomial

Φ_{p} (x + 1) = \frac{{(x + 1)}^{p} - 1}{x} = \sum_{i = 0}^{p - 1} (\binom{p}{i + 1}) x^{i},

which has coefficents divisible by $p$ but not $p^{2}$ except for its leading coefficient $a_{p - 1}$ , which is $1$ . Therefore, $Φ_{p} (x + 1)$ is Eisenstein, hence irreducible. But if $Φ_{p}$ were to factor into $g$ and $h$ , then $Φ_{p} (x + 1)$ would factor into $g (x + 1)$ and $h (x + 1)$ , which have the same leading coefficients as $g$ and $h$ , and hence are nonconstant if and only if $g$ and $h$ are. In other words, $Φ_{p}$ is irreducible as well.

Remark 5.3.5. §

The condition in the Eisenstein criterion that the constant coefficient not lie in the square of the prime ideal is in general necessary. For instance, $x^{2} - p^{2} \in ℤ [x]$ is never irreducible for a prime $p$ .

Often, we can tell if a polynomial is irreducible by considering its reductions modulo ideals.

Proposition 5.3.6. §

Let $R$ be an integral domain, and let $𝔭$ be a prime ideal of $R$ . Let $f \in R [x]$ with leading coefficient not in $𝔭$ . Let $\bar{f}$ denote the image of $f$ in $(R ∕ 𝔭) [x]$ given by reducing its coefficients modulo $𝔭$ .

a.: If $R$ is a UFD and $\bar{f}$ is irreducible in $Q (R ∕ 𝔭) [x]$ , then $f$ is irreducible in $Q (R) [x]$ .
b.: If $f$ is primitive and $\bar{f}$ is irreducible in $R ∕ 𝔭 [x]$ , then $f$ is irreducible in $R [x]$ .

Proof.

If $R$ is a UFD and $f$ is reducible in $Q (R) [x]$ , then by Proposition 5.2.9, we have that $f = 𝑔h$ for some nonconstant $g, h \in R [x]$ . Similarly, if $f$ is primitive and reducible in $R [x]$ , then $f = 𝑔h$ for nonconstant $g, h \in R [x]$ . In either case, since the leading coefficient of $f$ is not in $𝔭$ and $𝔭$ is prime, we have that the leading coefficients of $g$ and $h$ are not in $𝔭$ as well. That is, the images of $g$ and $h$ in $(R ∕ 𝔭) [x]$ are nonconstant, which means that $\bar{f}$ is a product of two nonconstant polynomials, hence reducible in $Q (R ∕ 𝔭) [x]$ . □

Remark 5.3.7. §

For $R = ℤ$ , Proposition 5.3.6 tells us in particular that if $f \in ℤ [x]$ is monic and its reduction $\bar{f} \in 𝔽_{p} [x]$ modulo $p$ is irreducible for any prime $p$ , then $f$ is irreducible.

Example 5.3.8. §

Let $f = x^{4} + x^{3} + 1001 \in ℤ [x]$ . We claim that $f$ is irreducible in $ℚ [x]$ . For this, consider its reduction modulo $2$ . The polynomial $\bar{f} = x^{4} + x^{3} + 1 \in (ℤ ∕ 2 ℤ) [x]$ is either irreducible, has a root in $(ℤ ∕ 2 ℤ) [x]$ , or is a product of two irreducible polynomials of degree $2$ . But $\bar{f} (0) = \bar{f} (1) = 1$ , and $x^{2} + x + 1$ is the only irreducible polynomial of degree $2$ in $(ℤ ∕ 2 ℤ) [x]$ , and ${(x^{2} + x + 1)}^{2} = x^{4} + x^{2} + 1 \neq \bar{f}$ , so $\bar{f}$ is irreducible. By Proposition 5.3.6, $f$ is irreducible in $ℚ [x]$ .

Example 5.3.9. §

The converse to Proposition 5.3.6 does not hold. For instance, $x^{2} + x + 1$ is irreducible in $ℚ [x]$ , but it has a root in $(ℤ ∕ 3 ℤ) [x]$ .

We also have the following simple test for the existence of roots of polynomials over UFDs.

Proposition 5.3.10. §

Let $R$ be a UFD and $f = \sum_{i = 0}^{n} a_{i} x^{i} \in R [x]$ with $a_{0}, a_{n} \neq 0$ . Suppose that $α \in Q (R)$ is a root of $f$ , and write $α$ in reduced form as $α = \frac{c}{d}$ for some $c, d \in R$ . Then $c$ divides $a_{0}$ and $d$ divides $a_{n}$ in $R$ .

Proof.

Since $x - \frac{c}{d}$ divides $f$ in $Q (R) [x]$ and $\frac{c}{d}$ is in reduced form, it follows from Proposition 5.2.9 that $f = (𝑑𝑥 - c) g$ for some $g \in R [x]$ . Writing $g = \sum_{i = 0}^{n - 1} b_{i} x^{i}$ , we see that $a_{0} = - c b_{0}$ and $a_{n} = d b_{n - 1}$ . □

Example 5.3.11. §

Let $f = 2 x^{3} - 3 x + 5 \in ℤ [x]$ . We check that $f (1) = 4$ , $f (- 1) = 6$ , $f (5) \equiv - 10 mod 25$ , $f (- 5) \equiv 20 mod 25$ , and $f (\frac{1}{2})$ , $f (- \frac{1}{2})$ , $f (\frac{5}{2})$ , and $f (- \frac{5}{2})$ are all represented by reduced fractions with denominators equal to $4$ . Proposition 5.3.10 therefore tells us that $f$ has no roots in $ℚ$ , hence is irreducible, being of degree $3$ .

5.4. Euclidean domains

Definition 5.4.1. §

A norm $f$ on an ring $R$ is a function $f : R \to ℤ_{\geq 0}$ with $f (0) = 0$ . We say that $f$ is positive if the only $a \in R$ for which $f (a) = 0$ is $a = 0$ .

Definition 5.4.2. §

Let $R$ be an integral domain. A Euclidean norm $ν$ on $R$ is a norm on $R$ such that for all nonzero $a, b \in R$ , one has

i.: $ν (a) \leq ν (𝑎𝑏)$ , and
ii.: there exist $q, r \in R$ with $a = 𝑞𝑏 + r$ and either $ν (r) < ν (b)$ or $r = 0$ .

Remark 5.4.3. §

Property (ii) of Definition 5.4.2 is known as the division algorithm.

Definition 5.4.4. §

A Euclidean domain $R$ is an integral domain such that there exists a Euclidean norm on $R$ .

Examples 5.4.5. §

a.: The integers $ℤ$ are a Euclidean domain with Euclidean norm $ν (a) = | a |$ for any nonzero $a \in ℤ$ .
b.: Every polynomial ring $F [x]$ over a field $F$ is a Euclidean domain, the degree function providing a Euclidean norm on $F [x]$ (if we take the degree of $0$ to be $0$ ). Note that this is not a positive norm.

Lemma 5.4.6. §

In a Euclidean domain $R$ with Euclidean norm $ν$ , the minimal value of $ν$ on all nonzero elements of $R$ is $ν (1)$ , and $ν (u) = ν (1)$ for $u \in R$ if and only if $u \in R^{\times}$ .

Proof.

By the definition of a Euclidean norm, we have $ν (1) \leq ν (a \cdot 1) = ν (a)$ for all nonzero $a \in R$ . If $u \in R^{\times}$ , then $ν (u) \leq ν (u \cdot u^{- 1}) = ν (1)$ , so $ν (u) = ν (1)$ . Conversely, if $b \in R$ with $ν (b) = ν (1)$ , then we may write $1 = 𝑞𝑏 + r$ for some $q, r \in R$ with either $ν (r) < ν (1)$ or $r = 0$ . By what we have shown, the latter holds, so $𝑞𝑏 = 1$ , and $b$ is a unit. □

Example 5.4.7. §

In $F [x]$ , the units are exactly the nonzero constant polynomials, i.e., those with degree $0$ .

While we will explain below that not every PID is a Euclidean domain, it is the case that every Euclidean domain is a PID.

Theorem 5.4.8. §

Every Euclidean domain is a PID.

Proof.

Let $I$ be a nonzero ideal in a Euclidean domain $R$ with Euclidean norm $ν$ . We must show that $I$ is principal. Let $b \in I$ be a nonzero element with minimal norm among all elements of $I$ . For any $a \in I$ , we may write $a = 𝑞𝑏 + r$ with $q, r \in R$ and either $ν (r) < ν (b)$ or $r = 0$ . Note that $a, b \in I$ , so $r \in I$ as well, which precludes the possibility of $ν (r) < ν (b)$ , since $ν (r)$ is minimal among norms of elements of $I$ . Therefore, we have $r = 0$ , so $a \in (b)$ . As $a$ was arbitrary and $b \in I$ , we have $I = (b)$ . □

The key property of Euclidean domains is the ability to perform the Euclidean algorithm, which we see in the following.

Theorem 5.4.9 (Euclidean algorithm). §

Let $R$ be a Euclidean domain with Euclidean norm $ν$ , and let $a, b \in R$ be nonzero elements. Let $r_{- 1} = a$ and $r_{0} = b$ . Suppose recursively that we are given elements $r_{j} \in R$ for $- 1 \leq j \leq i$ and some $i \geq 0$ . If $r_{i} \neq 0$ , write

r_{i - 1} = q_{i + 1} r_{i} + r_{i + 1}

(5.4.1)

with $q_{i + 1}, r_{i + 1} \in R$ and either $ν (r_{i + 1}) < ν (r_{i})$ or $r_{i + 1} = 0$ . If $r_{i + 1} \neq 0$ , repeat the process with $i$ replaced by $i + 1$ . The process terminates with $d = r_{n} \neq 0$ and $r_{n + 1} = 0$ for some $n \geq 1$ , and $(d)$ is the GCD of $a$ and $b$ . Moreover, we may use the formulas in (5.4.1) and recursion to write $d$ as $d = 𝑥𝑎 + 𝑦𝑏$ for some $x, y \in R$ .

Proof.

We note that the process must terminate, as the values of the $ν (r_{i})$ for $i \geq 0$ are decreasing. Moreover, the result $d = r_{n}$ satisfies $r_{n - 1} = q_{n + 1} r_{n}$ , so it divides $r_{n - 1}$ by definition, and then we see by downward recursion using (5.4.1) that $d$ divides every $r_{i - 1}$ . Finally, if $c$ is any common divisor of $a$ and $b$ , then it again recursively divides each $r_{i}$ (this time by upwards recursion and (5.4.1)), so $c$ divides $d$ . Therefore, $(d)$ is the GCD of $a$ and $b$ .

