Abstract Algebra — Chapter 3 Ring theory

Chapter 3
Ring theory

3.1. Rings

In this section, we define rings and fields. These are sets with two binary operations, known as addition and multiplication.

Definition 3.1.1. §

Let $R$ be a set with a pair $(+,⋅)$ of binary operations. We say that $R$ satisfies the left distributive law (with respect to $+$ and $\cdot$ ) if

a \cdot (b + c) = (a \cdot b) + (a \cdot c)

for all $a, b, c \in R$ , and we say that $R$ satisfies the right distributive law if

(a + b) \cdot c = (a \cdot c) + (b \cdot c)

for all $a, b, c \in R$ .

The distributive law being one of the standard axioms of arithmetic, it is satisfied by many common objects, such as $ℤ$ , $ℚ$ , $ℝ$ , $ℂ$ , and so on. We give one less standard example.

Example 3.1.2. §

The set $Maps (ℝ, ℝ)$ satisfies the left and right distributive laws with respect to the pair of operations $(+,⋅)$ . It satisfies the right distributive law with respect to $(+,∘)$ and $(⋅,∘)$ , where $\circ$ is composition.

We now define a ring.

Definition 3.1.3. §

A set $R$ with a pair $(+,⋅)$ of binary operations is a ring if

i.: $R$ is an abelian group under $+$ ,
ii.: the binary operation $\cdot$ is associative,
iii.: $R$ has an identity element $1$ under $\cdot$ , and
iv.: $R$ satisfies the left and right distributive laws.

Remark 3.1.4. §

When $+$ and $\cdot$ are used to denote the binary operations of a ring, we refer $+$ as addition and $\cdot$ as multiplication. Unless otherwise stated, the operations of $R$ will be denoted $+$ and $\cdot$ .

Remark 3.1.5. §

As in the case of groups, we often write $𝑎𝑏$ for $a \cdot b$ for $a, b$ in a ring $R$ . We also use $a \cdot b + c$ to denote $(a \cdot b) + c$ for $a, b, c \in R$ .

Examples 3.1.6. §

a.: The sets $ℤ$ , $ℚ$ , $ℝ$ , and $ℂ$ are all rings with respect to the usual operations of addition and multiplication.
b.: The sets $Maps (ℝ, ℝ)$ and $M_{n} (ℝ)$ for $n \geq 1$ are also rings with respect to addition and multiplication.
c.: The set $𝑛ℤ$ for $n \geq 1$ is a ring with respect to addition and multiplication.
d.: The set $ℤ ∕ 𝑛ℤ$ is a ring with respect to its operations of addition and multiplication.

Remark 3.1.7. §

Since the first binary operation on a ring $R$ is denoted $+$ , the identity element is denoted $0$ as usual, and the additive inverse of $a \in R$ is denoted $- a$ . The sum of $n$ copies of $a$ is denoted $𝑛𝑎$ for $n \geq 1$ , and $- (𝑛𝑎)$ is also denoted $- 𝑛𝑎$ .

We have the following properties in any ring.

Lemma 3.1.8. §

Let $R$ be a ring, and let $a, b \in R$ . Then we have

a.: $0 \cdot a = a \cdot 0 = 0$ ,
b.: $a \cdot (- b) = (- a) \cdot b = - 𝑎𝑏$ , and
c.: $(- a) \cdot (- b) = 𝑎𝑏$ .

Proof.

a.

We have

0 \cdot a + b \cdot a = (0 + b) \cdot a = b \cdot a

by the right distributive law and the fact that $0$ is an additive identity. Therefore, the Cancellation theorem tells us that $0 \cdot a = 0$ . Similarly, $a \cdot 0 = 0$ using the left distributive law instead of the right.

b.

We have

a \cdot (- b) + 𝑎𝑏 = a \cdot (- b + b) = a \cdot 0 = 0

by the left distributive law, the definition of the additive inverse, and part a. The other equality is similar.

c.

This follows from part b, which tells us that

(- a) \cdot (- b) = - (a \cdot (- b)) = - (- (𝑎𝑏)) = 𝑎𝑏 .

□

On a set with one element, there is only one possible binary operation, and using it as both addition and multiplication turns that set into a ring.

Definition 3.1.9. §

The zero ring is the ring ${0}$ . We say that a ring $R$ is a nonzero ring if $R$ has more than one element.

That a ring $R$ has an identity under $\cdot$ is to say exactly that there is an element $1 \in R$ with $1 \cdot a = a \cdot 1 = a$ for all $a \in R$ . By Lemma 2.1.4, the multiplicative identity $1$ in a ring is unique.

Examples 3.1.10. §

The rings $ℤ$ , $ℚ$ , $ℝ$ , $ℂ$ , $Maps (ℝ, ℝ)$ , $M_{n} (ℝ)$ for $n \geq 1$ , $ℤ ∕ 𝑛ℤ$ for $n \geq 1$ , and ${0}$ are all rings. However, $𝑛ℤ$ is not a ring for $n \geq 2$ .

Remark 3.1.11. §

One easily checks that $(𝑛𝑎) \cdot (𝑚𝑏) = (𝑛𝑚) 𝑎𝑏$ for $n, m \in ℤ$ and $a, b \in R$ for any ring $R$ . One has $(n \cdot 1) \cdot (m \cdot 1) = 𝑛𝑚 \cdot 1$ . We often denote $n \cdot 1$ by $n$ , though we remark that it is possible that $n \cdot 1 = m \cdot 1$ for $n \neq m$ , as will happen in any finite ring, for instance.

Remark 3.1.12. §

If $R$ is a ring with $1 = 0$ , then $x = 1 \cdot x = 0 \cdot x = 0$ for all $x \in R$ , so $R$ is the zero ring.

We now introduce the notion of a subring of a ring, which does not play quite as prominent of a role in ring theory as does the notion of a subgroup of a group in group theory.

Definition 3.1.13. §

A subring $S$ of a ring $R$ is a subset of $R$ containing $1$ that is a ring with respect to the restrictions to $S$ of the binary operations of addition and multiplication on $R$ .

We leave it to the reader to check the following.

Lemma 3.1.14. §

A subset $S$ of a ring $R$ is a subring if it is closed under the operations of addition and multiplication on $R$ , contains $0$ and $1$ , and contains $- a$ for all $a \in S$ .

Clearly, the property of being a subring is a transitive one.

Examples 3.1.15. §

a.: The ring $𝑛ℤ$ is not a subring of $ℤ$ , as it does not contain $1$ .
b.: The ring $ℤ$ is a subring of $ℚ$ , which is in turn a subring of $ℝ$ , which is in turn a subring of $ℂ$ .

Most of the study of ring theory is focused on commutative rings.

Definition 3.1.16. §

A ring $R$ is a commutative ring if multiplication on $R$ is commutative. We then say that the ring $R$ is commutative.

Definition 3.1.17. §

A ring $R$ that is not commutative is a noncommutative ring.

Examples 3.1.18. §

The rings $ℤ$ , $ℚ$ , $ℝ$ , $ℂ$ , $Maps (ℝ, ℝ)$ , $ℤ ∕ 𝑛ℤ$ for $n \geq 1$ , and ${0}$ are all commutative rings. However, $M_{n} (ℝ)$ is a noncommutative ring for all $n \geq 2$ .

The notion of a field is really just a special case of the notion of a ring, but it is an important one.

Definition 3.1.19. §

A field is a nonzero commutative ring for which every nonzero element has a multiplicative inverse.

In other words, a field is a nonzero commutative ring for which the nonzero elements form a group under multiplication (in fact, an abelian group).

Definition 3.1.20. §

A subfield of a field $F$ is a subring of $F$ that is a field.

Examples 3.1.21. §

a.: The rings $ℚ$ , $ℝ$ , and $ℂ$ are fields. Of course, $ℚ$ is a subfield of $ℝ$ and $ℂ$ , and $ℝ$ is a subfield of $ℂ$ .
b.: The ring $ℤ$ is not a field.

The analogous object to a field in the more general theory of possibly noncommutative rings is known as a division ring.

Definition 3.1.22. §

A division ring (or skew field) $D$ is a nonzero ring such that every nonzero element is invertible under multiplication.

Clearly, all fields are division rings. As with fields, we have multiplicative groups of division rings, which no longer need be abelian.

Definition 3.1.23. §

The group of nonzero elements in a division ring $D$ is known as the multiplicative group of $D$ and is denoted $D^{\times}$ .

We end this section with one example of a noncommutative division ring.

Definition 3.1.24. §

The ring of quaternions $ℍ$ is the set of distinct elements $a + 𝑏𝑖 + 𝑐𝑗 + 𝑑𝑘$ with $a, b, c, d \in ℝ$ , together with addition defined by

(a + 𝑏𝑖 + 𝑐𝑗 + 𝑑𝑘) + (a^{'} + b^{'} i + c^{'} j + d^{'} k) = (a + a^{'}) i + (b + b^{'}) j + (c + c^{'}) k

and multiplication defined by

\begin{matrix} (a + 𝑏𝑖 + 𝑐𝑗 + 𝑑𝑘) \cdot (a^{'} + b^{'} i + c^{'} j + d^{'} k) = (a a^{'} - b b^{'} - c c^{'} - d d^{'}) \\ + (a b^{'} + b a^{'} + c d^{'} - d c^{'}) i + (a c^{'} - b d^{'} + c a^{'} + d b^{'}) j + (a d^{'} + b c^{'} - c b^{'} + d a^{'}) k \end{matrix}

for $a, b, c, d, a^{'}, b^{'}, c^{'}, d^{'} \in ℝ$ .

Remark 3.1.25. §

The ring $ℍ$ is an $ℝ$ -vector space with basis $1$ , $i$ , $j$ , $k$ , where

α \cdot (a + 𝑏𝑖 + 𝑐𝑗 + 𝑑𝑘) = 𝛼𝑎 + (𝛼𝑏) i + (𝛼𝑐) j + (𝛼𝑑) k

for $α, a, b, c, d \in ℝ$ . Note that we have $𝑖𝑗 = k = - 𝑗𝑖$ , $𝑗𝑘 = i = - 𝑘𝑗$ , $𝑘𝑖 = j = - 𝑖𝑘$ , and $i^{2} = j^{2} = k^{2} = - 1$ in $ℍ$ .

Theorem 3.1.26. §

The quaternion algebra is a division ring.

Proof.

We give only a sketch. Distributivity is a direct consequence of the definitions of the operations of addition and multiplication. In fact, it is also easy to see that $α \cdot 𝑥𝑦 = x \cdot (𝛼𝑦) = (𝛼𝑥) \cdot y$ for $α \in ℝ$ and $x, y \in ℍ$ . Using the distributive law and the latter fact, associativity of multiplication follows from a check of associativity on the subset ${i, j, k}$ of $ℍ^{\times}$ . Finally, any nonzero $a + 𝑏𝑖 + 𝑐𝑗 + 𝑑𝑘 \in ℍ$ has inverse

{(a + 𝑏𝑖 + 𝑐𝑗 + 𝑑𝑘)}^{- 1} = {(a^{2} + b^{2} + c^{2} + d^{2})}^{- 1} (a - 𝑏𝑖 - 𝑐𝑗 - 𝑑𝑘),

so $ℍ$ is a division ring. □

3.2. Families of rings

In this section, we consider various sorts, or families, of rings one can construct out of other rings. We begin with matrix rings.

Definition 3.2.1. §

If $R$ is a nonzero ring, the matrix ring $M_{n} (R)$ consisting of $n$ -by- $n$ matrices with entries in $R$ is the set with the addition $(a_{𝑖𝑗}) + (b_{𝑖𝑗}) = (a_{𝑖𝑗} + b_{𝑖𝑗})$ and multiplication

(a_{𝑖𝑗}) \cdot (b_{𝑖𝑗}) = (\sum_{k = 1}^{n} a_{𝑖𝑘} b_{𝑘𝑗}) .

We leave the proof of the following to the reader.

Lemma 3.2.2. §

Let $R$ be a ring and $n \geq 1$ . Then $M_{n} (R)$ is a ring.

Lemma 3.2.3. §

The ring $M_{n} (R)$ is noncommutative if $R$ is a ring and $n \geq 2$ .

Proof.

Suppose $n = 2$ . Let $A = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ and $B = (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix})$ . Then $𝐴𝐵 = I$ , the identity matrix in $M_{2} (ℝ)$ , while $𝐵𝐴 = 0$ . The general case follows from the case $n = 2$ by taking matrices that contain the same entries as $A$ and $B$ in their upper lefthand corners and are zero in all other entries. □

Another important class of rings is the polynomial rings.

Definition 3.2.4. §

Let $R$ be a ring, and fix an indeterminate (i.e., a symbol) $x$ . The polynomial ring $R [x]$ with $R$ -coefficients is the set of finite formal (i.e., two are different if they are written differently) sums of powers of $x$ with coefficients in $R$ , i.e.,

R [x] = {\sum_{i = 0}^{\infty} a_{i} x^{i} ∣ a_{i} \in R for all i \geq 0, a_{i} = 0 for all i > N for some N \geq 0} .

together with the binary operations of addition and multiplication given by

\begin{array}{rcl} \sum_{i = 0}^{\infty} a_{i} x^{i} + \sum_{j = 0}^{\infty} b_{j} x^{j} = \sum_{i = 0}^{\infty} (a_{i} + b_{i}) x^{i}, \\ (\sum_{i = 0}^{\infty} a_{i} x^{i}) \cdot (\sum_{j = 0}^{\infty} b_{j} x^{j}) = \sum_{k = 0}^{\infty} (\sum_{i = 0}^{k} a_{i} b_{k - i}) x^{k} . \end{array}

An element $f = \sum_{i = 0}^{\infty} a_{i} x^{i}$ of $R [x]$ is called a polynomial, the $a_{i}$ are referred to as coefficients, and $x$ is called a variable.

Remark 3.2.5. §

If $a_{i} = 0$ for all $i > N$ , then we more commonly write $\sum_{i = 0}^{N} a_{i} x^{i}$ for $f = \sum_{i = 0}^{\infty} a_{i} x^{i}$ . We will also sometimes write

f = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n},

identifying $x^{0}$ with “ $1$ ” and $x^{1}$ with “ $x$ ”.

Definition 3.2.6. §

The degree $\deg f$ of a nonzero polynomial $f = \sum_{i = 0}^{\infty} a_{i} x^{i} \in R [x]$ is the smallest integer $N$ such that $a_{i} = 0$ for all $i > N$ . When needed, we consider the degree of $0$ to be $- \infty$ .

Remark 3.2.7. §

A polynomial of degree $0$ is said to be constant, a polynomial of degree $1$ is linear, a polynomial of degree $2$ is quadratic, followed by cubic, quartic, quintic, and so forth.

Definition 3.2.8. §

If $f$ is a polynomial of degree $n \geq 0$ , then its leading coefficent is the coefficient of $x^{n}$ in $f$ . If constant coefficient is the coefficient of $x^{0} = 1$ .

