Contents / Abstract Algebra / Ideals and Quotient Rings

Chapter 5

Ideals and Quotient Rings

Ideals and the absorption property that makes quotient multiplication well defined; Z and F[x] proved principal via the division algorithm; kernels of ring homomorphisms are exactly the ideals; the first isomorphism theorem for rings and the correspondence theorem; R/P a domain iff P is prime and R/M a field iff M is maximal, both directions; and the Chinese Remainder Theorem for comaximal ideals.

Introduction

Every construction in this chapter is an answer to one question: which subsets of a ring may be collapsed to zero? Collapsing a subset $I$ means declaring $a = b$ whenever $a - b \in I$ , and then asking the arithmetic of $R$ to survive the identification. For addition alone the answer is the one already known from Subgroups and quotient groups: any additive subgroup will do, because $(R, +)$ is abelian and every subgroup of an abelian group is normal. Multiplication is strictly more demanding, and the extra demand is the single axiom that gives this chapter its subject. If $a$ has been identified with $a^{'}$ and $b$ with $b^{'}$ , then $ab$ and $a^{'} b^{'}$ must be identified too; writing $a^{'} = a + s$ and $b^{'} = b + t$ with $s, t \in I$ gives

a^{'} b^{'} - ab = a t + s b + s t,

and there is no reason on earth for the right-hand side to lie in $I$ unless we insist that $I$ swallow every product of one of its own elements with an arbitrary element of $R$ . That insistence is the definition of an ideal, and the displayed identity is the whole chapter in one line. Everything else — the isomorphism theorems, the correspondence between ideals of $R / I$ and ideals of $R$ above $I$ , the dictionary translating "prime" into "the quotient is a domain" and "maximal" into "the quotient is a field" — is bookkeeping around that identity, carefully done.

The reward for the bookkeeping is enormous. Quotients are how new rings are built. The complex numbers are not a new axiom system; they are $R [x]$ with $x^{2} + 1$ set to zero. The field with $8$ elements is $F_{2} [x]$ with $x^{3} + x + 1$ set to zero. Modular arithmetic, which Congruences treats by hand, is $Z$ with $n$ set to zero. In each case a quotient turns a wish — "let there be a square root of $- 1$ ", "let there be a field of size $8$ " — into a theorem about an object we can compute with, and the well-definedness check above is what licenses the computation.

Throughout, ring means a ring with multiplicative identity $1$ , and from the section on quotients onward, unless a statement says otherwise, $R$ is commutative. The non-commutative case is not hard, but it doubles every definition into a left and a right version and obscures the ideas; the definitions below record the distinction once and then set it aside.

5.1Ideals and the ideal test

A subring is a subset closed under the ring operations. That is the wrong notion here, and it is worth seeing exactly why before the right one arrives: $Z$ is a subring of $Q$ , yet there is no sensible ring $Q / Z$ , because $\frac{1}{2} \cdot 1 = \frac{1}{2}$ escapes $Z$ and the identification breaks under multiplication. The correct notion trades closure under multiplication inside the subset for closure under multiplication from outside it.

Definition 5.1 (Left, right and two-sided ideal). Let $R$ be a ring and $I \subseteq R$ . Then $I$ is a left ideal if $(I, +)$ is a subgroup of $(R, +)$ and $r a \in I$ for every $r \in R$ and $a \in I$ ; a right ideal if instead $a r \in I$ for every such $r$ and $a$ ; and a two-sided ideal, written $I ⊴ R$ , if it is both. The condition $r a \in I$ is called absorption.

An ideal with $I \neq = R$ is called proper. When $R$ is commutative the three notions coincide and we say simply ideal.

Note what the definition does not require. An ideal need not contain $1$ — indeed a proper ideal never does, since $1 \in I$ forces $r = r \cdot 1 \in I$ for every $r$ . An ideal is therefore almost never a subring in the sense that includes the identity. This is a feature: the object we want to set to zero should not contain the element that multiplication fixes.

Intuition. Think of an ideal as a drain in the ring. Anything already in the drain stays there, and anything in the ring that touches it multiplicatively is pulled in too. Absorption is one-way and total: there is no product $r \cdot a$ with $a$ in the drain that escapes. Addition inside the drain is unrestricted, which is why the drain is an additive subgroup; it is only multiplication that has a direction to it.