Note that $d = r_{n - 2} - q_{n} r_{n - 1}$ , and suppose that we may write $d = z r_{j} + w r_{j + 1}$ for some $- 1 \leq j \leq n - 2$ . If $j = - 1$ , we are done. Otherwise, note that $r_{j + 1} = r_{j - 1} - q_{j + 1} r_{j}$ , so

d = z r_{j} + w (r_{j - 1} - q_{j + 1} r_{j}) = w r_{j - 1} + (z - q_{j + 1} w) r_{j},

and we have written $d$ as an $R$ -linear combination of $r_{j - 1}$ and $r_{j}$ . Repeat the process for $j - 1$ . The final result is the desired $R$ -linear combintation of $a$ and $b$ . □

Example 5.4.10. §

Take $ℤ$ and its usual Euclidean norm. We take $a = 550$ and $b = 154$ . Then $550 = 3 \cdot 154 + 88$ , so we set $r_{1} = 88$ . Then $154 = 88 + 66$ , so we set $r_{2} = 66$ , and $88 = 66 + 22$ , so we set $r_{3} = 22$ , and $66 = 3 \cdot 22$ , so we stop at $d = r_{3} = 22$ , which is therefore the greatest common divisor of $a$ and $b$ . Working backwards, we obtain

22 = 88 - 66 = 88 - (154 - 88) = 2 \cdot 88 - 154 = 2 \cdot (550 - 3 \cdot 154) - 154 = 2 \cdot 550 - 7 \cdot 154 .

That is, we have written $d$ as $2 a - 7 b$ .

Often Euclidean norms come in the form of multiplicative norms.

Definition 5.4.11. §

A multiplicative norm $N : R \to ℤ_{\geq 0}$ on a commutative ring $R$ is a positive norm such that for all $N (𝑎𝑏) = N (a) N (b)$ for all $a, b \in R$ .

Remark 5.4.12. §

Note that the existence of a multiplicative norm $N$ on a commutative ring $R$ forces $R$ to be an integral domain, for if $𝑎𝑏 = 0$ , then $N (a) N (b) = N (𝑎𝑏) = 0$ , so either $N (a) = 0$ or $N (b) = 0$ , and therefore either $a = 0$ or $b = 0$ .

Example 5.4.13. §

The absolute value on $ℤ$ is a multiplicative norm, as well as a Euclidean norm.

Example 5.4.14. §

The function $N$ on the Gaussian integers $ℤ [i]$ given by $N (a + 𝑏𝑖) = a^{2} + b^{2}$ is a multiplicative norm. Clearly, $a^{2} + b^{2} = 0$ if and only if $a + 𝑏𝑖 = 0$ . Given $a, b, c, d \in ℤ$ , we have

\begin{matrix} N ((a + 𝑏𝑖) (c + 𝑑𝑖)) = {(𝑎𝑐 - 𝑏𝑑)}^{2} + {(𝑎𝑑 + 𝑏𝑐)}^{2} = {(𝑎𝑐)}^{2} + {(𝑏𝑑)}^{2} + {(𝑎𝑑)}^{2} + {(𝑏𝑐)}^{2} \\ = (a^{2} + b^{2}) (c^{2} + d^{2}) = N (a + 𝑏𝑖) N (c + 𝑑𝑖) . \end{matrix}

Proposition 5.4.15. §

The ring $ℤ [i]$ of Gaussian integers is a Euclidean domain with respect to the Euclidean norm $N (a + 𝑏𝑖) = a^{2} + b^{2}$ for $a, b \in ℤ$ .

Proof.

Since $N$ is a multiplicative norm, we need only check the division algorithm. Extend $N$ to a function on $ℂ$ by defining $N (a + 𝑏𝑖) = a^{2} + b^{2}$ for $a, b \in ℝ$ . Let $a, b, c, d \in ℤ$ with $(c, d) \neq (0, 0)$ . Then we have

\frac{a + 𝑏𝑖}{c + 𝑑𝑖} = s + 𝑡𝑖

for some $s, t \in ℚ$ , and let $e, f \in ℤ$ be integers with $| s - e | \leq 1 ∕ 2$ and $| t - f | \leq 1 ∕ 2$ . Then we have

\begin{matrix} N (a + 𝑏𝑖 - (e + 𝑓𝑖) (c + 𝑑𝑖)) = N (c + 𝑑𝑖) N ((s - e) + (t - f) i) \\ \leq N (c + 𝑑𝑖) ({(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{2}) = N (c + 𝑑𝑖) ∕ 2 < N (c + 𝑑𝑖), \end{matrix}

so the division algorithm is satisfied: $a + 𝑏𝑖 = q (c + 𝑑𝑖) + r$ with $q = e + 𝑓𝑖$ and $N (r) < N (c + 𝑑𝑖)$ . □

Corollary 5.4.16. §

The units in $ℤ [i]$ are exactly $1, - 1, i, - i$ .

Proof.

Since $N$ is a Euclidean norm on $ℤ [i]$ , the units are exactly those nonzero elements of norm $N (1) = 1$ . We have $a^{2} + b^{2} = 1$ if and only if $(a, b) = (\pm 1, 0)$ or $(a, b) = (0, \pm 1)$ . □

Lemma 5.4.17. §

If $a, b, c, d \in ℤ$ and $c + 𝑑𝑖$ divides $a + 𝑏𝑖$ in $ℤ [i]$ , then $c - 𝑑𝑖$ divides $a - 𝑏𝑖$ in $ℤ [i]$ .

Proof.

Write $a + 𝑏𝑖 = (c + 𝑑𝑖) (e + 𝑓𝑖)$ for some $e, f \in ℤ$ . Then $a = 𝑐𝑒 - 𝑑𝑓$ and $b = 𝑐𝑑 + 𝑑𝑒$ , so

(c - 𝑑𝑖) (e - 𝑓𝑖) = (𝑐𝑒 - 𝑑𝑓) - (𝑐𝑓 + 𝑑𝑒) i = a - 𝑏𝑖 .

□

We can completely determine the irreducible elements in $ℤ [i]$ as follows.

Proposition 5.4.18. §

The irreducible elements in $ℤ [i]$ are, up to multiplication by a unit, $1 + i$ , primes $p \in ℤ$ with $p \equiv 3 mod 4$ , and $a + 𝑏𝑖$ for $a, b \in ℤ$ such that $p = a^{2} + b^{2} \equiv 1 mod 4$ is a prime in $ℤ$ . Moreover, the primes in $ℤ$ that can be written in the form $a^{2} + b^{2}$ are exactly $2$ and those that are $1$ modulo $4$ .

Proof.

First, note that if $a + 𝑏𝑖$ divides $c + 𝑑𝑖$ in $ℤ [i]$ for integers $a, b, c, d$ , then $N (a + 𝑏𝑖)$ divides $N (c + 𝑑𝑖)$ , since $N$ is multiplicative. So, $1 + i$ is irreducible since $N (1 + i) = 2$ .

Let $p$ be an odd prime in $ℤ$ . If $p$ is divisible by some irreducible element $π = a + 𝑏𝑖$ with $a, b \in ℤ$ , then since $p$ is prime, only one of two things can happen. Either $𝑎𝑏 = 0$ , or $a$ and $b$ are relatively prime in $ℤ$ , noting Corollary 5.4.16. Suppose $𝑎𝑏 \neq 0$ . By Lemma 5.4.17, we have that $a - 𝑏𝑖$ divides $p$ , and $\bar{π} = a - 𝑏𝑖$ is irreducible. If $\bar{π}$ were associate to $π$ , then $π$ would divide $2 a = (a + 𝑏𝑖) + (a - 𝑏𝑖)$ and $2 b = - i ((a + 𝑏𝑖) - (a - 𝑏𝑖))$ . Then $π$ divides $2$ , but that is impossible. Thus, $π$ and $\bar{π}$ both dividing $p$ implies that $p$ is divisible by $N (π) = a^{2} + b^{2}$ . As $p$ is prime, we have $p = a^{2} + b^{2}$ .

So, we have shown that either our odd prime $p$ is irreducible in $ℤ [i]$ or $p = a^{2} + b^{2}$ for some $a, b \in ℤ$ . Note that the squares in $ℤ ∕ 4 ℤ$ are $0$ and $1$ , so any integer of the form $a^{2} + b^{2}$ is $0$ , $1$ , or $2$ modulo $4$ . In particular, if $p \equiv 3 mod 4$ , then $p$ is irreducible in $ℤ [i]$ .

If $p \equiv 1 mod 4$ is prime in $ℤ$ , then ${(ℤ ∕ 𝑝ℤ)}^{\times}$ has order divisible by $4$ . As $ℤ ∕ 𝑝ℤ$ contains only two roots of $x^{2} - 1$ , which are $- 1$ and $1$ , so ${(ℤ ∕ 𝑝ℤ)}^{\times}$ contains an element of order $4$ . In particular, there exists $n \in ℤ$ such that $n^{2} \equiv - 1 mod p$ , which is to say that $p$ divides $n^{2} + 1$ . If $p$ were irreducible in $ℤ [i]$ , then $p$ would divide either $n + i$ or $n - i$ , but then it would divide both, being an integer. Thus $p$ would divide $2 i$ , which it does not. So, $p$ is reducible, which means equals $a^{2} + b^{2}$ for some $a, b \in ℤ$ . □

Lemma 5.4.19. §

Let $N$ be a multiplicative norm on an integral domain $R$ . Then $N (u) = 1$ for all $u \in R^{\times}$ .

Proof.

We have $N (1) = N {(1)}^{2}$ , and $R$ is an integral domain, so $N (1) = 1$ . Moreover, since

N (u^{- 1}) N (u) = N (1) = 1,

we have that $N (u^{- 1}) = N {(u)}^{- 1}$ , and therefore $N (u) = 1$ . □

Example 5.4.20. §

Consider the multiplicative norm $N$ on $ℤ [\sqrt{- 5}]$ given by

N (a + b \sqrt{- 5}) = a^{2} + 5 b^{2} .

We have $a^{2} + 5 b^{2} = 1$ if and only if $a = \pm 1$ and $b = 0$ , so the only units in $ℤ [\sqrt{- 5}]$ are $\pm 1$ . Now, if $2 = 𝛼𝛽$ for some nonunits $α, β \in ℤ [\sqrt{- 5}]$ , then $4 = N (2) = N (α) N (β)$ , so $N (α) = 2$ , but $2$ is clearly not a value of $N$ . Therefore, $2$ is irreducible, and so is $3$ . Also, we have that $N (1 \pm \sqrt{- 5}) = 6$ , and since $2$ and $3$ are not values of $N$ , we have that $1 \pm \sqrt{- 5}$ is irreducible as well. As these elements are all non-associates, the existence of the two factorizations

6 = 2 \cdot 3 = (1 + \sqrt{- 5}) (1 - \sqrt{- 5})

proves that $ℤ [\sqrt{- 5}]$ is not a UFD.

Not all principal ideal domains are Euclidean. We give most of the outline of how one produces an example.