Example 3.2.9. §

The polynomials $2 + 3 x - x^{2}$ and $1 + x$ are elements of $ℤ [x]$ . One has, as usual,

(2 + 3 x - x^{2}) \cdot (1 + x) = 2 + (2 + 3) x + (3 - 1) x^{2} - x^{3} = 2 + 5 x + 2 x^{2} - x^{3} .

The following is a direct consequence of the definitions of addition and multiplication in polynomial rings.

Lemma 3.2.10. §

Let $R$ be a ring, and let $f, g \in R [x]$ be polynomials. Then $\deg 𝑓𝑔 \leq \deg f \cdot \deg g$ . Moreover, we have

\deg (f + g) \leq \max {\deg f, \deg g},

and equality holds in the last statement if $\deg f \neq \deg g$ .

Definition 3.2.11. §

The polynomials $a = a + 0 \cdot x + 0 \cdot x^{2} + \dots$ for $a \in R$ are referred to as constant polynomials. These are exactly $0$ and the polynomials of degree $0$ . The set of constant polynomials forms a subring of $R [x]$ , which we also denote $R$ .

We leave it to the reader to check the following.

Lemma 3.2.12. §

Let $R$ be a ring. Then the polynomial ring $R [x]$ is in fact a ring.

Remark 3.2.13. §

The ring $R [x]$ is commutative if and only if $R$ is commutative. The $1$ is a multiplicative identity in $R$ , then $1$ is a multiplicative identity in $R [x]$ as well.

We may also consider polynomial rings in several variables.

Definition 3.2.14. §

Let $n \geq 1$ and $x_{1}, x_{2}, \dots, x_{n}$ be indeterminates. The polynomial ring in $n$ variables over a ring $R$ is defined to be

R [x_{1}, x_{2}, \dots, x_{n}] = (((R [x_{1}]) [x_{2}]) \dots) [x_{n}] .

We write an element of this ring as

\sum_{i_{1} = 0}^{N_{1}} \sum_{i_{2} = 0}^{N_{2}} \dots \sum_{i_{n} = 0}^{N_{n}} a_{i_{1} i_{2} \dots i_{N}} x_{1}^{i_{1}} x_{2}^{i_{2}} \dots x_{n}^{i_{n}},

where the coefficients lie in $R$ . The elements $x_{1}^{i_{1}} x_{2}^{i_{2}} \dots x_{n}^{i_{n}}$ are called monomials.

We will see below that this construction is independent, up to isomorphism, of the ordering of the variables.

Remark 3.2.15. §

In multiplying in $R [x_{1}, x_{2}, \dots, x_{n}]$ , the variables $x_{i}$ all commute with each other and the elements of $R$ . A quantity such as $x_{2} x_{1} x_{2}$ equals $x_{1} x_{2}^{2}$ .

Example 3.2.16. §

In the ring $ℤ [x, y]$ , we have polynomials like $x^{2} + 2 𝑥𝑦$ and $1 - x + y$ , and we have

(x^{2} + 2 𝑥𝑦) (1 - x + y) = x^{2} + 2 𝑥𝑦 - x^{3} + x^{2} y + 2 x y^{2} .

Finally, we consider direct products.

Definition 3.2.17. §

Let $I$ be an indexing set, and let ${R_{i} ∣ i \in I}$ be a nonempty collection of rings. Then the direct product $\prod_{i \in I} R_{i}$ of the $R_{i}$ over $i \in I$ is the binary structure is the direct product of the sets $R_{i}$ together with the binary operations of coordinate-wise addition and multiplication. If $I = X_{n} = {1, 2, \dots, n}$ , we write

\prod_{i \in I} R_{i} = R_{1} \times R_{2} \times \dots \times R_{n} .

That the direct product of rings is a ring is a simple consequence of its definition, and we state it without proof.

Lemma 3.2.18. §

Any direct product of rings is a ring.

Remarks 3.2.19. §

Let ${R_{i} ∣ i \in I}$ be a nonempty collection of rings, and set $R = \prod_{i \in I} R_{i}$ .

a.: The ring $R$ is commutative if and only if each $R_{i}$ is commutative.
b.: The zero element of $R$ is the element ${(0)}_{i \in I}$ .
c.: The element ${(1)}_{i \in I}$ is the multiplicative identity in $R$ .
d.: The element $e_{i}$ which is $0$ in every coordinate but the $i$ th, where it is $1$ , satisfies $e_{i}^{2} = e_{i}$ , but $e_{i}$ is not the multiplicative identity of $R$ (unless $I$ has only one element).

Example 3.2.20. §

If $R$ is any ring, then $R^{n}$ is the product of $n$ copies of $R$ .

3.3. Units

Not all rings are fields, but one can still ask which elements are invertible under multiplication. These elements are known as units.

Definition 3.3.1. §

A unit in a ring is a nonzero element $u \in R$ such that $u$ has a multiplicative inverse in $R$ . We also say that $u$ is invertible.

Examples 3.3.2. §

a.: The element $1$ is a unit in every nonzero ring.
b.: The units in a field $F$ are the elements of $F^{\times}$ .
c.: The only units in $ℤ$ are $1$ and $- 1$ .

Proposition 3.3.3. §

The units in a nonzero ring $R$ form a group under multiplication.

Proof.

Let $R^{\times}$ denote the set of units in $R$ . If $u, v \in R^{\times}$ , then let $u^{'}, v^{'} \in R^{\times}$ be multiplicative inverses to $u$ and $v$ respectively. We have

𝑢𝑣 \cdot (v^{'} u^{'}) = 1 = (v^{'} u^{'}) \cdot 𝑢𝑣,

so multiplication is a binary operation on $R^{\times}$ , which we already know to be associative. Clearly, $1$ is a unit and an identity in $R^{\times}$ , and by definition, every unit has an inverse in $R^{\times}$ , so $R^{\times}$ is a group. □

Definition 3.3.4. §

The group of units in a nonzero ring $R$ is denoted $R^{\times}$ .

Remark 3.3.5. §

If $F$ is a field, then its unit group and its multiplicative group coincide, and hence the notation $F^{\times}$ for both is unambiguous.

Example 3.3.6. §

The group of units in $M_{n} (R)$ for a ring $R$ is its subset ${GL}_{n} (R)$ of invertible matrices. E.g., if $R = ℝ$ , then these are the matrices with nonzero determinant.

Example 3.3.7. §

If $R = \prod_{i \in I} R_{i}$ is a direct product of rings $R_{i}$ over an indexing set $I$ , then

R^{\times} = \prod_{i \in I} R_{i}^{\times} .

Proposition 3.3.8. §

The units in $ℤ ∕ 𝑛ℤ$ for $n \geq 1$ are exactly the images of those $i \in ℤ$ relatively prime to $n$ .

Proof.

Let $i \in ℤ$ . By Proposition 2.3.14, we have $⟨ i ⟩ = ⟨ \gcd (i, n) ⟩$ as subgroups of $ℤ ∕ 𝑛ℤ$ . The set of $𝑖𝑗$ with $j \in ℤ ∕ 𝑛ℤ$ are exactly the elements of $⟨ i ⟩$ . Therefore, $i$ is a unit in $ℤ ∕ 𝑛ℤ$ if and only if $1$ is an integer multiple of $\gcd (i, n)$ in $ℤ ∕ 𝑛ℤ$ . Since $\gcd (i, n)$ is a divisor of $n$ , this can and will only happen if $\gcd (i, n) = 1$ , which is to say that $i$ is relatively prime to $n$ . □

Corollary 3.3.9. §

The group ${(ℤ ∕ 𝑛ℤ)}^{\times}$ has order $ϕ (n)$ , where $ϕ$ is the Euler $ϕ$ -function.

Corollary 3.3.10. §

For $n \geq 1$ , the ring $ℤ ∕ 𝑛ℤ$ is a field if and only if $n$ is prime.

We now have the following corollaries by the corollary of Lagrange’s theorem that the order of an element of a group divides the order of the group. What is remarkable is that they are nonobvious statements of simple arithmetic.

Corollary 3.3.11 (Euler’s theorem). §

Let $n \geq 1$ . Then

a^{ϕ (n)} \equiv 1 mod n

for every $a \in ℤ$ relatively prime to $n$ .

Note that every nonzero element of $ℤ ∕ 𝑝ℤ$ is relatively prime to $p$ . Hence we also also have the following special case of Euler’s theorem.

Corollary 3.3.12 (Fermat’s little theorem). §

Let $p$ be a prime number. Then

a^{p - 1} \equiv 1 mod p

for every $a \in ℤ$ not divisible by $p$ .

These raise the following questions. What is the order of a unit in $ℤ ∕ 𝑛ℤ$ ? We know it to be a divisor of $ϕ (n)$ , but is there a simple formula for it in terms of $a$ and $n$ ? This is one of many questions in the field of mathematics known as number theory. Let us give a few examples of arithmetic in $ℤ ∕ 𝑛ℤ$ .

Example 3.3.13. §

Suppose we wish to calculate $3^{362}$ in $ℤ ∕ 11 ℤ$ . Fermat’s little theorem tells us that $3^{10} \equiv 1 mod 11$ , so

3^{362} \equiv {(3^{10})}^{36} 3^{2} \equiv 3^{2} \equiv 9 mod 11 .

In other words, $3^{10} = 9$ in $ℤ ∕ 11 ℤ$ .

Example 3.3.14. §

What is the order of $2$ in ${(ℤ ∕ 101 ℤ)}^{\times}$ ? Since $101$ is prime, the order of $2$ must be a divisor of $100$ . We have $2^{5} < 101$ , and $2^{10} = 1024 \equiv 14 mod 101$ . Moreover, we have

\begin{array}{rcl} 2^{20} \equiv {(14)}^{2} \equiv 196 \equiv - 6 mod 101, \\ 2^{25} = 2^{20} 2^{5} \equiv - 6 \cdot 32 \equiv - 192 \equiv 10 mod 101, \\ 2^{50} \equiv 1 0^{2} \equiv - 1 mod 101 . \end{array}

Therefore, the order of $2$ in $ℤ ∕ 101 ℤ$ must be $100$ .

3.4. Integral domains

Definition 3.4.1. §

A left (resp., right) zero divisor in a ring $R$ is a nonzero element $a \in R$ such that there exists a nonzero element $b \in R$ with $𝑎𝑏 = 0$ (resp., $𝑏𝑎 = 0$ ). A zero divisor in a ring $R$ is an element that is either a left or a right zero divisor.

Remark 3.4.2. §

Note that $0$ is never considered to be a zero divisor (at least under our conventions). In fact, $1$ is never a zero divisor either, as $1 \cdot b = b$ for all $b \in R$ .

Example 3.4.3. §

The ring $M_{2} (ℝ)$ has zero divisors. For instance, we have

(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}) \cdot (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}) = (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}) \cdot (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}) = 0 .

Example 3.4.4. §

If $R = R_{1} \times R_{2}$ for some nonzero rings $R_{1}$ and $R_{2}$ , then $R$ has zero divisors, since if $a \in R_{1}$ is nonzero and $b \in R_{2}$ is nonzero, we have $(a, 0) \cdot (0, b) = (0, b) \cdot (a, 0) = 0$ . For instance, $ℤ^{n}$ has zero divisors for $n \geq 2$ , though $ℤ$ does not.

One might ask for a ring that contains a left zero divisor that is not a right zero divisor. For this, let us make the following general definition.

Definition 3.4.5. §

Let $A$ be an abelian group under addition. The endomorphism ring of $A$ is the set

End (A) = {f : A \to A ∣ f is a group homomorphism}

under addition and composition of functions.

Remark 3.4.6. §

If $A$ is an abelian group, then $End (A)$ is a ring, with $1$ being the identity function on $A$ . In general, $End (A)$ may be a noncommutative ring.

Example 3.4.7. §

Let $A = \prod_{i = 1}^{\infty} ℤ$ , an abelian group under addition. Define $L, R \in End (A)$ by

\begin{array}{rcl} L (a_{1}, a_{2}, a_{3}, \dots) = (a_{2}, a_{3}, a_{4}, \dots) & and & R (a_{1}, a_{2}, a_{3}, \dots) = (0, a_{1}, a_{2}, \dots) . \end{array}

Moreover, let $M \in End (A)$ be defined by

M (a_{1}, a_{2}, a_{3}, \dots) = (a_{1}, 0, 0, \dots) .

Then

\begin{array}{rcl} 𝐿𝑀 (a_{1}, a_{2}, a_{3}, \dots) = L (a_{1}, 0, 0, \dots) = 0 & and & 𝑀𝑅 (a_{1}, a_{2}, a_{3}, \dots) = M (0, a_{1}, a_{2}, \dots) = 0, \end{array}

so $L$ is a left zero divisor and $R$ is a right zero divisor. On the other hand,

𝐿𝑅 (a_{1}, a_{2}, a_{3}, \dots) = L (0, a_{1}, a_{2}, \dots) = (a_{1}, a_{2}, a_{3}, \dots),

so $𝐿𝑅 = 1$ . Therefore, $L$ cannot be a right zero divisor, for if $𝑋𝐿 = 0$ for some $X \in End (A)$ , then $0 = (𝑋𝐿) R = X (𝐿𝑅) = X$ . Similarly, $R$ is not a left zero divisor.

Example 3.4.8. §

In the ring $ℤ ∕ 6 ℤ$ , the elements $2$ , $3$ , and $4$ are zero divisors, since $2 \cdot 3 = 3 \cdot 4 = 0$ .

More generally, we have the following.

Lemma 3.4.9. §

For $n \geq 1$ , the zero divisors in $ℤ ∕ 𝑛ℤ$ are exactly its nonzero elements that are not relatively prime to $n$ .

Proof.

Let $\bar{a} \in ℤ ∕ 𝑛ℤ$ be nonzero, and let $b = n ∕ \gcd (a, n)$ . Then $\bar{a} \bar{b} = 0$ , and we know that $\bar{b} \neq 0$ if and only if $\gcd (a, n) \neq 1$ . On the other hand, if $\bar{a} \bar{b} = 0$ , then $𝑎𝑏$ is a multiple of $n$ , so $b$ is a multiple of $n ∕ \gcd (a, n)$ . Therefore, $a$ is a zero divisor if and only if $\gcd (a, n) \neq 1$ , which occurs if and only if $a$ is not relatively prime to $n$ □

As a corollary, if $p$ is a prime number, then $ℤ ∕ 𝑝ℤ$ has no zero divisors. In fact, we shall see momentarily that every field has no zero divisors.

Definition 3.4.10. §

A nonzero commutative ring $R$ is called an integral domain if $R$ contains no zero divisors.

Lemma 3.4.11. §

Every field is an integral domain.

Proof.

Let $F$ be a field, and let $a \in F$ be such that there exists a nonzero element $b \in F$ with $𝑎𝑏 = 0$ . Then $0 = (𝑎𝑏) b^{- 1} = a$ . □

By definition, any subring of an integral domain is also an integral domain.