Verifying the definition literally means checking four things — non-emptiness, closure under addition, closure under negation, absorption. Two suffice, and it is worth recording the shortcut, because every subsequent verification in this chapter uses it.

Theorem 5.2 (The ideal test). A subset $I$ of a ring $R$ is a two-sided ideal if and only if

$I \neq = \emptyset$ ;
$a - b \in I$ for all $a, b \in I$ ;
$r a \in I$ and $a r \in I$ for all $r \in R$ , $a \in I$ .

Proof. Necessity is immediate, since an additive subgroup is nonempty and closed under differences. For sufficiency, conditions (1) and (2) are exactly the subgroup test applied to $(R, +)$ : pick $a \in I$ by (1), so $0 = a - a \in I$ by (2); then for $b \in I$ we get $- b = 0 - b \in I$ ; and then for $a, b \in I$ we get $a + b = a - (- b) \in I$ . Hence $(I, +) \leq (R, +)$ . Condition (3) is absorption verbatim.∎

Example 5.3 (The even integers). Show that $2 Z = {2 k : k \in Z}$ is an ideal of $Z$ , and that the odd integers together with $0$ are not.

Solution.

$2 Z$ is nonempty, since $0 = 2 \cdot 0$ lies in it.
If $2 m, 2 n \in 2 Z$ then $2 m - 2 n = 2 (m - n) \in 2 Z$ .
For $k \in Z$ and $2 m \in 2 Z$ , $k (2 m) = 2 (km) \in 2 Z$ , and commutativity gives the right-sided condition for free.
By the theorem The ideal test, $2 Z ⊴ Z$ .
The set $S = {0} \cup {odd integers}$ fails at step (2) already: $3, 1 \in S$ but $3 - 1 = 2 \in / S$ . It also fails absorption, since $2 \cdot 3 = 6 \in / S$ . Either failure is fatal.

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Example 5.4 (A subring that is not an ideal). Show that $Z$ is a subring of $Q$ but not an ideal, and that the same phenomenon occurs for $R$ inside $C$ .

Solution.

$Z$ is closed under $+$ , $-$ and $\times$ and contains $1$ , so it is a subring of $Q$ .
Absorption fails: take $r = \frac{1}{3} \in Q$ and $a = 1 \in Z$ ; then $r a = \frac{1}{3} \in / Z$ .
Likewise $R \subseteq C$ is a subring, but $i \cdot 1 = i \in / R$ .
In both cases the ambient ring is a field, and the theorem Fields have no proper nonzero ideals below explains why no proper nonzero ideal could possibly have existed.

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Pitfall. "Closed under multiplication" and "absorbs multiplication" are different conditions and neither implies the other. $Z \subseteq Q$ is closed but does not absorb. The set ${0, 2, 4} \subseteq Z /6 Z$ both absorbs and is closed; the set $2 Z \subseteq Z$ absorbs and is closed but contains no identity. When you are asked to verify an ideal, verify absorption — multiplying by an element from the ambient ring, not from the subset.

Building new ideals from old

Ideals are stable under the natural set-theoretic and algebraic operations, and the proofs are short applications of the ideal test. These closure properties are what make the collection of ideals of $R$ a lattice, an observation the correspondence theorem will exploit.

Proposition 5.5 (Intersections of ideals). If ${I_{λ}}_{λ \in Λ}$ is any nonempty family of ideals of $R$ , then $⋂_{λ} I_{λ}$ is an ideal of $R$ .

Proof. Write $I = ⋂_{λ} I_{λ}$ . Every $I_{λ}$ contains $0$ , so $0 \in I$ and $I \neq = \emptyset$ . If $a, b \in I$ then $a, b \in I_{λ}$ for each $λ$ , so $a - b \in I_{λ}$ for each $λ$ , so $a - b \in I$ . If $r \in R$ and $a \in I$ then $r a \in I_{λ}$ and $a r \in I_{λ}$ for each $λ$ , hence both lie in $I$ . The theorem The ideal test applies.∎

The union of two ideals is essentially never an ideal — $2 Z \cup 3 Z$ contains $2$ and $3$ but not $5$ — so to combine ideals we take the smallest ideal containing them, which is the sum.

Definition 5.6 (Sum and product of ideals). For ideals $I, J ⊴ R$ define

I + J = {a + b : a \in I, b \in J}, I J = {\sum_{k = 1}^{n} a_{k} b_{k} : n \geq 0, a_{k} \in I, b_{k} \in J} .