Definition 5.4.21. §

An nonzero, non-unit element $b$ of an integral domain $R$ is called a universal side divisor if every element $a \in R$ may be written in the form $a = 𝑞𝑏 + r$ for some $q, r \in R$ with $r = 0$ or $r \in R^{\times}$ .

Lemma 5.4.22. §

Let $R$ be a Euclidean domain with Euclidean norm $ν$ . Let $b \in R$ be a nonzero, non-unit element such that $ν (b)$ is minimal among nonzero, non-unit elements of $R$ . Then $b$ is a universal side divisor of $R$ .

Proof.

Let $a \in R$ . By definition of $ν$ , we may write $a = 𝑞𝑏 + r$ with $ν (r) < ν (b)$ or $r = 0$ . By the minimality of $ν (b)$ , we must have that $r$ is a unit or $0$ . □

Example 5.4.23. §

We claim that the ring $R = ℤ [(1 + \sqrt{- 19}) ∕ 2]$ is not Euclidean. Suppose by contradiction that it is a Euclidean domain, and let $ν$ be a Euclidean norm on $R$ . We also have the multiplicative norm $N$ on $R$ given by

N (a + b \frac{1 + \sqrt{- 19}}{2}) = (a + b \frac{1 + \sqrt{- 19}}{2}) (a + b \frac{1 - \sqrt{- 19}}{2}) = a^{2} + 𝑎𝑏 + 5 b^{2} .

(5.4.2)

Note that if $α \in R - ℤ$ , then $N (α) \geq 5$ , so the only units in $R$ are $\pm 1$ .

Let $β \in R$ be a universal side divisor, which exists as $R$ is Euclidean, and write $2 = 𝑞𝛽 + r$ for $q \in R$ and $r \in {0, 1, - 1}$ . We then have that $N (β)$ divides $N (2 - r)$ as $N$ is multiplicative, so $N (β)$ divides $4$ or $9$ , and this implies $β \in {\pm 2, \pm 3}$ by the formula for $N$ . Now take $α = (1 + \sqrt{- 19}) ∕ 2$ , and set $α = q^{'} β + r^{'}$ with $q^{'} \in R$ and $r^{'} \in {0, 1, - 1}$ . We have $N (α) = N (α - 1) = 5$ and $N (α + 1) = 7$ , which are not multiples of $N (β) \in {4, 9}$ , so we obtain a contradiction.

Definition 5.4.24. §

A Dedekind-Hasse norm on an integral domain $R$ is a positive norm $μ$ on $R$ such that for every $a, b \in R$ , either $a \in (b)$ or there exists a nonzero element $c \in (a, b)$ such that $μ (c) < μ (b)$ .

Proposition 5.4.25. §

An integral domain $R$ is a PID if and only if there exists a Dedekind-Hasse norm on $R$ .

Proof.

Suppose first that $μ$ is a Dedekind-Hasse norm on $R$ . Let $I$ be a nonzero ideal of $R$ , and let $b \in I - {0}$ with minimal norm under $μ$ . If $a \in I$ , then since there does not exist a nonzero element $c \in (a, b) \subseteq I$ with $μ (c) < μ (b)$ by the minimality of $μ (b)$ , we have by definition of a Dedekind-Hasse norm that $a \in (b)$ . Thus $I = (b)$ .

Suppose on the other hand the $R$ is a PID. Define $μ : R \to ℤ_{\geq 0}$ by $μ (0) = 0$ , $μ (u) = 1$ for $u \in R^{\times}$ , and $μ (p_{1} p_{2} \dots p_{k}) = 2^{k}$ if $p_{1}, \dots, p_{k}$ are irreducible elements of $R$ . This is well-defined as $R$ is a UFD. Given $a, b \in R$ , we have $(a, b) = (d)$ for some $d \in R$ , since $R$ is a PID. Since $d$ divides $b$ , we have $μ (d) \leq μ (b)$ . If $μ (d) = μ (b)$ , then $a$ and $b$ have the same number of divisors as $d$ and therefore are associates, so $a \in (b)$ . Thus, $μ$ is a Dedekind-Hasse norm. □

Example 5.4.26. §

We have already seen that $R = ℤ [(1 + \sqrt{- 19}) ∕ 2]$ is not a Euclidean domain. To see that $R$ is a PID, it suffices to show that the multiplicative norm $N$ on $R$ given by (5.4.2) is a Dedekind-Hasse norm on $R$ . We outline the standard unenlightening verification.

Let $α, β \in R$ with $α \notin (β)$ . We claim that there exist $s, t \in R$ with $0 < N (𝑠𝛼 - 𝑡𝛽) < N (β)$ . Note that we can extend $N$ to a map $N : Q (R) \to ℤ_{\geq 0}$ by the formula (5.4.2), allowing $a, b \in ℚ$ . Our condition that $N$ on $R$ be a Dedekind-Hasse norm is then that $0 < N (s \frac{α}{β} - t) < 1$ . We will find $s$ and $t$ . For this, write

\frac{α}{β} = \frac{a + b \sqrt{- 19}}{c}

for $a, b, c \in ℤ$ with no common divisor and $c > 1$ .

First one considers the cases with $c \leq 4$ . If $c = 2$ , then either $a$ or $b$ is odd, then take $s = 1$ and $t = ((a - 1) + b \sqrt{- 19}) ∕ 2$ . If $c = 3$ , then $a^{2} + 19 b^{2} ≢ 0 mod 3$ , so $a^{2} + 19 b^{2} = 3 q + r$ with $r \in {1, 2}$ . Take $s = a - b \sqrt{- 19}$ and $t = q$ . If $c = 4$ , then again either $a$ or $b$ is odd. If only one is, then write $a^{2} + 19 b^{2} = 4 q + r$ with $1 \leq r \leq 4$ , and take $s = a - b \sqrt{- 19}$ and $t = q$ . If both are, write $a^{2} + 19 b^{2} = 8 q + 4$ , and take $s = \frac{1}{2} (a - b \sqrt{- 19})$ and $t = q$ .

Now suppose that $c \geq 5$ . Since $(a, b, c) = (1)$ , we have $x, y, z \in ℤ$ such that $𝑥𝑎 + 𝑦𝑏 + 𝑧𝑐 = 1$ . Write $𝑎𝑦 - 19 𝑏𝑥 = 𝑞𝑐 + r$ , with $q \in ℤ$ and $| r | \leq c ∕ 2$ . Take $s = y + x \sqrt{- 19}$ and $t = q - z \sqrt{- 19}$ . The reader will check that

N (s \frac{α}{β} - t) = c^{- 2} N (s (a + b \sqrt{- 19}) - 𝑡𝑐) = \frac{r^{2} + 19}{c^{2}},

which is at most $\frac{1}{4} + \frac{19}{36} = \frac{7}{9}$ if $c \geq 6$ and at most $\frac{4}{25} + \frac{19}{25} = \frac{23}{25}$ if $c = 5$ .

5.5. Vector spaces over fields

In this section, we give a very brief discussion of the theory of vector spaces over fields, as it shall be subsumed by the sections that follow it.

Definition 5.5.1. §

Let $F$ be a field. A vector space $V$ over $F$ is an abelian group under addition that is endowed with an operation $\cdot : F \times V \to V$ of scalar multiplication such that for all $a, b \in F$ and $v, w \in V$ , one has

i.: $1 \cdot v = v$ ,
ii.: $a \cdot (b \cdot v) = (𝑎𝑏) \cdot v$ ,
iii.: $(a + b) \cdot v = a \cdot v + b \cdot v$ ,
iv.: $a \cdot (v + w) = a \cdot v + a \cdot w$ .

Remark 5.5.2. §

In a vector space $V$ over a field $F$ , we typically write $𝑎𝑣$ for $a \cdot v$ , where $a \in F$ and $v \in V$ .

Example 5.5.3. §

If $F$ is a field, then $F^{n}$ is a vector space over $F$ under the operation

a \cdot (α_{1}, α_{2}, \dots, α_{n}) = (a α_{1}, a α_{2}, \dots, a α_{n})

for $a, α_{1}, α_{2}, \dots, α_{n} \in F$ .

Definition 5.5.4. §

An element of a vector space $V$ over a field $F$ is called a vector, and the elements of $F$ under in the operation $\cdot$ are referred to as scalars.

Example 5.5.5. §

In every vector space $V$ , there is an element $0$ , and it is called the zero vector.

Definition 5.5.6. §

The zero vector space $0$ is the vector space over any field $F$ that is the set ${0}$ with the operation $a \cdot 0 = 0$ for all $a \in F$ .

Example 5.5.7. §

If $F$ is a field, then $F [x]$ is a vector space over $F$ with $a \cdot f$ for $a \in F$ and $f \in F [x]$ defined to be the usual product of polynomials in $F [x]$ . I.e., the operation of scalar multiplication is just multiplication by a constant polynomial.

Example 5.5.8. §

The field $ℂ$ is an $ℝ$ -vector space, as well as a $ℚ$ -vector space. The field $ℝ$ is a $ℚ$ -vector space. The operations of scalar multiplication are just restrictions of the usual multiplication map on $ℂ$ .

The reader will easily check the following.

Lemma 5.5.9. §

If $V$ is a vector space over a field $F$ , then for $a \in F$ and $v \in V$ , we have

a.: $0 \cdot v = 0$ ,
b.: $a \cdot 0 = 0$ ,
c.: $- (𝑎𝑣) = (- a) v = a (- v)$ .

Definition 5.5.10. §

Let $V$ be a vector space over a field $F$ . A subspace $W$ of $V$ is a subset that is closed under the operations of addition and scalar multiplication to $W$ (i.e., to maps $W \times W \to V$ and $F \times W \to V$ , respectively) and is a vector space with respect to these operations.

The following is easily proven.

Lemma 5.5.11. §

A subset $W$ of a vector space $V$ is a subspace if and only if it is a subgroup under addition and closed under scalar multiplication.

Examples 5.5.12. §

a.: The zero subspace ${0}$ and $V$ are both subspaces of any vector space $V$ .
b.: The field $F$ is a subspace of $F [x]$ .

Definition 5.5.13. §

Let $V$ be a vector space over a field $F$ , and let $S$ be a subset of $V$ . A linear combination of elements of $S$ is any sum

\sum_{i = 1}^{n} a_{i} v_{i}

with $v_{1}, v_{2}, \dots, v_{n}$ distinct vectors in $S$ and $a_{1}, a_{2}, \dots, a_{n} \in F$ for some $n \geq 0$ . We say that such a linear combination is nontrivial if there exists a $j$ with $1 \leq j \leq n$ and $a_{j} \neq 0$ .

Definition 5.5.14. §

Let $V$ be a vector space over a field $F$ and $S$ be a set of vectors in $V$ . The subspace spanned by $S$ , also known as the span of $V$ , is the set of all linear combinations of elements of $S$ , or simply the zero subspace if $S$ is empty.