Examples 3.4.12. §

The fields $ℚ$ , $ℝ$ , $ℂ$ , and $ℤ ∕ 𝑝ℤ$ for any prime $p$ are all integral domains. That $ℤ$ is an integral domain is either an easy check or the fact that it is a subring of $ℚ$ . Since $ℤ ∕ 𝑛ℤ$ contains zero divisors for composite $n \geq 1$ , it is not an integral domain.

Proposition 3.4.13. §

Let $R$ be an integral domain. Then $R [x]$ is an integral domain. Moreover, if $f, g \in R [x]$ are nonzero, then $\deg 𝑓𝑔 = \deg f + \deg g$ , and the units in $R [x]$ are exactly the units in $R$ .

Proof.

Let $f, g \in R [x]$ be nonzero polynomials of degree $N$ and $M$ respectively. Write $f = \sum_{i = 0}^{N} a_{i} x^{i}$ and $g = \sum_{j = 0}^{M} b_{j} x^{j}$ . Then

𝑓𝑔 = \sum_{k = 0}^{N + M} c_{k} x^{k}, c_{k} = \sum_{i = 0}^{k} a_{i} b_{k - i} .

If $0 \leq i \leq N + M$ , then $a_{i} = 0$ if $i > N$ and $b_{M + N - i} = 0$ if $i < N$ , so $c_{N + M} = a_{N} b_{M}$ . Since $R$ is an integral domain, we then have $c_{N + M} \neq 0$ , so $𝑓𝑔 \neq 0$ . Therefore, we have $\deg 𝑓𝑔 = N + M$ . If $𝑓𝑔 = 1$ , then this forces $N = M = 0$ , and therefore $f = a_{0}$ , $g = b_{0}$ , and $a_{0} b_{0} = 1$ , which means that $f \in R^{\times}$ . □

One particularly nice use of integral domains is that they obey cancellation laws.

Lemma 3.4.14. §

Let $R$ be an integral domain, and let $a, b, c \in R$ be such that $𝑎𝑏 = 𝑎𝑐$ . Then either $a = 0$ or $b = c$ .

Proof.

If $𝑎𝑏 = 𝑎𝑐$ , then $a (b - c) = 0$ by the distributive law (and Lemma 3.1.8), so as $R$ contains no zero divisors, at least one of $a$ and $b - c$ must be $0$ . □

We have already seen that $ℤ ∕ 𝑛ℤ$ is an integral domain if and only if $n$ is prime, and so if and only if $ℤ ∕ 𝑛ℤ$ is a field. We have the following stronger result.

Theorem 3.4.15. §

If $R$ is a finite integral domain, then $R$ is a field.

Proof.

Let $a \in R$ be nonzero. Lemma 3.4.14 tells us that the elements $𝑎𝑏$ with $b \in R$ are all distinct. Since there are then $| R |$ of them, the set ${𝑎𝑏 ∣ b \in R}$ is $R$ itself. In particular, there exists $b \in R$ with $𝑎𝑏 = 1$ , proving that $a$ has a multiplicative inverse. □

Finally, we introduce the notion the characteristic of a ring.

Definition 3.4.16. §

Let $R$ be a ring. The characteristic $char (R)$ of $R$ is the smallest $n \geq 1$ such that $𝑛𝑎 = 0$ for all $a \in R$ if such an $n$ exists, and otherwise we set $char (R) = 0$ .

Examples 3.4.17. §

a.: The ring $ℤ ∕ 𝑛ℤ$ has characteristic $n$ , while $ℤ$ , $ℚ$ , $ℝ$ , and $ℂ$ all have characteristic $0$ .
b.: The characteristic of $M_{n} (R)$ for a ring $R$ is equal to the characteristic of $R$ for every $n \geq 1$ .

Lemma 3.4.18. §

The characteristic of a nonzero ring $R$ is the smallest $n > 1$ such that $n = 0$ in $R$ if such an $n$ exists, and is $0$ otherwise.

Proof.

We cannot have $1 \cdot a = 0$ unless $a = 0$ , so $char (R) \neq 1$ as $R$ is nonzero. Recall that $n \in R$ is considered to be $n \cdot 1$ . If $n = 0$ in $R$ , then clearly $𝑛𝑎 = 0$ for all $a \in R$ . On the other hand, that $n = 0$ is the special case of $𝑛𝑎 = 0$ with $a = 1$ . If $n = n \cdot 1 \neq 0$ for all $n \geq 1$ , then by definition, we have $char (R) = 0$ . □

Proposition 3.4.19. §

The characteristic of an integral domain is either $0$ or prime.

Proof.

We employ Lemma 3.4.18. If $R$ is an integral domain and $n = 0$ in $R$ for some composite $n > 1$ , then $n = m m^{'} = 0$ for some prime $m$ and $m^{'}$ dividing $n$ , which by the nonexistence of zero divisors implies that either $m$ or $m^{'}$ is zero. In other words, the smallest $n > 1$ with $n \neq 0$ in $R$ cannot be composite, so must be prime. □

3.5. Ring homomorphisms

In this section, we introduce the notion of a ring homomorphism, which is a function from one ring to another that is compatible with both addition and multiplication: in other words, it is a homomorphism of binary structures both for $(R, +)$ and for $(R, \cdot)$ .

Definition 3.5.1. §

Let $R$ and $Q$ be rings. A function $ϕ : R \to Q$ is a ring homomorphism if $ϕ (1) = 1$ and it satisfies

\begin{array}{rcl} ϕ (a + b) = ϕ (a) + ϕ (b) & and & ϕ (𝑎𝑏) = ϕ (a) ϕ (b) \end{array}

for all $a, b \in R$ .

We give some examples of ring homomorphisms.

Examples 3.5.2. §

a.: The reduction map $ϕ_{n} : ℤ \to ℤ ∕ 𝑛ℤ$ with $ϕ_{n} (a) = a$ is a surjective ring homomorphism.
b.: The multiplication-by- $n$ map $ψ_{n} : ℤ \to ℤ$ with $ψ_{n} (a) = 𝑛𝑎$ is not a ring homomorphism unless $n = 1$ .

Here are several standard ring homomorphisms.

Definition 3.5.3. §

Let $R$ be a ring.

a.: The identity homomorphism ${id}_{R} : R \to R$ is the ring homomorphism given by ${id}_{R} (a) = a$ for all $a \in R$ .
b.: If $R$ is a subring of a ring $Q$ , we have the inclusion map $ι_{R} : R \to Q$ with $ι_{R} (a) = a$ for all $a \in R$ .

An inclusion map is always injective, but will only be surjective if the subring is the whole ring. Here are some other examples.

Examples 3.5.4. §

Let $R$ be a nonzero ring.

a.: There is an injective ring homomorphism $ι : R \to R [x]$ that sends $a \in R$ to the constant polynomial $a \in R [x]$ .
b.: There is a surjective ring homomorphism $π : R [x] \to R$ that sends $f \in R [x]$ to its constant coefficient. Note that $π \circ ι = {id}_{R}$ , but $ι \circ π \neq {id}_{R}$ .

We mention another useful class of ring homomorphisms of polynomial rings, arising from maps on coefficients.

Examples 3.5.5. §

Let $R$ and $Q$ be rings, and let $ϕ : R \to Q$ be a ring homomorphism. This induces maps on polynomial rings and matrix rings, as follows.

a.

The map $\tilde{ϕ} : R [x] \to Q [x]$ induced by $ϕ$ on polynomial rings is given by the formula

\tilde{ϕ} (\sum_{i = 0}^{N} a_{i} x^{i}) = \sum_{i = 0}^{N} ϕ (a_{i}) x^{i}

for $a_{i} \in R$ for $0 \leq i \leq N$ for some $N \geq 0$ .

b.

The map $\tilde{ϕ} : M_{n} (R) \to M_{n} (Q)$ induced by $ϕ$ on matrix rings is given by the formula

\tilde{ϕ} ((a_{𝑖𝑗})) = (ϕ (a_{𝑖𝑗}))

for $(a_{𝑖𝑗}) \in M_{n} (R)$ .

Remark 3.5.6. §

If $R$ is a subring of $Q$ , then we may use the map of polynomial rings induced by the inclusion map of $R$ into $Q$ to view $R [x]$ as a subring of $Q [x]$ .

Remark 3.5.7. §

The product of ring homomorphisms $ϕ_{i} : R_{i} \to Q_{i}$ over an index set $i \in I$ is a ring homomorphism between the corresponding products.

Lemma 3.5.8. §

Let $R$ be a ring and $Q$ be an integral domain, and let $ϕ : R \to Q$ be a nonzero homomorphism. If $u \in R^{\times}$ , then $ϕ (u) \in Q^{\times}$ .

Proof.

Let $v$ be a multiplicative inverse to $u$ in $R$ . By the previous lemma

ϕ (u) ϕ (v) = ϕ (𝑢𝑣) = ϕ (1) = 1,

and, similarly, we have $ϕ (v) ϕ (u) = ϕ (𝑣𝑢) = 1$ . □

We also have the following.

Definition 3.5.9. §

If $R = \prod_{i \in I} R_{i}$ is a product of rings, then there are projection maps

π_{i} : R \to R_{i}, π_{i} ({(a_{i})}_{i \in I}) = a_{i}

which are ring homomorphisms.

Remark 3.5.10. §

If $R = \prod_{i \in I} R_{i}$ is a product of rings, the inclusion maps $ι_{i} : R_{i} \to R$ for $i \in I$ given by taking $a \in R_{i}$ to the element with $i$ th coordinate $a$ and $j$ th coordinate $0$ for $j \neq i$ are not ring homomorphisms if at least two $R_{i}$ are nonzero rings, since $ι (1) \neq 1 \in R$ .

As with group homomorphisms, we have notions of kernel and image of a ring homomorphism.

Definition 3.5.11. §

Let $ϕ : R \to Q$ be a ring homomorphism. Then the kernel of $ϕ$ is

\ker ϕ = {r \in R ∣ ϕ (r) = 0},

and the image of $ϕ$ is

im ϕ = {ϕ (r) ∣ r \in R} .

One can check very easily that $im ϕ$ is a subring of $Q$ for any ring homomorphism $ϕ : R \to Q$ . However, while $\ker ϕ$ is a subgroup of $R$ closed under multiplication, it will not contain $1$ unless $ϕ = 0$ .

Examples 3.5.12. §

Let $R$ be a ring. We consider the homomorphisms of Example 3.5.4.

a.: The inclusion $ι : R \to R [x]$ has $\ker ι = 0$ and $im ι$ the subring of constant polynomials in $R [x]$ , which we also denote $R$ .
b.: The projection $π : R [x] \to R$ has $im ι = R$ and kernel consisting of the polynomials with $0$ constant coefficient, which is the to say, the multiples of $x$ .

Note that since any ring homomorphism is, in particular, a homomorphism of abelian groups under addition, we have the following.

Lemma 3.5.13. §

A ring homomorphism $ϕ : R \to Q$ is injective if and only if $\ker ϕ = {0}$ .

We will have much more to say about kernels later. For now, let us finish with a corollary for fields.

Lemma 3.5.14. §

Let $ϕ : F \to F^{'}$ be a ring homomorphism, where $F$ and $F^{'}$ are fields. Then $ϕ$ is injective and $ϕ {(x)}^{- 1} = ϕ (x^{- 1})$ for all $x \neq 0$ .

Proof.

For any $x \in F^{\times}$ , we have

ϕ (1) = ϕ (x \cdot x^{- 1}) = ϕ (x) \cdot ϕ (x^{- 1}),

so $ϕ (x)$ is nonzero and has multiplicative inverse $ϕ (x^{- 1})$ . In particular, Lemma 3.5.13 tells us that $ϕ$ is injective. □

As usual, we can speak about injective and surjective ring homomorphisms, as well as isomorphisms.

Definition 3.5.15. §

A ring homomorphism $ϕ : R \to Q$ is an isomorphism if it is bijective.

For instance, let us check that a polynomial ring in two variables is independent of the ordering of the variables, up to an isomorphism. We leave it to the reader to treat the case of more than two variables using the following lemma and the construction in Example 1a.

Lemma 3.5.16. §

Let $x$ and $y$ be indeterminates. The map $σ : (R [x]) [y] \to (R [y]) [x]$ satisfying

σ (\sum_{j = 0}^{M} (\sum_{i = 0}^{N} a_{𝑖𝑗} x^{i}) y^{j}) = \sum_{i = 0}^{N} (\sum_{j = 0}^{M} a_{𝑖𝑗} y^{j}) x^{i},

(3.5.1)

where the $a_{𝑖𝑗}$ are elements of $R$ . is an isomorphism.

Proof.

Note that every element of $(R [x]) [y]$ may be expressed in the form on the left of (3.5.1), since a polynomial in $y$ with coefficients in $R [x]$ has finite degree (at most $M$ ), and each of the finitely many nonzero coefficients then has a degree, and we choose $N$ to be at least the maximum of these degrees. Similarly, every element of $(R [y]) [x]$ may be written in the form on the right of (3.5.1), so the map is onto. By definition, it is one-to-one, and we leave it to the reader to check that it is a ring homomorphism. □

As usual, the inverse of an isomorphism of rings is an isomorphism of rings.

3.6. Subrings generated by elements

Definition 3.6.1. §

Let $R$ be a subring of a ring $Q$ , and let $X$ be a set of elements of $Q$ . The subring of $Q$ generated over $R$ by $X$ is the smallest subring of $Q$ containing $R$ and $X$ .

Since the intersection of subrings containing a given set of elements is a subring, Definition 3.6.1 makes sense. When we have a finite set $X$ , we often speak of the subring generated over $R$ by the elements of $X$ , as opposed to $X$ itself. We will only be interested in a special case in which the elements we are adding to the subring commute with every element in that subring. We note the following, which we leave to the reader to verify.

Definition 3.6.2. §

Let $R$ be a subring of a ring $Q$ , and let $α \in Q$ commute with every element of $R$ . The ring given by adjoining $α$ to $R$ is

R [α] = {\sum_{i = 0}^{N} r_{i} α^{i} ∣ r_{i} \in R for all 0 \leq i \leq N for some N \geq 0} .

Remark 3.6.3. §

We often read $R [α]$ as “ $R$ adjoin $α$ .”

We leave it to the reader to check the following.

Lemma 3.6.4. §

Let $R$ be a subring of a ring $Q$ , and let $α \in Q$ commute with every element of $R$ . The $R [α]$ is the subring generated over $R$ by $α$ .

Definition 3.6.5. §

Let $R$ be a subring of $Q$ . If $α_{1}, α_{2}, \dots, α_{n} \in Q$ commute with each other and every element of $R$ , we set

R [α_{1}, α_{2} \dots, α_{n}] = (((R [α_{1}]) [α_{2}]) \dots) [α_{n}] .

Remark 3.6.6. §

The ring $R [α_{1}, \dots, α_{n}]$ in Definition 3.6.5 is the smallest subring of $Q$ containing $R$ and each $α_{i}$ , so generated over $R$ by the $α_{i}$ .