The product is the set of finite sums of products, not merely the set of products.

Proposition 5.7 (Sums and products are ideals). For ideals $I, J$ of $R$ , both $I + J$ and $I J$ are ideals, and

I J \subseteq I \cap J \subseteq I \subseteq I + J .

Moreover $I + J$ is the smallest ideal containing $I \cup J$ .

Proof. For $I + J$ : it contains $0 = 0 + 0$ ; if $a + b$ and $a^{'} + b^{'}$ lie in it, their difference is $(a - a^{'}) + (b - b^{'})$ , again of the required form; and $r (a + b) = r a + r b \in I + J$ by absorption in each summand, similarly on the right. For $I J$ : a difference of two finite sums of products is again a finite sum of products, after absorbing signs into the $a_{k}$ ; and $r \sum a_{k} b_{k} = \sum (r a_{k}) b_{k}$ is a finite sum of products because $r a_{k} \in I$ , with the right-hand version using $b_{k} r \in J$ .

For the chain, a typical generator $a_{k} b_{k}$ of $I J$ lies in $I$ by absorption (as $a_{k} \in I$ ) and in $J$ by absorption (as $b_{k} \in J$ ), so $a_{k} b_{k} \in I \cap J$ ; sums of such elements stay in the ideal $I \cap J$ , giving $I J \subseteq I \cap J$ . The remaining inclusions are immediate from $0 \in I$ and $0 \in J$ .

Finally, any ideal $K$ containing $I \cup J$ contains every sum $a + b$ with $a \in I$ , $b \in J$ , hence contains $I + J$ ; and $I + J$ does contain $I \cup J$ . So it is the smallest such ideal.∎

The inclusion $I J \subseteq I \cap J$ can be strict or an equality, and which occurs is a genuine arithmetic question. In $Z$ , $(m) (n) = (mn)$ while $(m) \cap (n) = (lcm (m, n))$ , so the two agree exactly when $g cd (m, n) = 1$ . This coprimality condition is the hypothesis of the Chinese Remainder Theorem, proved in the final section, and it is no coincidence.

Example 5.8 (Sum, product and intersection in $Z$ ). Compute $I + J$ , $I J$ and $I \cap J$ for $I = (4)$ and $J = (6)$ in $Z$ .

Solution.

$I + J = {4 a + 6 b}$ is the set of integer combinations of $4$ and $6$ , which by Bézout's identity (see The Euclidean algorithm) is $(g cd (4, 6)) = (2)$ .
$I J$ is generated by the single product $4 \cdot 6 = 24$ , so $I J = (24)$ .
$I \cap J$ consists of the integers divisible by both $4$ and $6$ , that is by $lcm (4, 6) = 12$ , so $I \cap J = (12)$ .
The chain reads $(24) ⊊ (12) ⊊ (4) ⊊ (2)$ , confirming $I J ⊊ I \cap J$ strictly. The failure is measured by $g cd (4, 6) = 2 \neq = 1$ , since $g cd \cdot lcm = 24$ .

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Ideals detect fields

The extreme cases ${0}$ and $R$ are ideals of every ring; they are called the trivial ideals. A ring with no others is as far from having interesting quotients as possible, and the following pair of results identifies exactly which rings those are.

Theorem 5.9 (Fields have no proper nonzero ideals). Let $F$ be a field and $I ⊴ F$ . Then $I = {0}$ or $I = F$ .

Proof. Suppose $I \neq = {0}$ and pick $a \in I$ with $a \neq = 0$ . Since $F$ is a field, $a^{- 1}$ exists in $F$ , and absorption gives $1 = a^{- 1} a \in I$ . Then for any $r \in F$ , absorption again gives $r = r \cdot 1 \in I$ . Hence $I = F$ .∎

The proof used only that the nonzero element $a$ happened to be a unit, which proves slightly more and is worth extracting.

Corollary 5.10 (A proper ideal contains no unit). If $I$ is a proper ideal of a ring $R$ and $u \in I$ , then $u$ is not a unit. Consequently $I \subseteq R ∖ R^{\times}$ , where $R^{\times}$ is the group of units.

Proof. If $u \in I$ were a unit, then $1 = u^{- 1} u \in I$ by absorption, whence $r = r \cdot 1 \in I$ for all $r$ and $I = R$ , contradicting properness.∎

The converse of Fields have no proper nonzero ideals is true in the commutative case, and it is the first place where an ideal-theoretic hypothesis is converted back into concrete arithmetic. It is also the engine of the maximal-ideal theorem later on.