Example 5.5.15. §

For any vector space $V$ , the set $V$ spans $V$ .

Definition 5.5.16. §

We say that a set $S$ of vectors in a vector space $V$ over a field $F$ spans $V$ if $V$ equals the subspace spanned by $S$ .

That is, $S$ spans an $F$ -vector space $V$ if, for every $v \in V$ , there exist $n \geq 0$ , $v_{i} \in V$ , and $a_{i} \in F$ for $1 \leq i \leq n$ such that

v = \sum_{i = 1}^{n} a_{i} v_{i} .

Definition 5.5.17. §

We say that a set of $S$ of vectors in a vector space $V$ over a field $F$ is linearly independent if every nontrivial linear combination of vectors in $S$ is nonzero. Otherwise, $S$ is said to be linearly dependent.

That is, a set $S$ of vectors in an $F$ -vector space $V$ is linearly independent if whenever $n \geq 1$ , $v_{i} \in V$ and $a_{i} \in F$ for $1 \leq i \leq n$ and

\sum_{i = 1}^{n} a_{i} v_{i} = 0,

then $a_{i} = 0$ for all $1 \leq i \leq n$ .

Lemma 5.5.18. §

Let $S$ be a linearly independent subset of a vector space $V$ over a field $F$ , and let $W$ be the span of $F$ . If $v_{0} \in V - W$ , then $S \cup {v_{0}}$ is also linearly independent.

Proof.

Let $v_{1}, v_{2}, \dots, v_{n} \in S$ and $c_{0}, c_{1}, \dots, c_{n} \in F$ for some $n \geq 1$ , and suppose that

\sum_{i = 0}^{n} c_{i} v_{i} = 0 .

We cannot have $c_{0} \neq 0$ , as then

v_{0} = - c_{0}^{- 1} \sum_{i = 1}^{n} c_{i} v_{i} \in W .

On the other hand, the fact that $c_{0} = 0$ implies that $c_{i} = 0$ for all $1 \leq i \leq n$ by the linear independence of $V$ . Thus, $S \cap {v_{0}}$ is linearly independent. □

Example 5.5.19. §

In any vector space $V$ , the empty set is linearly independent. If $v \in V$ is nonzero, then ${v}$ is also a linearly independent set.

Definition 5.5.20. §

A subset $B$ of a vector space $V$ over a field $F$ is said to be a basis of $V$ over $F$ if it is linearly independent and spans $V$ .

Example 5.5.21. §

The set ${e_{1}, e_{2}, \dots, e_{n}}$ of $F^{n}$ , where $e_{i}$ is the element of $F^{n}$ that has a $1$ in its $i$ th coordinate and $0$ in all others, is a basis of $F^{n}$ .

Example 5.5.22. §

The set ${x^{i} ∣ i \geq 0}$ is a basis of $F [x]$ . That is, every polynomial can be written as a finite sum of distinct monomials in a unique way.

Remark 5.5.23. §

For a field $F$ , it is very hard to write down a basis of $\prod_{i = 0}^{\infty} F$ . In fact, the proof that it has a basis uses the axiom of choice.

Definition 5.5.24. §

A vector space $V$ is said to be finite dimensional if it has a finite basis (i.e., a basis with finitely many elements). Otherwise $V$ is said to be infinite dimensional.

The following theorem employs Zorn’s lemma.

Theorem 5.5.25. §

Let $V$ be a vector space over a field $F$ . Every linearly independent subset of $V$ is contained in a basis of $V$ .

Proof.

Let $S$ be a linearly independent subset of $V$ , and let $X$ denote the set of linearly independent subsets of $V$ that contain $S$ . We order $X$ by containment of subsets. If $𝒞$ is a chain in $X$ , then its union $U = ⋃_{T \in 𝒞} T$ is linearly independent since if $v_{1}, v_{2}, \dots, v_{n} \in U$ for some $n \geq 1$ , then each $v_{i}$ is contained in some $T_{i} \in X$ for each $1 \leq i \leq n$ , and one of the sets $T_{j}$ contains the others, since $𝒞$ is a chain. Since $T_{j}$ is linearly independent, any nontrivial linear combination of the elements $v_{i}$ with $1 \leq i \leq n$ is nonzero. Therefore, $U$ is linearly independent as well, so is contained in $X$ .

By Zorn’s Lemma, $X$ now contains a maximal element $B$ , and we want to show that $B$ spans $V$ , so is a basis of $V$ containing $S$ . Let $W$ denote the span of $B$ . If $v \in V - W$ , then $B^{'} = B \cup {v}$ is linearly independent by Lemma 5.5.18, so an element of $X$ , which contradicts the maximality of $B$ . That is, $V = W$ , which is to say that $B$ spans $V$ . □

In particular, the empty set is contained in a basis of any vector space, so we have the following:

Corollary 5.5.26. §

Every vector space over a field contains a basis.

A similar argument yields the following.

Theorem 5.5.27. §

Let $V$ be a vector space over a field $F$ . Every subset of $V$ that spans $V$ contains a basis of $V$ .

Proof.

Let $S$ be a spanning subset of $V$ . Let $X$ denote the set of linearly independent subsets of $S$ , and order $X$ by containment. As seen in the proof of Theorem 5.5.25, any union of a chain of linearly independent subsets is linearly independent, so has an upper bound. Thus, Zorn’s lemma tells us that $X$ contains a maximal element $B$ . Again, we want to show that $B$ spans $V$ , so is a basis. If it were not, then there would exist some element of $V$ which is not in the span of $B$ , but is in the span of $S$ . In particular, there exists an element $v \in S$ that is not in the span of $B$ . The set $B \cup {v}$ is linearly independent, contradicting the maximality of $B$ . □

We also have the following, which can be generalized to a statement on cardinality.

Theorem 5.5.28. §

Let $V$ be a vector space over a field $F$ . If $V$ is finite dimensional, then every basis of $V$ contains the same number of elements, and otherwise every basis of $V$ is infinite.

Proof.

Let $B_{1} = {v_{1}, v_{2}, \dots, v_{n}}$ be a basis of $V$ with a minimal number $n$ of elements, and let $B = {w_{1}, w_{2}, \dots, w_{m}}$ be another basis of $V$ with $m \geq n$ . Then $B_{1}$ spans $V$ , so $w_{1}$ is a nontrivial linear combination of the $v_{i}$ for $1 \leq i \leq n$ :

w_{1} = \sum_{i = 1}^{n} a_{i} v_{i}

(5.5.1)

for some $a_{i} \in F$ . Letting $j$ be such that $a_{j} \neq 0$ , we may write $v_{j}$ as a linear combination of $w_{1}$ and the $v_{i}$ with $i \neq j$ . In other words, $B_{2} = (B_{1} - {v_{j}}) \cup {w_{1}}$ spans $V$ . Suppose

c_{j} w_{1} + \sum_{\begin{array}{c} i = 1 \\ i \neq j \end{array}}^{n} c_{i} v_{i} = 0

(5.5.2)

for some $c_{i} \in F$ . Using (5.5.1), we may rewrite the sum in (5.5.2) as a linear combination of the $v_{i}$ , the coefficient of $v_{j}$ in which is $a_{j} c_{j}$ , which forces $c_{j} = 0$ as $B_{1}$ is a linearly independent set. But then we see from (5.5.2) that all $c_{i} = 0$ as $B - {v_{j}}$ is linearly independent. So, $B_{2}$ is a basis of $V$ .

Suppose by recursion that, for $k \leq m$ , we have found a basis $B_{k}$ of order $n$ of $V$ that contains only $w_{1}, \dots, w_{k - 1}$ and elements of $B$ . Then $w_{k}$ is a nontrivial linear combination of the elements of $B_{k}$ , and the coefficient of some $v_{l}$ is nonzero in this linear combination by the linear independence of $B$ . We therefore have that $B_{k + 1} = (B_{k} - {v_{l}}) \cup {w_{k}}$ spans $V$ , and a similar argument to the above shows that it is a basis. Finally, we remark that the basis $B_{m + 1}$ must be $B_{1}$ itself, since it contains $B_{1}$ , so we have $m = n$ , as desired. □

Definition 5.5.29. §

The dimension of a finite-dimensional vector space $V$ over a field $F$ is the number of elements in a basis of $V$ over $F$ . We write $\dim_{F} (V)$ for this dimension.

Example 5.5.30. §

The space $F^{n}$ is of dimension $n$ over $F$ .

The maps between vector spaces that respect the natural operations on the spaces are called linear transformations.

Definition 5.5.31. §

A linear transformation $T : V \to W$ of $F$ -vector spaces is a function from $V$ to $W$ satisfying

\begin{array}{rcl} T (v + v^{'}) = T (v) + T (v^{'}) & and & T (𝑎𝑣) = 𝑎𝑇 (v) \end{array}

for all $a \in F$ and $v, v^{'} \in V$

Remark 5.5.32. §

In other words, a linear transformation is a homomorphism of the underlying groups that “respects scalar multiplication.”

Definition 5.5.33. §

A linear transformation $T : V \to W$ of $F$ -vector spaces is an isomorphism of $F$ -vector spaces if it is there exists an linear transformation $T^{- 1} : W \to V$ that is inverse to it.

Much as with group and ring homomorphisms, we have the following:

Lemma 5.5.34. §

A linear transformation is an isomorphism if and only if it is a bijection.

Examples 5.5.35. §

Let $V$ and $W$ be $F$ -vector spaces.

a.: The identity map ${id}_{V} : V \to V$ is an $F$ -linear transformation (in fact, isomorphism).
b.: The zero map $0 : V \to W$ is an $F$ -linear transformation.

5.6. Modules over rings

Definition 5.6.1. §

Let $R$ be a ring. A left $R$ -module, or left module over $R$ , is an abelian group $M$ together with an operation $\cdot : R \times M \to M$ such that for all $a, b \in R$ and $m, n \in M$ , one has

i.: $1 \cdot m = m$ ,
ii.: $(a \cdot b) \cdot m = (𝑎𝑏) \cdot m$ ,
iii.: $(a + b) \cdot m = a \cdot m + b \cdot n$ ,
iv.: $a \cdot (m + n) = a \cdot m + a \cdot n$ .

Definition 5.6.2. §

Let $R$ be a commutative ring. We refer more simply to a left $R$ -module as a $R$ -module, or module over $R$ .

Remark 5.6.3. §

When one speaks simply of a module over a ring $R$ , one means by default a left $R$ -module.

Notation 5.6.4. §

When an abelian group $M$ is seen as a left module over a ring $R$ via the extra data of some operation $R \times M \to M$ , we say that this operation endows $M$ with the additional structure of a left $R$ -module.

Example 5.6.5. §

The definition of a module over a field coincides with the definition of a vector space over a field. In other words, to say that $M$ a module over a field $F$ is exactly to say that $M$ is a vector space over $F$ .