Examples 3.6.7. §

a.

The ring $ℤ [i] = {a + 𝑏𝑖 ∣ a, b \in ℤ}$ is a subring of $ℂ$ known as the Gaussian integers. Note that since $i^{2} = - 1$ , it is unnecessary to consider polynomials of higher degree.

b.

The ring

ℤ [\sqrt[n]{2}] = {\sum_{i = 0}^{n - 1} a_{i} \sqrt[n]{2} ∣ a_{i} \in ℤ, 0 \leq i \leq n - 1}

is a subring of $ℝ$ .

c.

The ring

ℤ [i, \sqrt{2}] = {a + 𝑏𝑖 + c \sqrt{2} + d \sqrt{2} i ∣ a, b, c, d \in ℤ}

is a subring of $ℂ$ .

d.

The ring $ℚ [x^{2}]$ is a subring of $ℚ [x]$ consisting of polynomials of the form

\sum_{i = 0}^{N} a_{i} x^{2 i}

with every $a_{i} \in ℚ$ .

We may relate this to the evaluation of polynomial rings at ring elements.

Definition 3.6.8. §

Let $R$ be a subring of a ring $Q$ , and let $α \in Q$ commute with every element of $R$ . For $f = \sum_{i = 0}^{N} c_{i} x^{i} \in R [x]$ , we define the value of $f$ at $α \in R$ to be

f (α) = \sum_{i = 0}^{N} c_{i} α^{i} .

For any $a \in R$ , the evaluation-at- $α$ map is defined by

e_{α} : R [x] \to R [α], e_{α} (f) = f (α)

for all $f \in R [x]$ .

The following is a result of the definitions of addition and multiplication in $R [x]$ .

Lemma 3.6.9. §

Let $R$ be a subring of a ring $Q$ , and let $α \in Q$ commute with every element of $R$ . The evaluation-at- $α$ map $e_{α} : R [x] \to R [α]$ is a ring homomorphism.

Proof.

Let $f = \sum_{i = 0}^{N} c_{i} x^{i} \in R [x]$ , and let $g = \sum_{i = 0}^{M} d_{i} x^{i} \in R [x]$ for some $n \geq 0$ . Then we have

e_{α} (f + g) = (f + g) (α) = \sum_{i = 0}^{D} (c_{i} + d_{i}) α^{i} = \sum_{i = 0}^{D} c_{i} α^{i} + \sum_{i = 0}^{D} d_{i} α^{i} = e_{α} (f) + e_{α} (g),

where $D = \max {M, N}$ , and

e_{α} (𝑓𝑔) = \sum_{k = 0}^{M + N} (\sum_{i = 0}^{k} c_{i} d_{k - i}) α^{k} .

Since $α$ commutes with every element of $R$ , we have $c_{i} d_{k - i} α^{k} = c_{i} α^{i} d_{k - i} α^{k - i}$ for all $i \leq k$ , so the latter term equals

(\sum_{i = 0}^{N} c_{i} α^{i}) \cdot (\sum_{j = 0}^{M} d_{j} α^{j}) = e_{α} (f) \cdot e_{α} (g) .

□

Remark 3.6.10. §

The evaluation-at-zero map is none other than the ring homomorphism constructed in Example 3.5.4a that takes a polynomial to its constant term.

Example 3.6.11. §

If $X$ is a set and $R$ is a ring, then the set $Maps (X, R)$ of functions from $X$ to $R$ forms a ring under the usual operations of pointwise addition and multiplication on $R$ . Given $a \in X$ , we again have an evaluation-at- $a$ map

𝜖_{a} : Maps (X, R) \to R,

given by $𝜖_{a} (f) = f (a)$ for $f \in Maps (X, R)$ and $a \in X$ , which is a ring homomorphism.

Example 3.6.12. §

Let $R$ be a commutative ring, and let $a \in R$ . The evaluation map $e_{a}$ on $R [x]$ can be viewed as the composition $𝜖_{a} \circ κ$ , where

κ : R [x] \to Maps (R, R), κ (f) (a) = f (a)

for $f \in R [x]$ and $a \in R$ . In other words, $κ$ takes a polynomial to the function it defines. It is a ring homomorphism since $R$ is commutative.

Note that even if $R$ is commutative, $κ$ is not always injective. For instance, if $R = ℤ ∕ 𝑝ℤ$ for a prime number $p$ , then $f = x^{p} - x$ is a nonzero polynomial in $R [x]$ , but $p (a) = 0$ for all $a \in ℤ ∕ 𝑝ℤ$ , so $κ (p) = 0$ .

3.7. Ideals and quotient rings

In this section, we introduce the notion of an ideal of a ring. An ideal plays the role that a normal subgroup does in group theory, which is to say that we can take a quotient of a ring by an ideal and obtain another ring. The issue with simply using a subring can be seen in the following example.

Example 3.7.1. §

Consider the quotient group $ℚ ∕ ℤ$ under addition. The multiplication in $ℚ$ does not induce a well-defined multiplication on $ℤ$ . To see this, note that one would like

(a + ℤ) \cdot (b + ℤ) = (𝑎𝑏 + ℤ)

for any $a, b \in ℚ$ . But then we would have

0 + ℤ = (0 + ℤ) \cdot (\frac{1}{2} + ℤ) = (1 + ℤ) \cdot (\frac{1}{2} + ℤ) = \frac{1}{2} + ℤ,

which is clearly not the case.

To fix this, we introduce the notion of an ideal. We begin with left and right ideals.

Definition 3.7.2. §

A subset $I$ of a ring $R$ that is a subgroup under addition is called a left (resp., right) ideal if $R \cdot I \subseteq I$ (resp., $I \cdot R \subseteq I$ ).

Definition 3.7.3. §

A two-sided ideal, or more simply, an ideal, of a ring $R$ is any subset of $R$ that is both a left and a right ideal.

In other words, a left ideal $J$ of $R$ is an additive subgroup for which $r \cdot b \in J$ for all $r \in R$ and $b \in J$ , and a right ideal $K$ is one for which $c \cdot r \in K$ for all $r \in R$ and $c \in K$ . An ideal $I$ of $R$ is an additive subgroup for which both $r \cdot a \in I$ and $a \cdot r \in I$ for all $r \in R$ and $a \in I$ .

Remark 3.7.4. §

Note that $I \subseteq R \cdot I$ , so the condition that $R \cdot I \subseteq I$ (resp., $I \cdot R \subseteq I$ ) amounts to $R \cdot I = I$ (resp., $I \cdot R = I$ ).

In fact, we have the following simple criterion for a nonempty subset to be an ideal.

Lemma 3.7.5. §

Let $R$ be a ring, and let $I$ be a nonempty subset of $R$ . Then $I$ is a left (resp., right ideal) if and only if the following hold:

i.: $I$ is closed under addition: if $a, b \in I$ , then $a + b \in I$ , and
ii.: $I$ is closed under left (resp., right) multiplication by elements of $R$ : if $r \in R$ and $a \in I$ , then $𝑟𝑎 \in I$ (resp., $𝑎𝑟 \in I$ ).

Proof.

We need only see that a set $I$ satisfying (i) and (ii) is a subgroup. For this, we must show that it contains $0$ , which it does since $0 = 0 \cdot a$ for any $a \in I$ , and that it contains additive inverses, which it does since $- a = - 1 \cdot a$ for any $a \in I$ . □

Remark 3.7.6. §

Every left and every right ideal in a commutative ring $R$ is an ideal of $R$ .

Examples 3.7.7. §

a.: The subset $𝑛ℤ$ of $ℤ$ is an ideal of $ℤ$ for each $n \in ℤ$ . That is, any integer multiple of an integer multiple of $n$ is an integer multiple of $n$ .
b.: The subset $ℤ$ of $ℚ$ is not an ideal, as $1 ∕ 2 \cdot ℤ ⊈ ℤ$ , for instance.
c.: Let $R$ be a nonzero ring. Consider the set of matrices in $M_{n} (R)$ that are $0$ in all entries outside their first columns. This is a left ideal of $M_{n} (R)$ , but it is not a right ideal for $n \geq 2$ . Similarly, the set of matrices in $M_{n} (R)$ that are $0$ in all entries outside their first rows is a right ideal of $M_{n} (R)$ .
d.: Let $R$ be a ring. The set of all polynomials with zero constant coefficient is an ideal of $R$ , equal to the set of multiples of $x$ in $R [x]$ .

Definition 3.7.8. §

The zero ideal of a ring $R$ is the subset ${0}$ . The improper ideal of $R$ is the ring $R$ itself. An ideal is said to be nonzero if it is not equal to zero, and an ideal is said to be proper if it is not equal to $R$ .

We note the following.

Lemma 3.7.9. §

Let $R$ be a ring, and let $I$ be a left (or right) ideal of $R$ . Then $I = R$ if and only if $I$ contains a unit, and in particular if and only if $I$ contains $1$ .

Proof.

If $I = R$ , then clearly $I$ contains $1$ and therefore a unit. If $u \in I$ is a unit, then $u^{- 1} \in R$ , so $1 = u^{- 1} \cdot u \in I$ . And if $1 \in I$ , then $a = a \cdot 1 \in I$ for all $a \in R$ . □

The following classifies, as a special case, all ideals in a field.

Corollary 3.7.10. §

The only left and only right ideals in a division ring are ${0}$ and $D$ .

Proof.

If $I$ is a nonzero left or right ideal of $D$ , it then contains a unit, so is $D$ . □

We shall see later that the converse to Corollary 3.7.10 also holds. We give one more example.

Lemma 3.7.11. §

Let $R$ and $Q$ be rings. Then any left ideal of $R \times Q$ has the form $I \times J$ , where $I$ is a left ideal of $R$ and $J$ is a left ideal of $Q$ .

Proof.

Let $K$ be an ideal of $R \times Q$ . Let

\begin{array}{rcl} I = {a \in R ∣ (a, 0) \in K} & and & J = {b \in Q ∣ (0, b) \in K}, \end{array}

which are left ideals of $R$ and of $Q$ , respectively. If $(a, b) \in K$ , then $(1, 0) \cdot (a, b) = (a, 0)$ , so $a \in I$ and $(1, 0) \cdot (a, b) = (0, b)$ , so $b \in J$ . Therefore, $K \subseteq I \times J$ . Conversely, if $(a, b) \in I \times J$ , then $(a, 0) \in K$ and $(0, b) \in K$ , so $(a, b) = (a, 0) + (0, b) \in K$ , so $I \times J \subseteq K$ . □

The following is the ring-theoretic analogue of Proposition 2.12.11.

Proposition 3.7.12. §

Let $ϕ : R \to Q$ be a homomorphism of rings. Then $\ker ϕ$ is an ideal of $R$ .

Proof.

We know from Proposition 2.8.8 that $\ker ϕ$ is a subgroup of $R$ under addition. Moreover, if $r \in R$ and $a \in \ker ϕ$ , then

ϕ (𝑟𝑎) = ϕ (r) ϕ (a) = ϕ (r) \cdot 0 = 0,

so $𝑟𝑎 \in \ker ϕ$ . Similarly, we have $ϕ (𝑎𝑟) = 0$ , so $𝑎𝑟 \in \ker ϕ$ as well. □

We may now construct the analogue of a quotient group, known as a quotient ring.

Theorem 3.7.13. §

Let $R$ be a ring, and let $I$ be a two-sided ideal of $R$ . Then the quotient group $R ∕ I$ has a well-defined multiplication on it, given by

(r + I) \cdot (s + I) = (𝑟𝑠 + I)

for $r, s \in R$ . Moreover, with the usual addition of cosets and this multiplication, $R ∕ I$ becomes a ring.

Proof.

Suppose that $a, a^{'}, b, b^{'} \in R$ with $a + I = a^{'} + I$ and $b + I = b^{'} + I$ . Then there exist $x, y \in I$ with $a^{'} = a + x$ and $b^{'} = b + y$ . We have

(a^{'} b^{'} + I) = (a + x) (b + y) + I = 𝑎𝑏 + 𝑎𝑦 + 𝑥𝑏 + 𝑥𝑦 + I = 𝑎𝑏 + I,

since $𝑎𝑦, 𝑥𝑏, 𝑥𝑦 \in I$ in that $I$ is a two-sided ideal. Therefore, the multiplication on $R ∕ I$ is well-defined. That it is associative is a direct consequence of the associativity of multiplication on $R$ . Distributivity is again a consequence of distributivity on $R$ , but we write out the proof of the left distributive law:

\begin{matrix} (a + I) \cdot ((b + I) + (c + I)) = (a + I) \cdot (b + c + I) = a (b + c) + I = (𝑎𝑏 + 𝑎𝑐) + I \\ = (𝑎𝑏 + I) + (𝑎𝑐 + I) = (a + I) \cdot (b + I) + (a + I) \cdot (c + I) . \end{matrix}

□

Definition 3.7.14. §

The quotient of a ring $R$ by an ideal $I$ is the ring $R ∕ I$ defined by Theorem 3.7.13. We say that $R ∕ I$ is the quotient ring of $R$ by $I$ (or the factor ring of $R$ by $I$ ).

Examples 3.7.15. §

a.: The quotient of the ring $ℤ$ by the ideal $𝑛ℤ$ is the ring $ℤ ∕ 𝑛ℤ$ .
b.: The quotient of any ring $R$ by the zero ideal is isomorphic to $R$ . The quotient of any ring $R$ by $R$ is isomorphic to the zero ring.

The following is immediately verified.

Definition 3.7.16. §

The map $π_{I} : R \to R ∕ I$ defined by $π_{I} (a) = a + I$ is called the quotient map from $R$ to $R ∕ I$ .

Remark 3.7.17. §

For $R$ a ring, $I$ and ideal of $R$ , and $a, b \in R$ , we may sometimes write $a \equiv b 𝑚𝑜𝑑 I$ to mean that $a + I = b + I$ , or simply just $a = b$ when it is understood that we are working with the images of $a$ and $b$ under $π_{I}$ , i.e., in the ring $R ∕ I$ .

The following is easily verified.

Lemma 3.7.18. §

Let $I$ be an ideal in a ring $R$ . The quotient map $π_{I} : R \to R ∕ I$ is a surjective ring homomorphism with kernel $I$ .

We have the analogue of the first isomorphism theorem.

Theorem 3.7.19. §

Let $ϕ : R \to Q$ be a homomorphism of rings. Then the map

\bar{ϕ} : R ∕ \ker ϕ \to im ϕ

defined by $\bar{ϕ} (a + \ker ϕ) = ϕ (a)$ for all $a \in R$ is an isomorphism of rings.

Proof.