Theorem 5.11 (Trivial ideals characterise fields). Let $R$ be a commutative ring with $1 \neq = 0$ . Then $R$ is a field if and only if its only ideals are ${0}$ and $R$ .

Proof. ( $\Rightarrow$ ) is the theorem Fields have no proper nonzero ideals.

( $\Leftarrow$ ) Let $a \in R$ be nonzero; we must produce $a^{- 1}$ . Consider $(a) = {r a : r \in R}$ . This is an ideal: it is nonempty, $r a - r^{'} a = (r - r^{'}) a$ , and $s (r a) = (sr) a$ . It is nonzero, because $a = 1 \cdot a \in (a)$ and $a \neq = 0$ . By hypothesis the only nonzero ideal is $R$ itself, so $(a) = R$ . In particular $1 \in (a)$ , meaning $1 = ba$ for some $b \in R$ . By commutativity $ab = ba = 1$ , so $b = a^{- 1}$ . Every nonzero element is invertible and $1 \neq = 0$ , so $R$ is a field.∎

Remark. The hypothesis $1 \neq = 0$ excludes the zero ring, whose unique ideal is ${0} = R$ and which is not a field by convention. Commutativity is also essential to the converse as stated: the ring $M_{2} (R)$ of real $2 \times 2$ matrices has only the trivial two-sided ideals, yet it is not a division ring, since $(1000)$ is a nonzero non-unit. What is true in general is that a ring with only trivial two-sided ideals is called simple, and matrix rings over fields are the standard examples.

Exercises for §5.1 →

5.2Generators, principal ideals and $Z$

The proof of Trivial ideals characterise fields quietly constructed the smallest ideal containing a given element. That construction deserves a name and a general form.

Definition 5.12 (Generated ideal). Let $R$ be commutative with $1$ and let $a_{1}, \dots, a_{n} \in R$ . The ideal generated by $a_{1}, \dots, a_{n}$ is

(a_{1}, \dots, a_{n}) = {\sum_{k = 1}^{n} r_{k} a_{k} : r_{k} \in R} .

An ideal of the form $(a)$ is principal, and $a$ is a generator of it. An ideal admitting a finite generating set is finitely generated.

Proposition 5.13 (The generated ideal is the smallest one). The set $(a_{1}, \dots, a_{n})$ is an ideal containing each $a_{k}$ , and it is contained in every ideal of $R$ that contains all the $a_{k}$ .

Proof. It is an ideal by the ideal test: it contains $0$ (all $r_{k} = 0$ ); the difference of $\sum r_{k} a_{k}$ and $\sum r_{k}^{'} a_{k}$ is $\sum (r_{k} - r_{k}^{'}) a_{k}$ ; and $s \sum r_{k} a_{k} = \sum (s r_{k}) a_{k}$ . It contains $a_{j}$ by taking $r_{j} = 1$ and the rest $0$ , which uses the presence of $1$ in $R$ . If $J$ is an ideal with every $a_{k} \in J$ , then $r_{k} a_{k} \in J$ by absorption and the sum $\sum r_{k} a_{k} \in J$ because $J$ is closed under addition; hence $(a_{1}, \dots, a_{n}) \subseteq J$ .∎

Generators are not unique. In $Z$ , $(3) = (- 3)$ ; in any commutative ring $(a) = (u a)$ for a unit $u$ , since $a = u^{- 1} (u a)$ and $u a = u \cdot a$ . In an integral domain the converse holds too: $(a) = (b)$ with $a, b \neq = 0$ forces $a = u b$ for a unit $u$ , which is why associates are the right equivalence for divisibility, as in Integral domains and fields.

Definition 5.14 (Principal ideal domain). An integral domain in which every ideal is principal is a principal ideal domain, abbreviated PID.

The most important example is the one everybody learned in primary school, dressed in new clothes. Its proof is the division algorithm and nothing else, which is the model for every subsequent PID argument.

Theorem 5.15 (Every ideal of $Z$ is principal). Let $I ⊴ Z$ . Then $I = (d)$ where $d = 0$ if $I = {0}$ , and otherwise $d$ is the least positive element of $I$ . In particular $Z$ is a PID.