Example 5.6.6. §

The modules over $ℤ$ are exactly the abelian groups. That is, suppose that $A$ is a $ℤ$ -module, which by definition is an abelian group with an additional operation $\cdot : ℤ \times A \to A$ . We show that this additional operation satisfies $n \cdot a = 𝑛𝑎$ for $n \in ℤ$ and $a \in A$ , where $𝑛𝑎$ is the usual element of the abelian group $A$ . So, let $a \in A$ . By axiom (i), we have $1 \cdot a = a$ , and then the distributivity of axiom (iii) allows us to see that $n \cdot a = 𝑛𝑎$ for all $n \geq 1$ . Using axioms (iv) and (ii), we have

0 \cdot a = 0 \cdot (2 a - a) = 0 \cdot 2 a - 0 \cdot a = (0 \cdot 2) \cdot a - 0 \cdot a = 0 \cdot a - 0 \cdot a = 0,

and then finally we have

(- n) \cdot a + n \cdot a = (n - n) \cdot a = 0 \cdot a = 0,

so $(- n) \cdot a = - 𝑛𝑎$ for $n \geq 1$ .

Example 5.6.7. §

For a ring $R$ and $n \geq 1$ , the direct product $R^{n}$ is a left $M_{n} (R)$ -module via matrix multiplication $(A, v) \mapsto A \cdot v$ for $A \in M_{n} (R)$ and $v \in R^{n}$ ,viewing elements of $R^{n}$ as column vectors.

We also have the notion of a right $R$ -module.

Definition 5.6.8. §

Let $R$ be a ring. A right $R$ -module, or right module over $R$ , is an abelian group $M$ together with an operation $\cdot : M \times R \to R$ such that for all $a, b \in R$ and $m, n \in M$ , one has

i.: $m \cdot 1 = m$ ,
ii.: $m \cdot (a \cdot b) = m \cdot (𝑎𝑏)$ ,
iii.: $m \cdot (a + b) = m \cdot a + n \cdot b$ ,
iv.: $(m + n) \cdot a = m \cdot a + n \cdot a$ .

Example 5.6.9. §

Every left ideal $I$ over a ring $R$ is a left $R$ -module with respect to the restriction $R \times I \to I$ of the multiplication on $R$ . Every right ideal over $R$ is a right module with respect to the restriction $I \times R \to I$ of the multiplication on $R$ .

Definition 5.6.10. §

Let $R$ be a ring. The opposite ring $R^{op}$ to $R$ is the ring that is the abelian group $R$ together with the multiplication $\cdot^{op} : R \times R \to R$ given by $a \cdot^{op} b = 𝑏𝑎$ , where the latter product is taken in $R$ .

Remark 5.6.11. §

The identity map induces an isomorphism $R \to {(R^{op})}^{op}$ of rings.

The reader will easily check the following.

Lemma 5.6.12. §

A right module $M$ over $R$ also has the structure of a left module over $R^{op}$ , where the latter operation $\cdot^{op} : R^{op} : M \to M$ is given by $a \cdot^{op} m = 𝑚𝑎$ , where the latter product is that given by the right $R$ -module structure of $M$ .

Example 5.6.13. §

For a field $F$ , the map $T : M_{n} (F) \to M_{n} (F)$ given by transpose (that is, $A \mapsto A^{T}$ for $A \in M_{n} (F)$ ) is a ring isomorphism between $M_{n} (F)$ and $M_{n} {(F)}^{op}$ .

We also have the notion of a bimodule.

Definition 5.6.14. §

Let $R$ and $S$ be rings. An abelian group $M$ that is a left $R$ -module and a right $S$ -module is called an $R$ - $S$ -bimodule if

(r \cdot m) \cdot s = r \cdot (m \cdot s)

for all $r \in R$ , $s \in S$ , and $m \in M$ .

Examples 5.6.15. §

a.: Any left $R$ -module $M$ over a commutative ring $R$ is an $R$ - $R$ -bimodule with respect to given left operation and the (same) right operation $m \cdot r = 𝑟𝑚$ for $m \in M$ and $r \in R$ .
b.: A two-sided ideal of a ring $R$ is an $R$ - $R$ -bimodule with respect to the operations given by the usual multiplication on $R$ .
c.: For $m, n \geq 1$ , the abelian group $M_{𝑚𝑛} (R)$ of $m$ -by- $n$ matrices with entries in $R$ is an $M_{m} (R)$ - $M_{n} (R)$ -bimodule for the operations of matrix multiplication.

Let us return our focus to $R$ -modules, focusing on the case of left modules, as right modules are just left modules over the opposite ring by Lemma 5.6.12.

Definition 5.6.16. §

An $R$ -submodule (or, submodule) $N$ of a left module $M$ over a ring $R$ is a subset of $N$ that is closed under addition and the operation of left $R$ -multiplication and is an $R$ -module with respect to their restrictions $+ : N \times N \to N$ and $\cdot : R \times N \to N$ to $N$ .

Lemma 5.6.17. §

Let $R$ be a ring, $M$ be a left $R$ -module, and $N$ be a subset of $M$ . Then $N$ is an $R$ -submodule of $M$ if and only if it is nonempty, closed under addition, and closed under left $R$ -multiplication.

Proof.

Clearly, it suffices to check that if $N$ is nonempty and closed under addition and left $R$ -multiplication, then it is an $R$ -submodule. The condition of being closed under left $R$ -multiplication assures that $0$ and inverses of elements of $N$ lies in $N$ , so $N$ is an abelian group under $+$ on $M$ . The axioms for $N$ to be an $R$ -module under $\cdot$ are clearly satisfied as they are satisfied by elements of the larger set $M$ . □

Examples 5.6.18. §

a.

The subspaces of a vector space $V$ over a field $F$ are exactly the $F$ -submodules of $V$ .

b.

The subgroups of an abelian group are the $ℤ$ -submodules of that group.

c.

Any left ideal $I$ of $R$ is a left $R$ -submodule of $R$ viewed as a left $R$ -module.

d.

Any intersection of $R$ -submodules is an $R$ -submodule as well.

e.

For an $R$ -module $M$ and a left ideal $I$ , the abelian group

𝐼𝑀 = {\sum_{i = 1}^{n} a_{i} m_{i} ∣ a_{i} \in I, m_{i} \in M for 1 \leq i \leq n}

is an $R$ -submodule of $M$ .

We also have the following construction.

Definition 5.6.19. §

Let $M$ be an $R$ -module and ${N_{i} ∣ i \in I}$ be a collection of submodules for an indexing set $I$ . The sum of the submodules $N_{i}$ is the submodule $\sum_{i \in I} N_{i}$ of $M$ with elements $\sum_{i \in I} n_{i}$ for $n_{i} \in N_{i}$ and all but finitely many $n_{i}$ equal to $0$ .

If $M$ is an $R$ -module and $N$ is a submodule, we may speak of the quotient abelian group $M ∕ N$ . It is an $R$ -module under the action $r \cdot (m + N) = 𝑟𝑚 + N$ for $r \in R$ and $m \in M$ . This is well-defined, as a different representative $m + n$ of the coset $m + N$ for $n \in N$ will satisfy $r (m + n) + N = 𝑟𝑚 + 𝑟𝑛 + N = 𝑟𝑚 + N$ .

Definition 5.6.20. §

Let $M$ be a left $R$ -module and $N$ be an $R$ -submodule of $M$ . The quotient module $M ∕ N$ of $M$ by $N$ is the abelian group of cosets together with the multiplication $R \times M ∕ N \to M ∕ N$ given by $r \cdot (𝑛𝑁) = (𝑟𝑛) N$ .

Example 5.6.21. §

For an $R$ -module $M$ and a left ideal $I$ , we have the quotient module $M ∕ 𝐼𝑀$ . In particular, note that $R ∕ I$ is a left $R$ -module with respect to $r (s + I) = 𝑟𝑠 + I$ , even if it is not a ring (i.e., if $I$ is not two-sided).

We can also speak of homomorphisms of $R$ -modules.

Definition 5.6.22. §

Let $M$ and $N$ be left modules over a ring $R$ . A left $R$ -module homomorphism $ϕ : M \to N$ is a function such that $ϕ (r \cdot m) = 𝑟𝜙 (m)$ and $ϕ (m + n) = ϕ (m) + ϕ (n)$ for all $r \in R$ and $m, n \in M$ .

Notation 5.6.23. §

If $R$ is commutative (or it is understood that we are working with left modules), we omit the word “left” and speak simply of $R$ -module homomorphisms.

Remark 5.6.24. §

A right $R$ -module homomorphism $ϕ : M \to N$ is just a left $R^{op}$ -module homomorphism.

Definition 5.6.25. §

Let $M$ and $N$ be left modules over a ring $R$ .

a.: An isomorphism $f : M \to N$ of left $R$ -modules is a bijective homomorphism.
b.: An endomorphism of a left $R$ -module $M$ is a homomorphism $f : M \to M$ of left $R$ -modules.
c.: An automorphism of a left $R$ -modules $M$ is an isomorphism $f : M \to M$ of left $R$ -modules.

Notation 5.6.26. §

Sometimes, we refer to an $R$ -module homomorphism as an $R$ -linear map, and an endomorphism of $R$ -modules as an $R$ -linear endomorphism.

Examples 5.6.27. §

a.: The zero map $0 : M \to M$ and the identity map $id : M \to M$ are endomorphisms of an $R$ -module $M$ , with $id$ being an automorphism.
b.: Let $V$ and $W$ be vector spaces over a field $F$ . A left $F$ -module homomorphism $ϕ : V \to W$ is just an $F$ -linear transformation.
c.: Let $N$ be an $R$ -submodule of a left $R$ -module $M$ . The inclusion map $ι_{N} : N \to M$ is an $R$ -module homomorphism, as is the quotient map $π_{N} : M \to M ∕ N$ .
d.: If $M$ is an $R$ - $S$ -bimodule, then right multiplication $ψ_{s} : M \to M$ by an element $s \in S$ defines a left $R$ -module endomorphism. In particular, if $R$ is a commutative ring, then multiplication by $r \in R$ defines an $R$ -module endomorphism. Note that if $R$ is noncommutative, then the condition that left multiplication by $r \in R$ be a left module homomorphism $M \to M$ is that $r (𝑠𝑚) = s (𝑟𝑚)$ for all $r, s \in R$ and $m \in M$ , which need not hold.
e.: The identity map $F^{n} \to F^{n}$ provides an isomorphism between $F^{n}$ viewed as a left $M_{n} (F)$ -module via $(A, v) \mapsto 𝐴𝑣$ for $A \in M_{n} (F)$ and $v \in F^{n}$ (viewing $F^{n}$ as column vectors) and $F^{n}$ viewed as a left $M_{n} (F)$ -module via $(A, v) \mapsto v^{T} A^{T}$ (viewing $F^{n}$ as row vectors).