We know that $\bar{ϕ}$ is an isomorphism of additive groups by Theorem 2.13.11. Let $I = \ker ϕ$ . For $a, b \in R$ , we have

\bar{ϕ} ((a + I) (b + I)) = \bar{ϕ} (𝑎𝑏 + I) = ϕ (𝑎𝑏) = ϕ (a) ϕ (b) = \bar{ϕ} (a + I) \bar{ϕ} (b + I),

so $\bar{ϕ}$ is a ring homomorphism as well, therefore, a ring isomorphism. □

Example 3.7.20. §

The kernel of the homomorphism $π : R [x] \to R$ of Example 3.5.4a is the ideal $I$ of polynomials with zero constant coefficient. In that it is onto, $e_{0}$ induces an isomorphism between $R [x] ∕ I$ and $R$ .

Note that if $ϕ$ is a surjective map, then $ϕ$ carries ideals to ideals.

Proposition 3.7.21. §

Let $ϕ : R \to Q$ be a surjective homomorphism of rings. If $I$ is a left (resp., right) ideal of $R$ , then $ϕ (I)$ is a left (resp., right) ideal of $Q$ .

Proof.

We show this for left ideals $I$ of $R$ . Let $q \in Q$ . Then $q = ϕ (r)$ for some $r \in R$ , and if $a \in I$ , then $𝑠𝜙 (a) = ϕ (𝑟𝑎) \in ϕ (I)$ , so $ϕ (I)$ is a left ideal of $Q$ . □

Remark 3.7.22. §

If $ϕ : R \to Q$ is a ring homomorphism that is not surjective, $ϕ (R)$ is not an ideal of $Q$ , That is, since $ϕ (1) = 1 \in ϕ (R)$ , the ideal generated by $ϕ (R)$ is $Q$ , which is strictly larger than $ϕ (R)$ .

The following is a straightforward generalization of Proposition 3.7.12.

Proposition 3.7.23. §

Let $ϕ : R \to Q$ be a ring homomorphism. Let $J$ be a left (resp., right) ideal of $Q$ . Then $ϕ^{- 1} (J)$ is a left (resp., right) ideal of $R$ .

Proof.

We prove this for left ideals of $Q$ . Let $J$ be a left ideal of $Q$ . If $a, b \in R$ with $ϕ (a), ϕ (b) \in J$ , then $ϕ (a + b) = ϕ (a) + ϕ (b) \in J$ . If moreover $r \in R$ , then $ϕ (𝑟𝑎) = ϕ (r) ϕ (a) \in J$ , so part a holds. □

We can now classify the ideals in quotient rings.

Theorem 3.7.24. §

Let $R$ be a ring, and let $I$ be an ideal of $R$ . Then the quotient map $π_{I} : R \to R ∕ I$ induces a one-to-one correspondence between the left, right, and two-sided ideals of $R$ containing $I$ and the left, right, and two-sided ideals of $R ∕ I$ , respectively.

Proof.

We prove this for left ideals. If $J$ is a left ideal of $R$ containing $I$ , then $π_{I} (J)$ is a left ideal of $R ∕ I$ by Proposition 3.7.21. If $π_{I} (J) = π_{I} (K)$ for some left ideal of $R$ containing $K$ , then any $j \in J$ satisfies $j = k + i$ for some $k \in K$ and $i \in I$ , and therefore $j \in K$ since $I \subseteq K$ . We therefore have $J \subseteq K$ , and similarly $K \subseteq J$ , so $J = K$ . On the other hand, if $N$ is any left ideal of $R ∕ I$ , then $J = π^{- 1} (N)$ is a left ideal of $R$ , and it contains $I$ since $I = π^{- 1} ({0})$ . Since $π_{I} (J) = N$ , we are done. □

For later use, let us make the following definitions for left ideals, with the definitions for right (resp., two-sided) ideals coming from replacing the word “left” by “right” (resp., “two-sided”).

Definition 3.7.25. §

Let $ϕ : R \to Q$ be a ring homomorphism.

a.: For any left ideal $I$ of $R$ , the extension of $I$ by $ϕ$ is the left ideal generated by $ϕ (I)$ .
b.: For any left ideal $J$ of $Q$ , the contraction of $J$ by $ϕ$ is $ϕ^{- 1} (J)$ .
c.: We say that a left ideal of $Q$ is extended by $ϕ$ if it is the extension of some left ideal of $R$ by $ϕ$ .
d.: We say that a left ideal of $R$ is contracted by $ϕ$ if it is the contraction of some left ideal of $Q$ by $f$ .

We mention the following interesting proposition with a rather subtle proof.

Proposition 3.7.26. §

Let $ϕ : R \to Q$ be a homomorphism of rings. The maps given by contraction and extension of left ideals by $ϕ$ restrict to mutually inverse bijections

Extension and contraction of ideals. A full diagram description follows.

Diagram description: Extension and contraction of ideals

Extension and contraction restrict to mutually inverse bijections between these two collections of ideals.

Objects, listed by row and column:

Row 1, from left to right: column 1: left brace contracted left ideals of R right brace; column 3: left brace extended left ideals of Q right brace.

Arrows and lines:

An arrow from left brace contracted left ideals of R right brace to left brace extended left ideals of Q right brace, labelled extend.
An arrow from left brace extended left ideals of Q right brace to left brace contracted left ideals of R right brace, labelled contract.

The same holds with right ideals, or two-sided ideals, replacing left ideals.

Proof.

We prove this for left ideals. Let $I = ϕ^{- 1} (J)$ for some ideal $J$ of $Q$ . Note that $ϕ (ϕ^{- 1} (J)) = J \cap ϕ (R)$ . Then

ϕ^{- 1} (𝑄𝜙 (I)) = ϕ^{- 1} (𝑄𝜙 (ϕ^{- 1} (J))) = ϕ^{- 1} (Q (J \cap ϕ (R))),

and the latter set contains $ϕ^{- 1} (J \cap ϕ (R)) = I$ but also is contained in $I = ϕ^{- 1} (J)$ since

Q (J \cap ϕ (R)) \subseteq J .

Next, let $J = 𝑄𝜙 (I)$ for some ideal $I$ of $R$ . Then

𝑄𝜙 (ϕ^{- 1} (J)) = 𝑄𝜙 (ϕ^{- 1} (𝑄𝜙 (I)))

contains $𝑄𝜙 (I) = J$ since $ϕ^{- 1} (𝑄𝜙 (I)) \supseteq I$ , while

𝑄𝜙 (ϕ^{- 1} (J)) \subseteq 𝑄𝐽 = J

as well. □

3.8. Principal ideals and generators

Definition 3.8.1. §

A left ideal $J$ of a ring $R$ is said to be principal if there exists an element $a \in R$ such that

J = 𝑅𝑎 = {𝑟𝑎 ∣ r \in R} .

Similarly, a right ideal $K$ of a ring $R$ is principal if there exists an element $a \in R$ such that $K = 𝑎𝑅 = {𝑎𝑟 ∣ r \in R}$ . We then say that $J$ (resp., $K$ ) is the left (resp, right) ideal generated by $a$ .

Remark 3.8.2. §

Note that $𝑅𝑎$ for $a \in R$ is always a left ideal of $R$ , since $𝑟𝑎 - 𝑠𝑎 = (r - s) a$ for $r, s \in R$ , so $𝑅𝑎$ is an additive subgroup, and $s (𝑟𝑎) = (𝑠𝑟) a$ , so $𝑅𝑎$ is closed under left multiplication by elements of $R$ .

We also have the notion of a principal ideal.

Definition 3.8.3. §

An ideal $I$ of a ring $R$ is principal if there exists an element $a \in R$ such that

I = {\sum_{i = 1}^{N} r_{i} a s_{i} ∣ r_{i}, s_{i} \in R for 1 \leq i \leq N and N \geq 0} .

We then say that $I$ is generated by $a$ and write $I = (a)$ .

Remark 3.8.4. §

The set $𝑅𝑎𝑅 = {𝑟𝑎𝑠 ∣ r, s \in R}$ will not in general be a two-sided ideal, as $𝑟𝑎𝑠 + r^{'} a s^{'}$ for $r, r^{'}, s, s^{'} \in R$ need not itself be an element of $𝑅𝑎𝑅$ .

Examples 3.8.5. §

a.: For each $n \geq 1$ , the ideal $𝑛ℤ$ is the principal ideal $(n)$ .
b.: For every ring $R$ , the zero ideal is the principal ideal $(0)$ .
c.: For every ring $R$ , we have $R = (1)$ , so $R$ is a principal ideal of $R$ , known as the imp
d.: The ideal $(x)$ in $R [x]$ is the ideal consisting of all polynomials with nonconstant coefficient.
e.: The ideal generated by $(a, b) \in ℤ \times ℤ$ is equal to the set ${(𝑎𝑥, 𝑏𝑦) ∣ x, y \in ℤ}$ .

Example 3.8.6. §

Let $R$ be a nonzero ring, and let $n \geq 1$ . For integers $s, t$ with $1 \leq s, t \leq n$ , let $E_{𝑠𝑡} = (e_{𝑖𝑗}) \in M_{n} (R)$ be the matrix with $e_{𝑠𝑡} = 1$ and $e_{𝑖𝑗} = 0$ for $(i, j) \neq (s, t)$ . If $A = (a_{𝑖𝑗}) \in M_{n} (R)$ is any matrix, then the $(i, j)$ th entry of $A E_{𝑠𝑠}$ is $a_{𝑖𝑠}$ if $j = s$ and $0$ otherwise. Therefore, the left ideal generated by $E_{𝑠𝑠}$ is

M_{n} (R) E_{𝑠𝑠} = {(b_{𝑖𝑗}) \in M_{n} (R) ∣ b_{𝑖𝑗} = 0 for 1 \leq i, j \leq N, j \neq s},

the set of matrices that are zero outside of the $s$ th column. Similarly, the $(i, j)$ th entry of $E_{𝑠𝑠} A$ is $a_{𝑠𝑗}$ if $i = s$ and $0$ otherwise, so $E_{𝑠𝑠} M_{n} (R)$ is the right ideal of matrices that are zero outside of the $s$ th row.

The two-sided ideal $(E_{𝑠𝑠})$ of $M_{n} (R)$ is in fact all of $M_{n} (R)$ . To see this, note that $E_{𝑗𝑠} E_{𝑠𝑠} E_{𝑠𝑗} = E_{𝑗𝑗}$ for any $j \in ℤ$ . We then have

A = \sum_{j = 1}^{n} A E_{𝑗𝑗} = \sum_{j = 1}^{n} (A E_{𝑗𝑠}) E_{𝑠𝑠} E_{𝑠𝑗} \in (E_{𝑠𝑠}) .

Note, however, that the set $X = M_{n} (R) E_{𝑠𝑠} M_{n} (R)$ is not $M_{n} (R)$ , since each column of a matrix in $X$ has entries which are all equal to each other.

Definition 3.8.7. §

A nonzero ring $R$ is simple if its only ideals are $0$ and $R$ .

Remark 3.8.8. §

The reader can check using Example 3.8.6 that if $D$ is a division ring and $A \in M_{n} (D)$ is nonzero, then the ideal $(A)$ is all of $M_{n} (D)$ . So, $M_{n} (D)$ is simple, but note that it is not a division ring if $n \geq 2$ , and it does have proper, nonzero left ideals.

The following three results also clearly have analogues for right ideals that we leave unstated.

Proposition 3.8.9. §

Let $D$ be a nonzero ring that contains no nonzero, proper left ideals. Then $D$ is a division ring.

Proof.

Let $u \in D$ be nonzero. By assumption, we have $𝐷𝑢 = D$ , so there exists $v \in D$ such that $𝑣𝑢 = 1$ . Then $𝐷𝑣 = D$ , so there exists $w \in D$ such that $𝑤𝑣 = 1$ . We then have $w = 𝑤𝑣𝑢 = u$ , so $u = v^{- 1} \in D^{\times}$ . □

Lemma 3.8.10. §

Let $R$ be a ring, and let $a, b \in R$ . Then $𝑅𝑎 \subseteq 𝑅𝑏$ if and only if there exists $r \in R$ such that $a = 𝑟𝑏$ .

Proof.

If $𝑅𝑎 \subseteq 𝑅𝑏$ , then since $a \in 𝑅𝑏$ , we have $a = 𝑟𝑏$ for some $r \in R$ . Conversely, if $a = 𝑟𝑏$ and $r^{'} \in R$ , then $r^{'} a = (r^{'} r) b \in 𝑅𝑏$ , so $𝑅𝑎 \subseteq 𝑅𝑏$ . □

Lemma 3.8.11. §

Let $R$ be a ring that has no zero divisors. Let $a, b \in R$ . Then $𝑅𝑎 = 𝑅𝑏$ if and only if $b = 𝑢𝑎$ for some $u \in R^{\times}$ .

Proof.

Note that $a = 0$ if and only if $b = 0$ , so we may suppose that $a$ and $b$ are nonzero with $𝑅𝑎 = 𝑅𝑏$ . Since $b \in 𝑅𝑎$ , we have that there exists $u \in R$ with $b = 𝑢𝑎$ . Similarly, there exists $v \in R$ with $a = 𝑣𝑏$ . But then $a = 𝑣𝑢𝑎$ and $b = 𝑢𝑣𝑏$ , so $(1 - 𝑣𝑢) a = (1 - 𝑢𝑣) b$ . Since $R$ has no left zero divisors, we have $𝑢𝑣 = 𝑣𝑢 = 1$ . Conversely, if $b = 𝑢𝑎$ , then clearly $b \in 𝑅𝑎$ , so $𝑅𝑏 \subseteq 𝑅𝑎$ . On the other hand, $a = u^{- 1} b$ , so $𝑅𝑎 \subseteq 𝑅𝑏$ as well. □

Example 3.8.12. §

In $ℚ [x]$ , we have $(f) = (g)$ if and only if $f = 𝑐𝑔$ for some $c \in ℚ^{\times}$ , since $ℚ {[x]}^{\times} = ℚ^{\times}$ .

We have various operations that can be performed on ideals.

Lemma 3.8.13. §

Let $I$ and $J$ be left ideals (resp., right ideals) of a ring $R$ .

a.

The set

I + J = {a + b ∣ a \in I, b \in J}

is a left ideal (resp., right ideal) of $R$ .

b.

The intersection $I \cap J$ is a left (resp., right ideal) of $R$ .

Proof.

a.: If $a \in I$ , $b \in J$ , and $r \in R$ , then $r (a + b) = 𝑟𝑎 + 𝑟𝑏$ , and $𝑟𝑎 \in I$ , $𝑟𝑏 \in J$ since $I$ and $J$ are ideals, so $r (a + b) \in I + J$ . Moreover, $I + J$ is a subgroup of $R$ under addition by Lemma 4.1.4.
b.: If $a, b \in I \cap J$ and $r \in R$ , then clearly $a - b \in I \cap J$ and $𝑟𝑎 \in I \cap J$ , so $I \cap J$ is a left ideal of $R$ .

□

Remark 3.8.14. §

The argument of Lemma 3.8.13b carries over to show that an arbitrary intersection of left (resp., right) ideals of a ring $R$ is a left (resp., right) ideal of $R$ .

Clearly, addition of ideals forms an associative and commutative binary operation on the set of ideals of a ring. More generally, we have the following result.