Proof. If $I = {0}$ then $I = (0)$ and we are done, so assume $I$ contains some $a \neq = 0$ . Since $I$ is an additive subgroup it contains $- a$ as well, so $I$ contains a positive integer, and the set $I \cap Z_{> 0}$ is a nonempty set of positive integers. By the well-ordering principle it has a least element $d$ .

We claim $I = (d)$ . Since $d \in I$ , absorption gives $k d \in I$ for every $k \in Z$ , so $(d) \subseteq I$ . Conversely let $n \in I$ . By the division algorithm there are $q, r \in Z$ with

n = q d + r, 0 \leq r < d .

Now $q d \in I$ and $n \in I$ , so $r = n - q d \in I$ because $I$ is closed under subtraction. If $r > 0$ then $r$ is a positive element of $I$ strictly smaller than $d$ , contradicting minimality. Hence $r = 0$ and $n = q d \in (d)$ . Therefore $I \subseteq (d)$ and the two sets are equal. Since $Z$ is an integral domain, it is a PID.∎

Corollary 5.16 (Bézout from the ideal picture). For $m, n \in Z$ not both zero, $(m, n) = (g cd (m, n))$ . In particular there exist $x, y \in Z$ with $g cd (m, n) = m x + n y$ .

Proof. By the theorem Every ideal of $Z$ is principal, $(m, n) = (d)$ for the least positive $d$ in $(m, n)$ . Since $m, n \in (d)$ , $d$ divides both $m$ and $n$ , so $d$ is a common divisor. Conversely $d \in (m, n)$ means $d = m x + n y$ for some integers $x, y$ , and any common divisor $c$ of $m$ and $n$ divides $m x + n y = d$ . So $d$ is a common divisor divisible by every common divisor: $d = g cd (m, n)$ . The representation $d = m x + n y$ is Bézout's identity.∎

This is a clean example of the general strategy of the subject. The gcd was originally defined by a divisibility condition, an awkward thing to manipulate; the ideal $(m, n)$ is defined by closure properties, easy to manipulate; and the theorem identifies them. Bézout's identity, which requires a construction when proved by the Euclidean algorithm, falls out of the identification for free.

Example 5.17 (A generator of an ideal of $Z$ ). Find a single generator for $I = (18, 30, 42) ⊴ Z$ , and determine whether $7 \in I$ .

Solution.

Extending Bézout from the ideal picture to three generators, $(18, 30, 42) = (g cd (18, 30, 42))$ , since the set of integer combinations of the three is again an ideal and hence principal.
$g cd (18, 30) = 6$ and $g cd (6, 42) = 6$ , so $I = (6)$ .
A check: $6 = 18 \cdot 2 + 30 \cdot (- 1) + 42 \cdot 0$ , so $6$ genuinely lies in $I$ .
Since $6 ∤ 7$ , we have $7 \in / (6) = I$ .

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Theorem 5.18 (Polynomials over a field form a PID). Let $F$ be a field and $I ⊴ F [x]$ a nonzero ideal. Then $I = (g)$ where $g$ is any nonzero element of $I$ of least degree, and $g$ is unique if required to be monic. Hence $F [x]$ is a PID.

Proof. Choose $g \in I ∖ {0}$ of least degree; scaling by the inverse of its leading coefficient — legitimate because $F$ is a field and by absorption keeps us inside $I$ — we may take $g$ monic. As $g \in I$ , absorption gives $(g) \subseteq I$ .

Let $f \in I$ . Since $g$ is monic, the division algorithm in $F [x]$ (see Polynomial rings) yields $f = q g + r$ with $r = 0$ or $de g r < de g g$ . As before $r = f - q g \in I$ . If $r \neq = 0$ then $r$ is a nonzero element of $I$ of degree smaller than $de g g$ , contradicting the choice of $g$ . So $r = 0$ and $f \in (g)$ , giving $I \subseteq (g)$ .

For uniqueness, if $(g) = (h)$ with both monic then $g ∣ h$ and $h ∣ g$ , so $h = c g$ with $c \in F^{\times}$ ; comparing leading coefficients gives $c = 1$ . Finally $F [x]$ is an integral domain because $F$ is, so it is a PID.∎

Both PID proofs have the same skeleton: a notion of size (absolute value, degree) taking values in the non-negative integers, a division algorithm producing a remainder strictly smaller than the divisor, and well-ordering to force the remainder to vanish. A ring carrying such a size function is a Euclidean domain, and the argument above proves in general that every Euclidean domain is a PID. The converse fails, but the counterexample $Z [\frac{1 + - 19}{2}]$ is well beyond this chapter.