Note that we may speak of the kernel and the image of a left $R$ -module, as an $R$ -module homomorphism is in particular a group homomorphism. The reader will easily verify the following.

Lemma 5.6.28. §

Let $ϕ : M \to N$ be a left $R$ -module homomorphism. Then $\ker ϕ$ and $im ϕ$ are $R$ -submodules of $M$ and $N$ , respectively.

We also have analogues of all of the isomorphism theorems for groups. Actually, these are virtually immediate consequences of said isomorphism theorems, as the fact that one has isomorphisms of groups follows immediately from them, and then one need only note that these isomorphisms are actually homomorphisms of $R$ -modules.

Theorem 5.6.29. §

Let $R$ be a ring. Let $ϕ : M \to N$ be an homomorphism of left $R$ -modules. Then there is an isomorphism $\bar{ϕ} : M ∕ \ker ϕ \to im ϕ$ given by $\bar{ϕ} (m + \ker ϕ) = ϕ (m)$ .

Theorem 5.6.30. §

Let $R$ be a ring, and let $N$ be left $R$ -submodules of an $R$ -module $M$ . Then there is an isomorphism of $R$ -modules

M ∕ (M \cap N) \to_{}^{\sim} (M + N) ∕ N, m + (M \cap N) \mapsto m + N .

Theorem 5.6.31. §

Let $R$ be a ring, let $M$ be an $R$ -module, and let $Q \subseteq N$ be $R$ -submodules of $M$ . Then there is an isomorphism

M ∕ N \to_{}^{\sim} (M ∕ Q) ∕ (N ∕ Q), m + N \mapsto (m + Q) + (N ∕ Q) .

We also have the following analogue of Theorems 2.13.10 and 3.7.24.

Theorem 5.6.32. §

Let $R$ be a ring, let $M$ be an $R$ -module, and let $N$ be an $R$ -submodule of $M$ . Then the map $P \mapsto P ∕ N$ gives a bijection between submodules $P$ of $M$ containing $N$ and submodules of $M ∕ N$ . This bijection has inverse $Q \mapsto π_{N}^{- 1} (Q)$ on submodules $Q$ of $M ∕ N$ , where $π_{N} : M \to M ∕ N$ is the quotient map.

5.7. Free modules and generators

Definition 5.7.1. §

Let $S$ be a subset of an $R$ -module $M$ .

a.: The submodule of $M$ generated by $S$ is the intersection of all submodules of $M$ containing $S$ .
b.: We say that $S$ generates $M$ , or is a set of generators or generating set of $M$ , if no proper $R$ -submodule of $M$ contains $S$ .

Remark 5.7.2. §

The $R$ -submodule of $M$ generated by $S$ consists of the elements $\sum_{i = 1}^{n} a_{i} m_{i}$ with $m_{i} \in S$ and $a_{i} \in R$ for $1 \leq i \leq n$ and some $n \geq 1$ . The proof is much as before.

Remark 5.7.3. §

The sum $\sum_{i \in I} N_{i}$ of submodules $N_{i}$ of $M$ is the submodule generated by $\cup_{i \in I} N_{i}$ .

Notation 5.7.4. §

The $R$ -submodule of an $R$ -module $M$ generated by for a single element $m \in M$ (or, more precisely, by ${m}$ ) is denoted $R \cdot m$ .

Definition 5.7.5. §

We say that an $R$ -module is finitely generated if it has a finite set of generators.

Definition 5.7.6. §

We say that an $R$ -module is cyclic if it can be generated by a single element.

Example 5.7.7. §

A cyclic $R$ -submodule of $R$ is just a principal left ideal.

Lemma 5.7.8. §

An $R$ -module is cyclic if and only if it is isomorphic to $R ∕ I$ for some left ideal $I$ of $R$ .

Proof.

Let $M$ be a cyclic $R$ -module, generated by an element $m$ . Let $I = {r \in R ∣ 𝑟𝑚 = 0}$ . The surjective $R$ -module homomorphism $R \to M$ sending $r \in R$ to $𝑟𝑚$ has kernel the left ideal $I$ , so by the first isomorphism theorem for modules, we have $R ∕ 𝐼≅𝑀$ . Conversely, if $I$ is an left ideal of $R$ , then the quotient $R$ -module $R ∕ I$ is generated by the coset of $1$ . □

We can define direct sums and direct products of modules.

Definition 5.7.9. §

Let ${(M_{i})}_{i \in I}$ be a collection of left modules over a ring $R$ .

a.: The direct product $\prod_{i \in I} M_{i}$ is the $R$ -module that is the direct product of the abelian groups $M_{i}$ together with the left $R$ -multiplication $r \cdot {(m_{i})}_{i \in I} = {(r m_{i})}_{i \in I}$ for $r \in R$ and $m_{i} \in M_{i}$ for all $i \in I$ .
b.: The direct sum $\oplus_{i \in I} M_{i}$ is the $R$ -module that is the direct sum of the abelian groups $M_{i}$ together with the left $R$ -multiplication $r \cdot {(m_{i})}_{i \in I} = {(r m_{i})}_{i \in I}$ for $r \in R$ and $m_{i} \in M_{i}$ for all $i \in I$ with all but finitely many $m_{i} = 0$ .

Remark 5.7.10. §

If $I$ is a finite set, then the canonical injection

⨁_{i \in I} M_{i} \to \prod_{i \in I} M_{i}

is an isomorphism. In this case, the two concepts are often used interchangeably.

Notation 5.7.11. §

A direct sum (resp., product) of two $R$ -modules $M$ and $N$ is denoted $M \oplus N$ .

Definition 5.7.12. §

We say that an $R$ -submodule $A$ of an $R$ -module $B$ is a direct summand of $C$ if there exists an $R$ -module $C$ such that $B = A \oplus C$ . In this case, $C$ is called a complement to $A$ in $B$ .

Definition 5.7.13. §

Let $R$ be a ring.

a.: An $R$ -module $M$ is free on a subset $X$ of $M$ if for any $R$ -module $N$ and function $\bar{ϕ} : X \to N$ of elements of $N$ , there exists a unique $R$ -module homomorphism $ϕ : M \to N$ such that $ϕ (x) = \bar{ϕ} (x)$ for all $x \in X$ .
b.: A basis of an $R$ -module $M$ is a subset of $M$ on which it is free.

Remark 5.7.14. §

An abelian group $A$ is free on a set $X$ if and only if it is a free $ℤ$ -module on $X$ , as follows from Proposition 4.4.11.

In fact, we have the following alternative definition of a free $R$ -module. The proof is nearly identical to Proposition 4.4.11, so omitted.

Proposition 5.7.15. §

An $R$ -module $M$ is free on a basis $X$ if and only if the set $X$ generates $M$ and, for every $n \geq 1$ and $x_{1}, x_{2}, \dots, x_{n} \in X$ , the equality

\sum_{i = 1}^{n} c_{i} x_{i} = 0

for some $c_{1}, c_{2}, \dots, c_{n} \in R$ implies that $c_{i} = 0$ for all $i$ .

Remark 5.7.16. §

We might refer to the property that a set $X$ generates an $R$ -module $M$ as saying that $M$ is the $R$ -span of $X$ . The property that $\sum_{i = 1}^{n} c_{i} x_{i} = 0$ implies $c_{i} = 0$ , where $c_{i} \in R$ and $x_{i} \in X$ for $1 \leq i \leq n$ and some $n \geq 1$ can be referred to as saying that the set $X$ is $R$ -linearly independent.

Corollary 5.7.17. §

For any set $X$ , the $R$ -module $\oplus_{x \in X} R$ is free on the standard basis ${e_{x} ∣ x \in X}$ , where $e_{x}$ for $x \in X$ is the element which is nonzero only in its $x$ -coordinate, in which it is $1$ .

Proof.

The $e_{x}$ span $\oplus_{x \in X} R$ by its definition and are clearly $R$ -linearly independent. □

Corollary 5.7.18. §

Every $R$ -module is a quotient of a free $R$ -module.

Proof.

Let $M$ be an $R$ -module, and choose a generating set $X$ of $M$ (e.g., $M$ itself). Take the unique $R$ -module homomorphism

ψ : ⨁_{x \in X} R \to M

which satisfies $ψ (e_{x}) = x$ for all $x \in X$ . It is onto as $X$ generates $M$ . □

Noting Corollary 5.5.26, we also have the following.

Corollary 5.7.19. §

Every vector space over a field $F$ is a free $F$ -module.

The following is also a consequence of the universal property.

Theorem 5.7.20. §

Let $R$ be a commutative ring. A free module $M$ on a set $X$ is isomorphic to a free module $N$ on a set $Y$ if and only if $X$ and $Y$ have the same cardinality.

Proof.

If $X$ and $Y$ have the same cardinality, then any bijection $f : X \to Y$ gives an injection $X \to N$ which extends uniquely to a homomorphism $ϕ : M \to N$ . Similarly, the inverse of $f$ extends uniquely to a homomorphism $ψ : N \to M$ , and $ψ \circ ϕ$ (resp., $ϕ \circ ψ$ ) is then the unique extension to a homomorphism of the inclusion $X \to M$ (resp., $Y \to N$ ), therefore the identity. That is, $ϕ$ and $ψ$ are inverse isomorphisms.

For the converse, we first suppose that $Y$ is infinite and that there is an isomorphism $M \to N$ . Let $B$ denote the image of $X$ in $N$ , which is then necessarily an $R$ -basis of $N$ . Each element $y \in Y$ is contained in the span of a finite subset $B_{y}$ of $B$ . The union $B^{'}$ of these sets $B_{y}$ spans $N$ . For any $v \in B - B^{'}$ , the set $B^{'} \cup {v}$ is $R$ -linearly dependent, which cannot happen as $B$ is a basis. Thus, $B = B^{'}$ . Now, the cardinality $| B |$ of $B$ is at most the cardinality of the disjoint union of the sets $B_{y}$ for $y \in Y$ , each of which is finite. In particular, we have

| X | = | B | \leq | Y \times ℤ | = | Y |,

the latter equality holding as $Y$ is infinite. If $X$ is also infinite, then by reversing the roles of $X$ and $Y$ , this forces $| X | = | Y |$ .

Finally, suppose that $Y$ is finite, without loss of generality. Let $𝔪$ be a maximal ideal of $R$ . Consider the field $F = R ∕ 𝔪$ , and observe that

M ∕ 𝔪𝑀≅ (⨁_{x \in X} R) ∕ 𝔪 (⨁_{x \in X} R) ≅ ⨁_{x \in X} F,

and similarly for $Y$ . An isomorphism $M \to_{}^{\sim} N$ induces an isomorphism of $F$ -vector spaces $M ∕ 𝔪𝑀 \to_{}^{\sim} N ∕ 𝔪𝑁$ , which by the above isomorphisms have bases of cardinality $| X |$ and $| Y |$ respectively. Since $Y$ is finite, Theorem 5.5.28 tells us that $X$ must be finite of order $| Y |$ . □

The following is immediate.