Lemma 3.8.15. §

Let $T$ be an indexing set, and let ${I_{t} ∣ t \in T}$ be a collection of left (resp., right ideals) of a ring $R$ . Then the set

\sum_{t \in T} I_{t} = {\sum_{i = 1}^{N} a_{t_{i}} ∣ t_{i} \in T, a_{t_{i}} \in I_{t_{i}} for each 1 \leq i \leq N for some N \geq 0}

of finite sums of elements of the ideals $I_{t}$ is an ideal of $R$ , equal to the intersection of all ideals of $R$ containing $I_{t}$ for every $t \in T$ .

Proof.

Note that $\sum_{t \in T} I_{t}$ consists exactly of finite sums of elements in the union $\cup_{t \in T} I_{t}$ . It is a subgroup, as the sum of two finite sums is a finite sum, and the negative of two finite sums is as well. Moreover, it is an ideal, as for any $N \geq 0$ , $t_{i} \in T$ and $a_{t_{i}} \in I_{t_{i}}$ for $1 \leq i \leq N$ , we have

r \cdot \sum_{i = 1}^{N} a_{t_{i}} = \sum_{i = 1}^{N} r a_{t_{i}},

and $r a_{t_{i}} \in I_{t_{i}}$ since $I_{t_{i}}$ is a left ideal of $R$ . Therefore $\sum_{t \in T} I_{t}$ is a left ideal, and similarly, it is a right ideal.

Finally, note that if $J$ is any ideal of $R$ containing each $I_{t}$ , then it must contain any finite sum of elements in these ideals, i.e., in $\cup_{t \in T} I_{t}$ . Therefore, $J$ contains $\sum_{t \in T} I_{t}$ . Therefore, the intersection of all ideals of $R$ containing each $I_{t}$ is an ideal of $R$ containing $\sum_{t \in I_{t}} I_{t}$ , and $\sum_{t \in I_{t}} I_{t}$ is itself an ideal of $R$ containing each $I_{t}$ , so it equals the intersection. □

Definition 3.8.16. §

Let $R$ be a ring, and let ${I_{t} ∣ t \in T}$ be a collection of left (resp., right) ideals. The sum of the ideals $I_{t}$ with $t \in T$ is the left (resp., right) ideal $\sum_{t \in T} I_{t}$ of $R$ .

We will define generators solely for two-sided ideals, though they have obvious analogues for left and right ideals.

Definition 3.8.17. §

Let $X$ be a subset of a ring $R$ . The ideal $(X)$ generated by $X$ is the sum of the ideals $(x)$ for $x \in X$ . If $I$ is an ideal of $R$ and $I = (X)$ , we say that $X$ is a set of generators of $I$ , and $X$ generates $I$ . The elements of $X$ are called generators. If $X = {a_{1}, a_{2}, \dots, a_{n}}$ is a finite set, then we write $(a_{1}, a_{2}, \dots, a_{n})$ for $(X)$ .

Remarks 3.8.18. §

a.: Every ideal is generated by the set of all of its elements.
b.: Using Lemma 1.2.24, we could equivalently have defined $(X)$ to be the smallest ideal containing $X$ .

Since the set-theoretic product of two ideals will not in general be closed under addition, we depart from earlier notation to make the following definition.

Definition 3.8.19. §

Let $I$ and $J$ be ideals of a ring $R$ . Then the product $𝐼𝐽$ of $I$ and $J$ is the ideal of $R$ generated by all $𝑎𝑏$ with $a \in I$ and $b \in J$ .

In particular, we may speak of powers $I^{n} = 𝐼𝐼 \dots I$ of an ideal $I$ for any $n \geq 1$ . Products are easily calculated in terms of generators, as seen in the following examples.

Examples 3.8.20. §

a.: If $R$ is a ring and $x, y \in R$ , then $(x) \cdot (y) = (𝑥𝑦)$ .
b.: In the ring $ℚ [x, y]$ , we have
$(x, y) \cdot (x^{2}, x + y) = (x^{3}, x^{2} y, x^{2} + 𝑥𝑦, 𝑥𝑦 + y^{2}) .$

Definition 3.8.21. §

We say that two ideals $I$ and $J$ of a ring $R$ are coprime if $I + J = R$ .

Definition 3.8.22. §

For $k \geq 1$ , we say that ideals $I_{1}, \dots, I_{k}$ of a ring $R$ are pairwise coprime if $I_{i} + I_{j} = R$ for all $1 \leq i < j \leq k$ .

We prove a general form of the Chinese Remainder Theorem.

Theorem 3.8.23 (Chinese Remainder Theorem). §

Let $I_{1}, \dots, I_{k}$ be pairwise coprime two-sided ideals of a ring $R$ for some $k \geq 1$ . Then there is an isomorphism

R ∕ (I_{1} \cap I_{2} \cap \dots \cap I_{k}) \to_{}^{\sim} R ∕ I_{1} \times R ∕ I_{2} \times \dots \times R ∕ I_{k}

that sends the coset of $a \in R$ to $(a + I_{1}, a + I_{2}, \dots, a + I_{k})$ .

Proof.

The kernel of the map $R \to \prod_{i = 1}^{k} R ∕ I_{i}$ induced by the diagonal map is clearly $I_{1} \cap I_{2} \cap \dots \cap I_{k}$ . We need only see that it is surjective. Consider the case that $k = 2$ . Let $a, b \in R$ . Then there exist $d \in I_{1}$ and $c \in I_{2}$ such that $a + I_{1} = c + I_{1}$ and $b + I_{2} = d + I_{2}$ . If we set $x = c + d$ , then $x + I_{1} = c + I_{1}$ and $x + I_{2} = d + I_{2}$ , so $x$ maps to $(a + I_{1}, b + I_{2})$ .

For any $k \geq 3$ , suppose by induction we know the result for $k - 1$ , so $R ∕ (I_{2} \cap \dots \cap I_{k}) ≅𝑅 ∕ I_{2} \times \dots R ∕ I_{k}$ . We therefore need only see that $I_{1}$ and $I_{2} \cap \dots \cap I_{k}$ are coprime. Note that $I_{2} \cap \dots \cap I_{k}$ contains the product $I_{2} \dots I_{k}$ . For each $2 \leq i \leq k$ , let $a_{i} \in I_{1}$ and $b_{i} \in I_{i}$ be such that $a_{i} + b_{i} = 1$ . Then $1 = (a_{2} + b_{2}) \dots (a_{k} + b_{k})$ is an element of $I_{1}$ plus $b_{1} \dots b_{k} \in I_{2} \dots I_{k}$ , as needed. □

Definition 3.8.24. §

An ideal $I$ of a ring $R$ is said to be finitely generated if it has a finite set of generators, which is to say that $I = (a_{1}, a_{2}, \dots, a_{n})$ for some $n \geq 1$ and $a_{1}, a_{2}, \dots, a_{n} \in I$ .

Example 3.8.25. §

If $R [x, y]$ , the ideal $(x, y)$ is the ideal of elements with $0$ constant term, as every monomial other than $1$ is either divisible $x$ or $y$ . It is not principal, since no element of $R [x, y]$ not in $R^{\times}$ divides both $x$ and $y$ , but it is finitely generated.

Example 3.8.26. §

Let $n \geq 2$ . The ideal $(n, x)$ of $ℤ [x]$ is the set of all sums $𝑛𝑓 + 𝑥𝑔$ with $f, g \in ℤ [x]$ , which is equal to the set of polynomials with $ℤ$ -coefficients and constant coefficient divisible by $n$ . This is not principal, since $n$ and $x$ are both multiples only of $\pm 1$ , which are not contained in $(n, x)$

Example 3.8.27. §

Consider the ideal $(4, 6)$ of $ℤ$ . It contains $2 = 6 - 4$ , so $(2) \subseteq (4, 6)$ and we have $4, 6 \in (2)$ , so $(4, 6) \subseteq (2)$ . Therefore, $(4, 6)$ is a principal ideal of $ℤ$ , equal to the ideal $(2)$ .

In fact, note the following.

Lemma 3.8.28. §

The ideals of $ℤ$ are exactly the subgroups of $ℤ$ under addition, i.e., the $𝑛ℤ$ with $n \geq 0$ . In particular, every ideal of $ℤ$ is principal.

Proof.

Ideals are by definition subgroups under addition, and if $I$ is an ideal of $ℤ$ , the condition that $ℤ \cdot I \subseteq I$ is a consequence of this, since it merely says that $ℤ$ -multiples of elements of $I$ are contained in $I$ . That the subgroups of $ℤ$ have the form $𝑛ℤ$ is Corollary 2.3.12. □

This leads to the following definition.

Definition 3.8.29. §

An integral domain $R$ is a principal ideal domain, or PID, if every ideal in $R$ is principal.

So far, we have the following examples.

Examples 3.8.30. §

a.: The ring $ℤ$ is a principal ideal domain.
b.: Every field is a principal ideal domain.
c.: If $R$ and $Q$ are principal ideal domains, then every ideal $R \times Q$ is principal, though it is not a domain.

3.9. Polynomial rings over fields

We consider polynomial rings over a field. One of the key properties of polynomial rings over a field is that we can divide any polynomial by any nonzero polynomial, obtaining a remainder. More generally, we can divide a polynomial over a commutative ring by another so long as the leading coefficient of the divisor is a unit.

Theorem 3.9.1 (Division algorithm). §

Let $R$ be a commutative ring. Suppose that $f, g \in R [x]$ are polynomials and that the leading coefficient of $g$ is a unit in $R$ . Then there exist unique polynomials $q, r \in R [x]$ such that $f = 𝑞𝑔 + r$ and $\deg r < \deg g$ .

Proof.

The case that $f = 0$ is trivial, so we assume that $f$ is nonzero. We verify this by induction on the degree $n$ of $f$ . Note that if $n \leq \deg g$ , and in particular if $n = 0$ , then we may take $q = 0$ and $r = f$ if $\deg g > 0$ and $q = f g^{- 1}$ and $r = 0$ if $\deg g = 0$ (recalling that we consider the degree of $0$ to be less than that of every nonzero polynomial). So suppose that $n \geq m = \deg g$ . Let $a_{n}$ be the nonzero coefficient of $x^{n}$ in $f$ and $b_{m} \in R^{\times}$ be the coefficient of $x^{m}$ in $g$ . Then $f^{'} = f - a_{n} b_{m}^{- 1} x^{n - m} g$ has degree at most $n$ , and the coefficient of $x^{n}$ is $a_{n} - a_{n} b_{m}^{- 1} \cdot b_{m} = 0$ , so in fact we have $\deg f^{'} < n$ . By induction, therefore, there exist $q^{'}$ and $r$ in $R [x]$ such that $f^{'} = q^{'} g + r$ and $\deg r < m$ . Setting $q = a_{n} b_{m}^{- 1} + q^{'}$ , we have

f = a_{n} b_{m}^{- 1} g + f^{'} = (a_{n} b_{m}^{- 1} + q^{'}) g + r = 𝑞𝑔 + r,

as desired.

If $𝑞𝑔 + r = q^{'} g + r^{'}$ for some $q^{'}, r^{'} \in R [x]$ with $\deg r^{'} < \deg g$ , then we have

(q - q^{'}) g + (r - r^{'}) = 0 .

(3.9.1)

If $q \neq q^{'}$ , we would have

\deg (q - q^{'}) g \geq \deg g > \deg (r - r^{'}),

in contradiction to (3.9.1). So, we must have $q = q^{'}$ , and then (3.9.1) yields $r = r^{'}$ , establishing uniqueness. □

We next show that polynomial rings in one variable over a field form another class of principal ideal domains.

Theorem 3.9.2. §

Let $F$ be a field. Then $F [x]$ is a principal ideal domain. In fact, any nonzero ideal $I$ of $F$ is generated by any nonzero polynomial that has minimal degree among all polynomials in $I$ .

Proof.

By Theorem 3.4.13, $F [x]$ is an integral domain. Let $I$ be a nonzero ideal in $F [x]$ , and let $g$ be a nonzero polynomial in $F [x]$ of minimal degree. We claim that $I = (g)$ . Let $f \in I$ . Using the division algorithm, we write $f = 𝑞𝑔 + r$ with $q, r \in F [x]$ with $\deg r < \deg g$ . Then $r = f - 𝑞𝑔 \in I$ , which by the minimality of the degree of $g$ forces $r = 0$ . Thus $f \in (g)$ , and as $f$ was arbitrary, we have $I = (g)$ . □

Definition 3.9.3. §

Let $F$ be a field. A nonconstant polynomial $f \in F [x]$ is irreducible if there does not exist any $g \in F [x]$ with $0 < \deg g < \deg f$ that divides $f$ . A nonconstant polynomial that is not irreducible is called reducible. A noncontant divisor of a polynomial is referred to as a factor.

Example 3.9.4. §

By definition, any polynomial of degree $1$ is irreducible in $F [x]$ . The polynomial $x^{2} + 1$ is irreducible in $ℚ [x]$ but not in $ℂ [x]$ , where we have

x^{2} + 1 = (x + i) (x - i) .

On the other hand, $x^{2}$ is reducible for any $F$ , since $x^{2} = x \cdot x$ .

Definition 3.9.5. §

Let $R$ be a ring. We say that $a \in R$ is a root (or zero) of a polynomial $f \in R [x]$ if $f (a) = 0$ .

Definition 3.9.6. §

In a commutative ring $R$ , we say that an element $b$ divides an element $a$ in $R$ if there exists some $c \in R$ such that $a = 𝑏𝑐$ . Equivalently, $b$ divides $a$ if $a \in (b)$ . We sometimes write $b ∣ a$ to denote that $b$ divides $a$ .

We note the following.

Proposition 3.9.7. §

Let $F$ be a field, and let $f \in F [x]$ . Then $a \in F$ is a root of $f$ if and only if $x - a$ divides $f$ .

Proof.

If $x - a$ divides $f$ , then there exists $g \in F [x]$ with $f = (x - a) g$ . We then have $f (a) = (a - a) g (a) = 0$ , noting Lemma 3.6.9. Conversely, if $a$ is a zero of $f$ , then the division algorithm implies that there exists some $q \in F [x]$ and $c \in F$ such that $f = q (x - a) + c$ . We then have

0 = f (a) = q (a) (a - a) + c = c,

so $x - a$ divides $f$ . □

We obtain the following corollaries.

Corollary 3.9.8. §

Let $F$ be a field and $f \in F [x]$ be a polynomial of degree greater than $1$ . If $f$ has a root in $F$ , then $f$ is reducible.

Proof.

If $a \in F$ is a root of $f$ , then Proposition 3.9.7 implies that $f = g (x - a)$ for some $g \in F [x]$ with $\deg g = \deg f - 1 > 0$ , so $f$ is not irreducible. □

Since a reducible polynomial of degree 2 or 3 must have a linear factor, we therefore have the following.