Corollary 5.7.21. §

Let $R$ be a commutative ring, and let $M$ be a free $R$ -module on a set of $n$ elements. Then every basis of $M$ has $n$ elements.

By Theorem 5.7.23, we may make the following definition.

Definition 5.7.22. §

The rank of a free module $M$ over a commutative ring $R$ is the unique $n \geq 0$ such that $𝑀≅ R^{n}$ if it exists. Otherwise, $M$ is said to have infinite rank.

For an integral domain, we can do somewhat better with a bit of work. In fact, the following result does not require this assumption, but the proof we give does.

Theorem 5.7.23. §

Let $R$ be an integral domain. Let $M$ be a free $R$ -module on a set of $n$ elements, and let $Y$ be a subset of $M$ . Then:

i.: if $Y$ generates $M$ , then $Y$ has at least $n$ elements,
ii.: if $Y$ is $R$ -linearly independent, then $Y$ has at most $n$ elements, and
iii.: $Y$ is a basis if and only if it generates $M$ and has exactly $n$ elements.

Moreover, a free module on an infinite set cannot be generated by a finite set of elements.

Proof.

Suppose that $M$ is free on $n$ elements. A choice of basis defines an isomorphism $M \to_{}^{\sim} R^{n}$ of $R$ -modules, so we may assume that $M = R^{n}$ . Note that $R^{n}$ is contained in the $Q (R)$ -module $Q {(R)}^{n}$ via the canonical inclusion, and any generating set $Y$ of $R^{n}$ spans $Q {(R)}^{n}$ . But by Theorems 5.5.28 and 5.5.27, this forces $Y$ to have at least $n$ elements. If $Y$ has $n$ elements, then $Y$ would similarly be a basis of $Q {(R)}^{n}$ . So, if we had $\sum_{i = 1}^{n} c_{i} y_{i} = 0$ for some $c_{i} \in R$ and distinct $y_{i} \in Y$ , then each $c_{i} = 0$ , which means that $Y$ is an $R$ -basis of $R^{n}$ .

On the other hand, if $Y$ has more than $n$ elements, then by Theorem 5.5.25, the set $Y$ cannot be linearly independent in $Q {(R)}^{n}$ . That is, there exist $α_{i} \in Q {(R)}^{n}$ and distinct $y_{i} \in Y$ for $1 \leq i \leq m$ and $m \geq 1$ with $\sum_{i = 1}^{m} α_{i} y_{i} = 0$ and not all $α_{i} = 0$ . For each $i$ , write $α_{i} = c_{i} d_{i}^{- 1}$ with $c_{i}, d_{i} \in R$ and $d_{i} \neq 0$ . Taking $d$ to be the product of the $d_{i}$ , we then have $a_{i} = d α_{i} \in R$ and not all $a_{i} = 0$ . Since $\sum_{i = 1}^{m} a_{i} y_{i} = 0$ , it follows that $Y$ is not a basis.

Finally, if $N$ is a free module on an infinite set $X$ , then $𝑁≅ \oplus_{x \in X} R$ , and so we take $N$ to be the latter module. We then have that $\oplus_{x \in X} Q (R)$ is a $Q (R)$ -vector space with an infinite basis. But then Theorem 5.5.28 tells us that every basis is infinite, which by Theorem 5.5.27 tells us that a finite set cannot span. □

Remark 5.7.24. §

The full analogues of Theorems 5.5.25 and 5.5.27 do not hold for modules over arbitrary rings, over even abelian groups. That is, take the free $ℤ$ -module $ℤ$ . The set ${2}$ does not span it and is not contained in a basis of $ℤ$ , and the set ${2, 3}$ does span it and does not contain a basis.

Example 5.7.25. §

The polynomial ring $R [x]$ is a free $R$ -module on the basis ${x^{i} ∣ i \in ℤ_{\geq 0}}$ .

Remark 5.7.26. §

Consider the ideal $I = (2, x)$ of $ℤ [x]$ . It is not a free $ℤ [x]$ -module. To see this, first note that it is not a principal ideal so cannot be generated by a single element. As $I$ can be generated by the two elements $2$ and $x$ , if $I$ were free, then it would follow from Theorem 5.7.23 that ${2, x}$ would be a basis for $I$ . On the other hand, $x \cdot 2 - 2 \cdot x = 0$ , which would contradict Proposition 5.7.15.

Proposition 5.7.27. §

Let $M$ be an $R$ -module, and let $π : M \to F$ be a surjective $R$ -module homomorphism, where $F$ is $R$ -free. Then there exists an injective $R$ -module homomorphism $ι : F \to M$ such that $π \circ ι = {id}_{F}$ . Moreover, we have $M = \ker (π) \oplus ι (F)$ .

Proof.

Let $X$ be an $R$ -basis of $F$ , and for each $x \in X$ , choose $m_{x} \in M$ with $π (m_{x}) = x$ . We take $ι : F \to M$ to be the unique $R$ -module homomorphism with $ι (x) = m_{x}$ for all $x \in X$ , which exists as $F$ is free. Then $π \circ ι (x) = x$ for all $x \in X$ , so $π \circ ι = {id}_{F}$ by uniqueness, and $ι$ must be injective.

Finally, let $A = \ker π$ . Note that any $m \in M$ satisfies $m - ι \circ π (m) \in A$ , so $M = A + ι (F)$ . If $m \in A \cap ι (F)$ , then $m = ι (n)$ for some $n \in F$ and $n = π \circ ι (n) = π (m) = 0$ , so $m = 0$ . In other words, we have $M = A \oplus ι (F)$ . □

In particular, every free quotient of an $R$ -module $M$ is isomorphic to a direct summand of $M$ .

5.8. Matrix representations

We work in this section with (nonzero) homomorphisms of free modules over a ring $R$ . Most of the time, the case of interest is that of linear transformations of vector spaces over fields, but there is no additional restriction caused by working is full generality.

Lemma 5.8.1. §

Let $R$ be a ring. Let $A \in M_{𝑚𝑛} (R)$ be a matrix for some $m, n \geq 1$ . Then there is a unique $R$ -module homomorphism $T : R^{n} \to R^{m}$ satisfying $T (v) = 𝐴𝑣$ for all $v \in R^{n}$ , where $𝐴𝑣$ is matrix multiplication, viewing elements of $R^{m}$ and $R^{n}$ as column vectors.

Proof.

Define $T (e_{j}) = \sum_{i = 1}^{m} a_{𝑖𝑗} f_{i}$ , where $e_{j}$ (resp., $f_{i}$ ) is the $j$ th (resp., $i$ th) standard basis element of $R^{n}$ (resp., $R^{m}$ ). If $v = \sum_{j = 1}^{n} c_{j} e_{j}$ for some $c_{j} \in F$ with $1 \leq j \leq n$ , then

T (v) = \sum_{j = 1}^{n} c_{j} T (e_{j}) = \sum_{i = 1}^{m} (\sum_{j = 1}^{n} a_{𝑖𝑗} c_{j}) f_{i} = 𝐴𝑣 .

The uniqueness follows from the fact that $R^{n}$ is free, so any $R$ -module homomorphism from it is determined by its values on a basis □

Definition 5.8.2. §

An ordered basis is a basis of a free $R$ -module together with a total ordering on the basis.

Remark 5.8.3. §

We refer to a finite (ordered) basis on a free $R$ -module as a set ${v_{1}, v_{2}, \dots, v_{n}}$ and take this implicitly to mean that the set has cardinality $n$ and that the basis is ordered in the listed order (i.e., by the ordering $v_{i} \leq v_{i + 1}$ for all $1 \leq i < n$ ).

Example 5.8.4. §

The standard basis ${e_{1}, e_{2}, \dots, e_{n}}$ on $R^{n}$ is ordered in the order of positions of the nonzero coordinate of its elements.

Notation 5.8.5. §

If $B = {v_{1}, v_{2}, \dots, v_{n}}$ is an ordered basis of a free $R$ -module $V$ , then we let $φ_{B} : R^{n} \to V$ denote the $R$ -module isomorphism satisfying $φ_{B} (e_{i}) = v_{i}$ for all $i$ .

Given ordered bases of free $R$ -modules $V$ and $W$ , an $R$ -module homomorphism $T : V \to W$ can be described by a matrix.

Definition 5.8.6. §

Let $V$ and $W$ be free modules over a ring $R$ with ordered bases $B = {v_{1}, v_{2}, \dots, v_{n}}$ and $C = {w_{1}, w_{2}, \dots, w_{m}}$ , respectively. Let $T : V \to W$ be an $R$ -module homomorphism. We say that a matrix $A = (a_{𝑖𝑗}) \in M_{𝑛𝑚} (R)$ represents $T$ with respect to the bases $B$ and $C$ if

T (v_{j}) = \sum_{i = 1}^{m} a_{𝑖𝑗} w_{i}

for all $1 \leq j \leq n$ .

Remark 5.8.7. §

Given ordered bases $B = {v_{1}, \dots, v_{n}}$ of a free module $V$ and $C = {w_{1}, \dots, w_{m}}$ of a free module $W$ , the composition

φ_{C}^{- 1} \circ T \circ φ_{B} : R^{n} \to_{}^{φ_{B}} V \to_{}^{T} W \to_{}^{φ_{C}^{- 1}} R^{m},

is given by multiplication by a matrix $A$ by Lemma 5.8.1. This $A$ is the matrix representing $T$ with respect to $B$ and $C$ .

Terminology 5.8.8. §

Let $V$ be a free $R$ -module with finite basis $B$ , and let $T : V \to V$ be an $R$ -module homomorphism. We say say that a matrix $A$ represents $T$ with respect to $B$ if $A$ represents $T$ with respect to $B$ and $B$ . If $V = R^{n}$ and $B$ is the standard basis, we simply say that $A$ represents $T$ .

Lemma 5.8.9. §

Let $T^{'} : U \to V$ and $T : V \to W$ be homomorphisms of finite rank free $R$ -modules. Let $B$ , $C$ , and $D$ be bases of $U$ , $V$ , and $W$ , respectively. Suppose that $A^{'}$ represents $T^{'}$ with respect to $B$ and $C$ and that $A$ represents $T$ with respect to $C$ and $D$ . Then $A A^{'}$ represents $T \circ T^{'} : U \to W$ with respect to $B$ and $D$ .

Proof.