Corollary 3.9.9. §

Let $F$ be a field and $f \in F [x]$ be a polynomial of degree $2$ or $3$ . Then $f$ is reducible if and only if it has a root in $F$ .

Corollary 3.9.10. §

Let $F$ be a field, and let $f \in F [x]$ be a nonzero polynomial. Then $f$ has at most $\deg f$ distinct roots in $F$ .

Proof.

Suppose that $f = (x - a_{1}) \dots (x - a_{m}) g$ , where $g \in F [x]$ has no roots, and $a_{1}$ , $a_{2}$ , $…$ , $a_{m} \in F$ . Clearly, we may write $f$ in this form, as otherwise we can factor out from $g$ a linear term $x - b$ for some $β$ with $g (b) = 0$ . Moreover, we must have $m \leq n$ by degree considerations. Finally, if $f (c) = 0$ for some $c \in F$ , then since $F$ is an integral domain, we must have $c - a_{i} = 0$ for some $i$ , which is to say that the $a_{i}$ are the only roots of $f$ . □

Examples 3.9.11. §

a.

The polynomial $x^{2}$ has $0$ as its only root.

b.

The polynomial $x^{2} + 1$ has no roots in $ℚ$ , but it has two roots, $\pm 1$ , in $ℂ$ .

c.

The polynomial

x^{4} - x^{2} - 2 x - 1 = (x^{2} + x + 1) (x^{2} - x - 1)

is not irreducible in $ℚ [x]$ , but it has no roots in $ℚ$ .

3.10. Maximal and prime ideals

Recall that $ℤ ∕ 𝑛ℤ$ is a field for $n$ prime, but $ℤ ∕ 𝑛ℤ$ is not a field for $n$ composite. In this section, we shall see how we can interpret this as a property of the ideal $𝑛ℤ$ .

Definition 3.10.1. §

An ideal $𝔪$ of a ring $R$ is maximal if it is a proper ideal of $R$ that is not properly contained in any proper ideal of $R$ .

In other words, a proper ideal $𝔪$ of $R$ is maximal if there does not exist an ideal $N$ of $R$ such that $𝔪 ⊊ N ⊊ R$ .

Examples 3.10.2. §

a.: The maximal ideals of $ℤ$ are exactly the $𝑝ℤ$ for $p$ prime, as $𝑚ℤ$ contains $𝑛ℤ$ if and only if $m$ divides $n$ . In particular, as $p$ is a prime number, $𝑝ℤ$ is not contained in $𝑛ℤ$ for any $n \geq 2$ with $n \neq p$ .
b.: In a field, the unique maximal ideal is $(0)$ .
c.: In $ℤ \times ℤ$ , the maximal ideals have either the form $𝑝ℤ \times ℤ$ or $ℤ \times 𝑝ℤ$ for some prime number $p$ .

Proposition 3.10.3. §

Let $F$ be a field. The maximal ideals of $F [x]$ are exactly the ideals of the form $(f)$ with $f \in F [x]$ irreducible.

Proof.

Let $f \in F [x]$ . If $f = 0$ , then $(f) = 0$ , which is not maximal. If $f$ is a nonzero constant, then $(f) = (1) = F [x]$ . If $f$ is reducible, then $f = 𝑔h$ with $g, h \in F [x]$ nonconstant, and then $(f) \subseteq (g)$ , but $g \notin (f)$ since $\deg g < \deg f$ , so $(f)$ is not maximal.

If $f$ is irreducible and $I$ is an ideal containing $(f)$ , then $I = (g)$ for some $g \in F [x]$ as $F [x]$ is a PID. There then exists $h \in F [x]$ such that $f = 𝑔h$ . Since $f$ is irreducible, we then have that either $g$ or $h$ is constant, which is to say that $I = (g) = F [x]$ or $I = (g) = (f)$ . In other words, $(f)$ is maximal. □

The following gives an alternate characterization of maximal ideals of rings.

Theorem 3.10.4. §

A proper ideal $𝔪$ in a commutative ring $R$ is maximal if and only if $R ∕ 𝔪$ is a field.

Proof.

By Theorem 3.7.24, the ideals in $R ∕ 𝔪$ are in one-to-one correspondence with the ideals in $R$ containing $𝔪$ , which are just $𝔪$ and $R$ . Since $R ∕ 𝔪$ has just two ideals, they must be $0$ and $R ∕ 𝔪$ . Therefore, every nonzero element of $R ∕ 𝔪$ generates the ideal $R ∕ 𝔪$ , so is a unit. It follows that $R ∕ 𝔪$ is a field. □

Remark 3.10.5. §

The same argument can be applied to noncommutative rings $R$ to conclude that if $𝔪$ is maximal then $R ∕ 𝔪$ has no nonzero proper ideals. However, as we have remarked above, this does not imply that $R ∕ 𝔪$ is a division ring.

Example 3.10.6. §

Recall that $ℤ ∕ 𝑛ℤ$ is a field if and only if $n \geq 2$ is prime, which is to say if and only if $𝑛ℤ$ is a maximal ideal of $ℤ$ .

Example 3.10.7. §

Since $x^{2} + 1$ is irreducible over $ℚ$ , the ideal $𝔪 = (x^{2} + 1)$ is maximal in $ℚ [x]$ . Clearly, $𝔪$ is contained in the kernel of the evaluation map $e_{i} : ℚ [x] \to ℚ (i)$ defined by $e_{i} (f) = f (i)$ , but then it must be the entire kernel as the kernel is proper and $𝔪$ is maximal. By the first isomorphism theorem for rings, the field $ℚ [x] ∕ (x^{2} + 1)$ is isomorphic to $ℚ [i]$ . In particular, $ℚ [i]$ is equal to the subfield $ℚ (i)$ of $ℂ$ consisting of fractions $\frac{a + 𝑏𝑖}{c + 𝑑𝑖}$ with $a, b, c, d \in ℚ$ and $(c, d) \neq (0, 0)$ . One can also see this directly: the multiplicative inverse of $c + 𝑑𝑖$ is $\frac{c}{c^{2} + d^{2}} - \frac{d}{c^{2} + d^{2}} i$ .

Example 3.10.8. §

The ring $ℚ [x] ∕ (x^{2})$ is not a field, or even an integral domain, since $x \cdot x \in (x^{2})$ .

Example 3.10.9. §

In $ℤ [x]$ , the ideals $(p, x)$ , where $p$ is a prime number, are maximal. To see this, consider the homomorphism

ϕ : ℤ [x] \to ℤ ∕ 𝑝ℤ

given by $ϕ (f) = f (0) + 𝑝ℤ$ . This is surjective with kernel consisting of those $f$ with constant coefficient a multiple of $p$ , which is to say the ideal $(p, x)$ .

Given a proper ideal $I$ of a ring $R$ : is $I$ necessarily even contained in a maximal ideal? Assuming the axiom of choice, the answer is yes. We require a preliminary lemma.

Lemma 3.10.10. §

Let $𝒞$ be a chain of ideals in a ring $R$ , ordered with respect to inclusion of subsets of $R$ . Then the ideal

N = ⋃_{J \in 𝒞} J

is an ideal of $R$ .

Proof.

If $x, y \in N$ , then $x \in J$ and $y \in K$ for some $J, K \in 𝒞$ . Then $J \cup K$ is either $J$ or $K$ , so is in $𝒞$ , and we then have $x - y \in J \cup K$ , so $x - y \in N$ . Thus, $N$ is a subgroup of $R$ under addition. For $a \in R$ and $x \in N$ , we have that $x \in J$ for some $J \in 𝒞$ , and then $𝑎𝑥$ and $𝑥𝑎$ are elements of $J$ , since $J$ is an ideal. In particular, they are also elements of $N$ . Therefore, $N$ is an ideal. □

Theorem 3.10.11. §

Let $I$ be a proper ideal of a ring $R$ . Then there exists a maximal ideal $𝔪$ of $R$ that contains $I$ .

Proof.

Let $X$ be the set of proper ideals of $R$ containing $I$ , which we endow with the usual partial ordering $\subseteq$ . Suppose that $𝒞 \subset X$ is a chain. Consider the ideal

N = ⋃_{J \in 𝒞} J

of $R$ . Note that $1 \notin N$ since $1 \notin J$ for all $J \in 𝒞$ , so $N$ is proper. In other words, $N \in X$ , and it is an upper bound for $𝒞$ . Zorn’s lemma then tells us that $X$ contains a maximal element, which is necessarily a maximal ideal of $R$ . □

In commutative rings, maximal ideals are part of a broader class of ideals known as prime ideals.

Definition 3.10.12. §

Let $R$ be a commutative ring. A proper ideal $𝔭$ of $R$ is said to be a prime ideal (or prime) if for all $b, c \in R$ with $𝑏𝑐 \in 𝔭$ , either $b \in 𝔭$ or $c \in 𝔭$ .

Examples 3.10.13. §

a.: If $A$ is an integral domain, then $(0)$ is a prime ideal.
b.: In $ℤ$ , the prime ideals are exactly $(0)$ and the $𝑝ℤ$ for $p$ prime. That is, if $𝑎𝑏 \in (p)$ with $p$ prime, then $p$ divides $𝑎𝑏$ , so $p$ divides $a$ or $p$ divides $b$ , and hence either $a \in (p)$ or $b \in (p)$ .

We have the following analogue of Theorem 3.10.4.

Theorem 3.10.14. §

Let $R$ be a commutative ring. Then a proper ideal $𝔭$ of $R$ is prime if and only if $R ∕ 𝔭$ is an integral domain.

Proof.

The ideal $𝔭$ is prime if and only if $𝑎𝑏 \in 𝔭$ implies than $a \in 𝔭$ or $b \in 𝔭$ , which translates to the fact that $𝑎𝑏 = 0$ imples $a = 0$ or $b = 0$ in the ring $R ∕ 𝔭$ . □

Corollary 3.10.15. §

Let $R$ be a commutative ring. Then every maximal ideal of $R$ is prime.

Proof.

If $𝔪$ is a maximal ideal of $R$ , then Theorem 3.10.4 then tells us that $R ∕ 𝔪$ is a field. Theorem 3.10.14 yields that $𝔪$ is prime. □

As for polynomial rings over fields, we have the following theorem.

Proposition 3.10.16. §

Let $F$ be a field. The prime ideals in $F [x]$ are exactly $(0)$ and those $(f)$ such that $f \in F [x]$ is irreducible.

Proof.

Note that if $f$ is nonconstant and reducible, then $f = 𝑔h$ for some nonconstant $g, h \in F [x]$ of degree less than $\deg f$ , so $g, h \notin (f)$ . Therefore, $(f)$ is not prime.

On the other hand, if $f$ is nonconstant and irreducible, then Proposition 3.10.3 tells us that $(f)$ is maximal, and Corollary 3.10.15 then tells us that $(f)$ is prime. □

Example 3.10.17. §

In $ℤ [x]$ , the ideal $(x)$ is prime, since $ℤ [x] ∕ (x) ≅ℤ$ , but $(x)$ is not maximal. This follows either from the fact that $ℤ$ is not a field, or the fact that $(x)$ is properly contained in $(p, x)$ for any prime $p$ .

Prime ideals are preserved under contraction.

Lemma 3.10.18. §

Let $ϕ : R \to Q$ be a homomorphism of commutative rings. Then any contraction of a prime ideal of $Q$ by $ϕ$ is a prime ideal of $R$ .

Proof.

For a prime ideal $𝔮$ of $Q$ , this amounts to the fact that if $a, b \in R$ with $ϕ (𝑎𝑏) \in 𝔮$ , then $ϕ (a) \in 𝔮$ or $ϕ (b) \in 𝔮$ , since $𝔮$ is prime. □

The following lemma is also very useful.

Lemma 3.10.19. §

a.: Let $𝔭_{1}, \dots, 𝔭_{k}$ be prime ideals of $R$ . If an ideal $I$ is contained in $⋃_{i = 1}^{k} 𝔭_{i}$ , then $I$ is contained in some $𝔭_{i}$ .
b.: Let $𝔞_{1}, \dots, 𝔞_{k}$ be ideals of $R$ . If a prime ideal $𝔭$ contains $\prod_{i = 1}^{k} 𝔞_{i}$ (resp., equals $⋂_{i = 1}^{k} 𝔞_{i}$ ), then $𝔭$ contains (resp., equals) some $𝔞_{i}$ .

Proof.

We prove part a by induction on $k$ , it being clearly true for $k = 1$ . Suppose that $I$ is not contained in any $𝔭_{i}$ but $I$ is contained in the union of the $𝔭_{i}$ . By induction, for each $i$ , we can find $a_{i} \in I$ such that $a_{i} \notin 𝔭_{j}$ for all $j \neq i$ . By assumption, we then have $a_{i} \in 𝔭_{i}$ for each $i$ . The element

b = \sum_{i = 1}^{k} \prod_{\begin{array}{c} j = 1 \\ j \neq i \end{array}}^{k} a_{i}

of $I$ has image in $R ∕ 𝔭_{i}$ equal to the image of its $i$ th term, which is nonzero by the primality of $i$ . That is, $b \notin ⋃_{i = 1}^{k} 𝔭_{i}$ , which is a contradiction.

As for part b, let $𝔞 = \prod_{i = 1}^{k} 𝔞_{i}$ . Suppose that $𝔭$ does not contain any $𝔞_{i}$ , and choose $a_{i} \in 𝔞_{i}$ with $a_{i} \notin 𝔭$ for each $i$ . Then $a = \prod_{i = 1}^{k} a_{i} \in 𝔞$ , but $a \notin 𝔭$ by primality of $𝔭$ . Therefore, $𝔭$ does not contain $𝔞$ . If on the other hand $𝔭 = ⋂_{i = 1}^{k} 𝔞_{i}$ , then $𝔭$ contains some $𝔞_{i}$ by what we have shown, so must equal it in that $𝔭 \subseteq ⋂_{i = 1}^{k} 𝔞_{i}$ by assumption. □

3.11. Fields of fractions

As is seen by the most basic case of the integers $ℤ$ , not all rings are fields. Yet, $ℤ$ is contained in many fields, the smallest being $ℚ$ , the rational numbers. The field $ℚ$ consists exactly of fractions $\frac{a}{b}$ , where $a$ and $b$ are integers and $b$ is nonzero. One can ask more generally, given an ring $R$ , does one have a good notion of a fraction $\frac{a}{b}$ with $a, b \in R$ and $b \neq 0$ ? And, if so, can one form a field out of them? As we shall, see in the case of an integral domain, the answer is yes.

Lemma 3.11.1. §

Let $R$ be an integral domain, and set

X = {(a, b) \in R \times R ∣ b \neq 0} .

The relation $\sim$ on $X$ given by $(a, b) \sim (c, d)$ if and only if $𝑎𝑑 = 𝑏𝑐$ is an equivalence relation.

Proof.

For $(a, b) \in X$ , we have $𝑎𝑏 = 𝑏𝑎$ , so $(a, b) \sim (a, b)$ , so $\sim$ is reflexive. If $(c, d) \in X$ with $(a, b) \sim (c, d)$ , then $𝑎𝑑 = 𝑏𝑐$ implies $𝑐𝑏 = 𝑑𝑎$ , so $(c, d) \sim (b, a)$ as well, and $\sim$ is symmetric. Finally, if $(e, f) \in X$ as well and $(a, b) \sim (c, d)$ while $(c, d) \sim (e, f)$ , we have $𝑎𝑑 = 𝑏𝑐$ and $𝑐𝑓 = 𝑑𝑒$ . Multiplying the former equality by $f$ and then applying the latter, we obtain

𝑎𝑑𝑓 = 𝑏𝑐𝑓 = 𝑏𝑑𝑒 .

Since $d \neq 0$ and $R$ is an integral domain, this implies $𝑎𝑓 = 𝑏𝑒$ , which means that $(a, b) \sim (e, f)$ , and therefore $\sim$ is transitive. □

Note that the last step shows the need for having an integral domain in order to have an equivalence relation in Lemma 3.11.1.

Definition 3.11.2. §

Let $R$ be an integral domain. We let $Q (R)$ denote the set of equivalence classes of elements of $X$ under the relation $\sim$ of Lemma 3.11.1. The equivalence class of $(a, b)$ with $a, b \in R$ and $b \neq 0$ will be denoted $\frac{a}{b}$ , and it is called the quotient of $a$ by $b$ . By using the symbol $\frac{a}{b}$ , we are implicitly representing the quotient by $(a, b)$ , and this representative is called a fraction. We then refer to $a$ as the numerator of $\frac{a}{b}$ and $b$ as the denominator of $\frac{a}{b}$ .

The following is immediate.

Lemma 3.11.3. §

Let $R$ be an integral domain, and let $a, b, x \in R$ with $b$ and $x$ nonzero. Then we have

\frac{a}{b} = \frac{𝑎𝑥}{𝑏𝑥}

in $Q (R)$ .

Lemma 3.11.4. §

Let $R$ be an integral domain. There are well-defined operations $+$ and $\cdot$ on $Q (R)$ given by

\frac{a}{b} + \frac{c}{d} = \frac{𝑎𝑑 + 𝑏𝑐}{𝑏𝑑}

and

\frac{a}{b} \cdot \frac{c}{d} = \frac{𝑎𝑐}{𝑏𝑑} .

Proof.

Let $X$ be as in Lemma 3.11.1. Define $+$ on $X$ by

(a, b) + (c, d) = (𝑎𝑑 + 𝑏𝑐, 𝑏𝑑)

and $\cdot$ on $X$ by

(a, b) \cdot (c, d) = (𝑎𝑐, 𝑏𝑑) .

To prove the proposition, we must show that if $(a, b) \sim (a^{'}, b^{'})$ and $(c, d) \sim (c^{'}, d^{'})$ , we have

\begin{array}{rcl} (a, b) + (c, d) \sim (a^{'}, b^{'}) + (c^{'}, d^{'}) & and & (a, b) \cdot (c, d) = (a^{'}, b^{'}) \cdot (c^{'}, d^{'}) . \end{array}

We check that

(𝑎𝑑 + 𝑏𝑐) b^{'} d^{'} = a b^{'} d d^{'} + c d^{'} b b^{'} = b a^{'} d d^{'} + d c^{'} b b^{'} = 𝑏𝑑 (a^{'} d^{'} + b^{'} c^{'})

and

𝑎𝑐 b^{'} d^{'} = a b^{'} c d^{'} = b a^{'} c^{'} d = 𝑏𝑑 a^{'} c^{'},

as desired. □

Corollary 3.11.5. §

Let $R$ be an integral domain and $a, a^{'}, b \in R$ with $b \neq 0$ . In $Q (R)$ , one has

\frac{a}{b} + \frac{a^{'}}{b} = \frac{a + a^{'}}{b} .

Proof.

Noting Lemma 3.11.3, we have

\frac{𝑎𝑏 + a^{'} b}{b^{2}} = \frac{(a + a^{'}) b}{b^{2}} = \frac{a + a^{'}}{b} .

□

Theorem 3.11.6. §

Let $R$ be an integral domain. Under the operations $+$ and $\cdot$ of Lemma 3.11.4, the ring $Q (R)$ is a field.

Proof.

First, we note that addition is commutative since

\frac{a}{b} + \frac{c}{d} = \frac{𝑎𝑑 + 𝑏𝑐}{𝑏𝑑} = \frac{𝑐𝑏 + 𝑑𝑎}{𝑑𝑏} = \frac{c}{d} + \frac{a}{b},

and it is associative since

\begin{matrix} (\frac{a}{b} + \frac{c}{d}) + \frac{e}{f} = \frac{𝑎𝑑 + 𝑏𝑐}{𝑏𝑑} + \frac{e}{f} = \frac{(𝑎𝑑 + 𝑏𝑐) f + 𝑏𝑑𝑒}{𝑏𝑑𝑓} \\ = \frac{𝑎𝑑𝑓 + b (𝑐𝑓 + 𝑑𝑒)}{𝑏𝑑𝑓} = \frac{a}{b} + \frac{𝑐𝑓 + 𝑑𝑒}{𝑑𝑓} = \frac{a}{b} + (\frac{c}{d} + \frac{e}{f}) . \end{matrix}

Next, we note that

\frac{0}{1} + \frac{a}{b} = \frac{a \cdot 1 + 0 \cdot b}{1 \cdot b} = \frac{a}{b},

so $0 = \frac{0}{1}$ in $Q (R)$ . We also have

\frac{- a}{b} + \frac{a}{b} = \frac{- 𝑎𝑏 + 𝑎𝑏}{b^{2}} = \frac{0}{b^{2}} = \frac{0}{1},

the latter step by noting that $0 \cdot 1 = b^{2} \cdot 0 = 0$ . Hence, $Q (R)$ is an abelian group under addition.

We note that multiplication is associative, as

(\frac{a}{b} \cdot \frac{c}{d}) \cdot \frac{e}{f} = \frac{𝑎𝑐}{𝑏𝑑} \cdot \frac{e}{f} = \frac{𝑎𝑐𝑒}{𝑏𝑑𝑓} = \frac{a}{b} \cdot \frac{𝑐𝑒}{𝑑𝑓} = \frac{a}{b} \cdot (\frac{c}{d} \cdot \frac{e}{f}) .

We check distributivity as follows:

\frac{a}{b} \cdot (\frac{c}{d} + \frac{e}{f}) = \frac{a}{b} \cdot \frac{𝑐𝑓 + 𝑑𝑒}{𝑑𝑓} = \frac{𝑎𝑐𝑓 + 𝑎𝑑𝑒}{𝑏𝑑𝑓} = \frac{𝑎𝑐𝑓}{𝑏𝑑𝑓} + \frac{𝑎𝑑𝑒}{𝑏𝑑𝑓} = \frac{𝑎𝑐}{𝑏𝑑} + \frac{𝑎𝑒}{𝑏𝑓} = \frac{a}{b} \cdot \frac{c}{d} + \frac{a}{b} \cdot \frac{e}{f} .

Note that

\frac{1}{1} \cdot \frac{a}{b} = \frac{a}{b},

so $1 = \frac{1}{1}$ in $Q (R)$ . Finally, note that $\frac{a}{b} \neq 0 = \frac{0}{1}$ if and only if $a \neq 0$ , and in this case we can form $\frac{b}{a}$ . We then have

\frac{b}{a} \cdot \frac{a}{b} = \frac{𝑎𝑏}{𝑎𝑏} = \frac{1}{1} = 1,

so $\frac{b}{a} = {(\frac{a}{b})}^{- 1}$ . Therefore, $Q (R)$ is a field. □

Definition 3.11.7. §

Let $R$ be an integral domain. The field $Q (R)$ is called the quotient field, or the field of fractions, of $R$ .

Remark 3.11.8. §

The field $Q (R)$ is not a quotient of $R$ in the sense it is the set of equivalence classes for an equivalence relation on $R$ itself. Rather, it is a set of quotients of elements of $R$ in the sense of division, and in fact it contains $R$ . That is, quotient rings and quotient fields are quite different should not be confused with each other.

Definition 3.11.9. §

Let $F$ be a field. The field $F (x)$ of fractions of $F [x]$ is called the field of rational functions in one variable over $F$ .

Example 3.11.10. §

The fraction $\frac{x + 1}{x^{2} + 1}$ is an element of $ℚ (x)$ , as is $\frac{x^{2} - x}{x^{2} + 1}$ , and

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

The following theorem says, in essence, that $Q (R)$ is the smallest field containing $R$ .

Theorem 3.11.11. §

Let $R$ be an integral domain.

a.: The map $ι_{R} : R \to Q (R)$ given by $ι_{R} (r) = \frac{r}{1}$ is an injective ring homomorphism. We use it to identify $R$ with a subring of $Q (R)$ , setting $r = \frac{r}{1}$ .
b.: If $F$ is any field containing $R$ , then is a unique injective ring homomorphism $Q (R) \to F$ that restricts to the inclusion map $R \to F$ .

Proof.

That $ι_{R}$ is a ring homomorphism is easily checked, and it is injective since $\frac{r}{1} = \frac{0}{1}$ implies by definition that $r = 0$ . Now, suppose that $R$ is contained a field $F$ . Define $𝜃 : Q (R) \to F$ by

𝜃 (\frac{a}{b}) = a b^{- 1} .

This is well-defined, as if $𝑎𝑑 = 𝑏𝑐$ for some $c, d \in R$ with $d \neq 0$ , then $a b^{- 1} = c d^{- 1}$ . Moreover, for any quotients $\frac{a}{b}$ and $\frac{c}{d}$ in $Q (R)$ , we have

𝜃 (\frac{a}{b} + \frac{c}{d}) = (𝑎𝑑 + 𝑏𝑐) {(𝑏𝑑)}^{- 1} = (𝑎𝑑 + 𝑏𝑐) \cdot {(𝑏𝑑)}^{- 1} = a b^{- 1} + c d^{- 1} = 𝜃 (\frac{a}{b}) + 𝜃 (\frac{c}{d})

and

𝜃 (\frac{a}{b} \cdot \frac{c}{d}) = 𝑎𝑐 \cdot b d^{- 1} = a b^{- 1} c d^{- 1} = 𝜃 (\frac{a}{b}) \cdot 𝜃 (\frac{c}{d}),

so $𝜃$ is a ring homomorphism. If $𝜃 (\frac{a}{b}) = 0$ , then $a b^{- 1} = 0$ , which implies that $a = 0$ , and hence $\frac{a}{b} = 0$ . Therefore, $𝜃$ is injective. Also, note that

𝜃 (a) = 𝜃 (\frac{a}{1}) = a \cdot 1^{- 1} = a .

Finally, if $χ : Q (R) \to F$ is any homomorphism with which restricts to the inclusion map $R \to F$ , then we have

χ (\frac{a}{b}) = χ (\frac{a}{1} \cdot \frac{1}{b}) = χ (a) \cdot χ {(b)}^{- 1} = a b^{- 1} = 𝜃 (\frac{a}{b}),

so $χ = 𝜃$ . □

Let us make our comment prior to the theorem more precise.

Corollary 3.11.12. §

Let $R$ be an integral domain and $K$ a field containing it. Then there is a smallest subfield $F$ of $K$ containing $R$ , and it is isomorphic to the field of fractions of $R$ via a map $Q (R) \to K$ that extends the identity map on $R$ .

Proof.

The smallest field $F$ containing $R$ is simply the intersection of all fields contained in $K$ and containing $R$ . We then apply Theorem 3.11.11 to the inclusion map $ι_{R} : R \to F$ . The image of the induced map $𝜃 : Q (R) \to F$ is a field containing $R$ and contained in $K$ , so must be $F$ itself. □

Corollary 3.11.12 allows us think more concretely about fields of fractions by speaking of fields of fractions inside a given field.

Definition 3.11.13. §

Let $R$ be an integral domain and $K$ a field containing it. The field of fractions of $R$ in $K$ is the smallest subfield $F$ of $K$ containing $R$ .

The next corollary tells us that it’s okay to think of elements of a field $F$ of the form $a b^{- 1}$ with $a \in F$ and $b \in F^{\times}$ as fractions $\frac{a}{b}$ .

Corollary 3.11.14. §

If $F$ is a field, then it is isomorphic to its own field of fractions.

Proof.

By Corollary 3.11.12, there is a field containing $F$ in $F$ , which of course is $F$ itself, that is isomorphic to the field of fractions $Q (F)$ . □

In other words, the field of fractions of $F$ is $F$ .

Corollary 3.11.15. §

Let $R$ and $Q$ be integral domains, and let $ϕ : R \to R^{'}$ be an injective ring homomorphism. Then there is a unique homomorphism $Q (ϕ) : Q (R) \to Q (R^{'})$ such that $Q (ϕ) (a) = ϕ (a)$ for all $a \in R$ .

Proof.

As the composite map $ι_{R^{'}} \circ ϕ : R \to Q (R^{'})$ is injective, Theorem 3.11.11 tells us that there is a unique injective homomorphism $Q (ϕ) : Q (R) \to Q (R^{'})$ with $Q (ϕ) \circ ι_{R} = ι_{R^{'}} \circ ϕ$ , as desired. □

Example 3.11.16. §

The quotient field of $ℤ [i]$ is isomorphic to $ℚ (i)$ . To see this, note that $ℚ (i)$ is a field containing $ℤ [i]$ , and so there is an inclusion homomorphism $Q (ℤ [i]) \to ℚ (i)$ that takes a fraction of the form $\frac{a + 𝑏𝑖}{c + 𝑑𝑖}$ with $a, b, c, d \in ℤ$ and $(c, d) \neq (0, 0)$ to itself, but every element in $ℚ (i)$ has the form $q + 𝑟𝑖$ with $q, r \in ℚ$ , and any such element can be written as such a fraction with $d = 0$ .

Example 3.11.17. §

The quotient field of $ℤ [x]$ is $ℚ (x)$ , the quotient field of $ℚ [x]$ . To see this, note that the inclusion map $α : ℤ [x] \to ℚ [x]$ sending a polynomial to itself induces an injective homomorphism $Q (α) : Q (ℤ [x]) \to ℚ (x)$ by Corollary 3.11.15. Moreover, for $f, g \in ℤ [x]$ , we have $Q (α) (\frac{f}{g}) = \frac{f}{g}$ by definition. If $f, g \in ℚ [x]$ , then there exists a nonzero $a \in ℤ$ such that $𝑎𝑓, 𝑎𝑔 \in ℤ [x]$ . (Here, $a$ is the least common multiple of the denominators of the coefficients of $P$ and $Q$ , written as fractions in lowest terms.) Then $\frac{P}{Q} = \frac{𝑎𝑃}{𝑎𝑄}$ in $ℚ (x)$ , so $\frac{P}{Q}$ is in the image of the map $Q (α)$ . Therefore, $Q (α)$ is an isomorphism.