We have that $A$ represents $φ_{D}^{- 1} \circ T \circ φ_{C}$ and $A^{'}$ represents $φ_{C}^{- 1} \circ T^{'} \circ φ_{B}$ . In other words, the maps are left multiplication by the corresponding matrices. The map

φ_{D}^{- 1} \circ T \circ T^{'} \circ φ_{B} = (φ_{D}^{- 1} \circ T \circ φ_{C}) \circ (φ_{C}^{- 1} \circ T^{'} \circ φ_{B}),

is then left multiplication by $A A^{'}$ , which is to say that it is represented by $A A^{'}$ . □

Definition 5.8.10. §

Let $B = {v_{1}, \dots, v_{n}}$ and $B^{'} = {v_{1}^{'}, \dots, v_{n}^{'}}$ be ordered bases of a free $R$ -module $V$ . The change-of-basis matrix from $B$ to $B^{'}$ is the matrix $Q_{B, B^{'}} = (q_{𝑖𝑗})$ that represents the $R$ -module homomorphism $T_{B, B^{'}} : V \to V$ with $T_{B, B^{'}} (v_{i}) = v_{i}^{'}$ for $1 \leq i \leq n$ with respect to $B$ .

Remark 5.8.11. §

If $v_{j}^{'} = \sum_{i = 1}^{n} q_{𝑖𝑗} v_{i}$ for all $i$ , then the change-of-basis matrix $Q_{B, B^{'}}$ of Definition 5.8.10 is the matrix $(q_{𝑖𝑗})$ . It is invertible, and $Q_{B^{'}, B} = Q_{B, B^{'}}^{- 1}$ .

Remark 5.8.12. §

Let $V$ be free of rank $n$ with bases $B$ and $B^{'}$ . By definition, the change-of-basis matrix $Q_{B, B^{'}}$ represents $φ_{B}^{- 1} \circ T_{B, B^{'}} \circ φ_{B}$ . On the other hand, we also have that that $φ_{B^{'}} = T_{B, B^{'}} \circ φ_{B}$ . Thus, see that

φ_{B}^{- 1} \circ φ_{B^{'}} = φ_{B}^{- 1} \circ T_{B, B^{'}} \circ φ_{B},

is represented by $Q_{B, B^{'}}$ .

Theorem 5.8.13 (Change of basis theorem). §

Let $T : V \to W$ be a linear transformation of free $R$ -modules of finite rank. Let $B$ and $B^{'}$ be ordered bases of $V$ and $C$ and $C^{'}$ be ordered bases of $W$ . If $A$ is the matrix representing $T$ with respect to $B$ and $C$ , then $Q_{C, C^{'}}^{- 1} A Q_{B, B^{'}}$ is the matrix representing $T$ with respect to $B^{'}$ and $C^{'}$ .

Proof.

We have that $A$ represents $φ_{C}^{- 1} \circ T \circ φ_{B}$ , and we wish to compute the matrix representing $φ_{C^{'}}^{- 1} \circ T \circ φ_{B^{'}}$ . We have

φ_{C^{'}}^{- 1} \circ T \circ φ_{B^{'}} = (φ_{C^{'}}^{- 1} \circ φ_{C}) \circ (φ_{C}^{- 1} \circ T \circ φ_{B}) \circ (φ_{B}^{- 1} \circ φ_{B^{'}}),

and these three matrices are represented by $Q_{C, C^{'}}^{- 1}$ , $A$ , and $Q_{B, B^{'}}$ , respectively. □

5.9. Associative algebras

Much as with groups, we may speaker of the center of a ring.

Definition 5.9.1. §

The center $Z (R)$ of a ring $R$ is the subring of $R$ given by the subset

Z (R) = {a \in R ∣ 𝑎𝑏 = 𝑏𝑎 for all b \in R} .

We now define the notion of an algebra over a commutative ring.

Definition 5.9.2. §

Let $R$ be a commutative ring. An (associative) $R$ -algebra $A$ is the pair of an $R$ -module $A$ and a binary operation $\cdot$ on $A$ which, together with the addition on $A$ , makes $A$ into a ring, and which satisfies

r (a \cdot b) = (𝑟𝑎) \cdot b = a \cdot (𝑟𝑏)

for all $r \in R$ and $a, b \in A$ .

Remark 5.9.3. §

An $R$ -algebra $A$ comes endowed with a homomorphism $ϕ : R \to Z (A)$ , given by $ϕ (r) = r \cdot 1$ for $r \in R$ and the element $1 \in A$ . In fact, to give an $R$ -algebra $A$ is to give a ring $A$ and a ring homomorphism $ϕ : R \to Z (A)$ with $ϕ (1) = 1$ , for then this provides an $R$ -module structure on $A$ given by $r \cdot a = ϕ (r) a$ , which makes $A$ into an $R$ -algebra.

Remark 5.9.4. §

Often, it is supposed that the map $ϕ : R \to Z (A)$ that defines the $R$ -algebra structure on $A$ is injective. This is automatically the case if $R$ is a field.

Examples 5.9.5. §

i.: The polynomial ring $R [x_{1}, \dots, x_{n}]$ over a commutative ring $R$ is an $R$ -algebra for any $n \geq 1$ .
ii.: The matrix ring $M_{n} (R)$ over a commutative ring $R$ is an $R$ -algebra for any $n \geq 1$ .
iii.: If $F$ is a field and $E$ is a field containing $F$ as a subring, then $F$ is an $E$ -algebra.
iv.: Every ring $R$ is a $ℤ$ -algebra.

Example 5.9.6. §

The ring $ℍ = {a + 𝑏𝑖 + 𝑐𝑗 + 𝑑𝑘 ∣ a, b, c, d \in ℝ}$ of quaternions has center $Z (ℍ) = ℝ$ , hence is an algebra over $ℝ$ . Note that $ℂ = {a + 𝑏𝑖 ∣ a, b \in ℝ}$ is contained in $ℍ$ , but $ℍ$ is not a $ℂ$ -algebra, as $ℂ$ is not contained in the center of $ℍ$ .

Definition 5.9.7. §

An $R$ -algebra homomorphism $f : A \to B$ is a ring homomorphism of $R$ -algebras $A$ and $B$ that is also a homomorphism of $R$ -modules.

Example 5.9.8. §

For any $R$ -algebra $A$ , the structure homomorphism $ϕ : R \to A$ with image in $Z (A)$ is a homomorphism of $R$ -algebras.

We provide three other classes of examples.

Definition 5.9.9. §

Let $R$ be a commutative ring and $M$ an $R$ -module. Then endomorphism ring ${End}_{R} (M)$ of $M$ over $R$ is the $R$ -algebra of $R$ -linear endomorphisms of $M$ . It is a ring under addition and composition of endomorphisms, and has the $R$ -module structure given by multiplication of scalars: that is,

(r \cdot f) (m) = r \cdot f (m)

for all $f \in {End}_{R} (M)$ , $r \in R$ , and $m \in M$ .

Remark 5.9.10. §

The map $ϕ : R \to Z ({End}_{R} (M))$ defining the $R$ -algebra structure on ${End}_{R} (M)$ takes $r \in R$ to left multiplication by $r$ on $M$ .

Definition 5.9.11. §

The automorphism group ${Aut}_{R} (M)$ of an $R$ -module $M$ is the group of $R$ -automorphisms of $M$ under composition.

Remark 5.9.12. §

The unit group of ${End}_{R} (M)$ is ${Aut}_{R} (M)$ .

Example 5.9.13. §

We have an isomorphism of $R$ -algebras ${End}_{R} (R^{n}) \to_{}^{\sim} M_{n} (R)$ by taking $T \in {End}_{R} (R^{n})$ to the matrix $A$ representing it with respect to the standard basis of $R^{n}$ . We have that $T \in {Aut}_{R} (R^{n})$ if and only if $A$ is invertible.

Proposition 5.9.14. §

Let $M$ be an $R$ -module and $A$ be an $R$ -algebra with structure map $ϕ : R \to Z (A)$ . There is a bijection between operations $\cdot : A \times M \to M$ which make $M$ into a left $A$ -module such that $ϕ (r) \cdot m = r \cdot m$ for all $r \in R$ and $m \in M$ and $R$ -algebra homomorphisms $ψ : A \to {End}_{R} (M)$ determined by $ψ (a) (m) = a \cdot m$ for all $a \in A$ and $m \in M$ .

We mention another class of algebras known as group rings. The reader will easily check the following.

Lemma 5.9.15. §

Let $R$ be a commutative ring an $G$ be a group. The set $R [G]$ of elements $\sum_{g \in G} a_{g} g$ with $a_{g} \in R$ for all $g \in G$ and almost all $a_{g} = 0$ and addition and multiplication of multiplication are given respectively by the formulas

\begin{array}{rcl} \sum_{g \in G} a_{g} g + \sum_{g \in G} b_{g} g = \sum_{g \in G} (a_{g} + b_{g}) g & and & (\sum_{g \in G} a_{g} g) \cdot (\sum_{g \in G} b_{g} g) = \sum_{g \in G} (\sum_{h \in G} a_{h} b_{h^{- 1} g}) g \end{array}

is an $R$ -algebra with $R$ -module structure given by the scalar multiplication

r \cdot \sum_{g \in G} a_{g} g = \sum_{g \in G} (r a_{g}) g .

Remark 5.9.16. §

As an $R$ -module, $R [G]$ is in bijection with the free $R$ -module $\oplus_{g \in G} R \cdot g$ with basis $G$ . The multiplication on $R [G]$ is the unique multiplication that restricts to the multiplication on $G$ and makes $R [G]$ into an $R$ -algebra. The identity element $1$ in $R [G]$ is the identity of $G$ .

Definition 5.9.17. §

The group ring $R [G]$ of a group $G$ with coefficients in a commutative ring $R$ is the unique $R$ -algebra that is free as an $R$ -module with basis $G$ and has multiplication that restricts to the multiplication on $G$ .

Example 5.9.18. §

For $n \geq 1$ , there is an isomorphism

R [x] ∕ (x^{n} - 1) \to_{}^{\sim} R [ℤ ∕ 𝑛ℤ], \sum_{i = 0}^{n - 1} a_{i} x^{i} \mapsto \sum_{i = 0}^{n - 1} a_{i} [i],

of $R$ -modules, where $[i]$ denotes the group element corresponding to $i \in ℤ ∕ 𝑛ℤ$ . Similarly, one has $R [x, x^{- 1}] ≅𝑅 [ℤ]$ .

We also have the notion of a free $R$ -algebra over a commutative ring $R$ .

Definition 5.9.19. §

Let $R$ be a commutative ring, and let $X$ be a set. The free $R$ -algebra $R ⟨ X ⟩$ on $X$ is the $R$ -algebra with underlying additive group the free $R$ -module on the words in $X$ and multiplication the unique $R$ -bilinear map given on words in $X$ by concatenation.

Remark 5.9.20. §

Another way of describing the free $R$ -algebra on a set $X$ is that it’s a noncommutative polynomial ring with variables in $X$ .

Notation 5.9.21. §

If $X = {x_{1}, \dots, x_{n}}$ has $n$ elements, then we write $R ⟨ x_{1}, \dots, x_{n} ⟩$ for $R ⟨ X ⟩$ .