Not every ring is a PID, and the smallest interesting failure is one ring away from $Z$ .

Proposition 5.19 (A non-principal ideal). In $Z [x]$ the ideal $I = (2, x)$ is not principal.

Proof. First, $I$ is proper: every element of $I$ has the form $2 f (x) + xg (x)$ , whose constant term is $2 f (0)$ , an even integer. So no polynomial with odd constant term lies in $I$ ; in particular $1 \in / I$ and $I \neq = Z [x]$ .

Suppose $I = (h)$ for some $h \in Z [x]$ . Since $2 \in (h)$ we have $2 = h \cdot u$ for some $u$ , and comparing degrees in the integral domain $Z [x]$ forces $de g h = 0$ , so $h$ is a constant $c$ with $c ∣ 2$ in $Z$ , i.e. $c \in {\pm 1, \pm 2}$ . Since $x \in (h) = (c)$ we would need $x = c k (x)$ , which for $c = \pm 2$ is impossible because the coefficient of $x$ on the right is even. Hence $c = \pm 1$ and $I = (1) = Z [x]$ , contradicting properness. So no generator exists.∎

Remark. The obstruction is dimensional. Geometrically, $Z [x]$ behaves like a two-dimensional object and $(2, x)$ cuts it down by two independent conditions — "reduce mod $2$ " and "set $x = 0$ " — which no single equation can express. The same phenomenon occurs in $F [x, y]$ , where $(x, y)$ is not principal for exactly the same reason. By contrast, in a PID every ideal is a single condition, which is why divisibility in $Z$ and $F [x]$ is so much simpler than in $Z [x]$ .

Exercises for §5.2 →

5.3Quotient rings and the well-definedness crux

We can now carry out the construction promised in the introduction. Let $I$ be an ideal of $R$ . Since $(I, +)$ is a subgroup of the abelian group $(R, +)$ , the quotient group $R / I$ exists and consists of the additive cosets $r + I = {r + a : a \in I}$ , with $r + I = s + I$ precisely when $r - s \in I$ . All of that is Subgroups and quotient groups applied to $(R, +)$ ; nothing about multiplication has been used yet.

Notation. We write $r + I$ for the coset, $\overset{r}{ˉ}$ when the ideal is clear, and $r \equiv s (mod I)$ for the relation $r - s \in I$ . For $R = Z$ and $I = (n)$ this is exactly the congruence notation of Congruences.

Definition 5.20 (Quotient ring). Let $I ⊴ R$ . The quotient ring $R / I$ is the set of cosets ${r + I : r \in R}$ equipped with

(r + I) + (s + I) = (r + s) + I, (r + I) (s + I) = rs + I .

The definition as written is not yet legitimate, because both formulas are stated in terms of representatives $r$ and $s$ , while the objects being combined are the cosets. A different choice of representative must give the same answer, or the formula defines nothing at all. This is the crux of the chapter and deserves to be proved slowly.

Theorem 5.21 (The quotient operations are well defined). Let $I$ be a two-sided ideal of $R$ and suppose $r + I = r^{'} + I$ and $s + I = s^{'} + I$ . Then

(r + s) + I = (r^{'} + s^{'}) + I and rs + I = r^{'} s^{'} + I .

Moreover, if $I$ is merely an additive subgroup that is not an ideal, the multiplication formula fails to be well defined.

Proof. The hypotheses say $r^{'} - r \in I$ and $s^{'} - s \in I$ ; write $r^{'} = r + u$ and $s^{'} = s + v$ with $u, v \in I$ .

Addition. $(r^{'} + s^{'}) - (r + s) = u + v \in I$ , because $I$ is closed under addition. Hence $(r^{'} + s^{'}) + I = (r + s) + I$ . Only the subgroup property was used.

Multiplication. Expand:

r^{'} s^{'} = (r + u) (s + v) = rs + r v + u s + uv,

so that

r^{'} s^{'} - rs = r v + u s + uv .

Now examine the three terms one at a time. The term $r v$ has $v \in I$ and $r \in R$ , so $r v \in I$ by left absorption. The term $u s$ has $u \in I$ and $s \in R$ , so $u s \in I$ by right absorption. The term $uv$ has $u \in I$ and $v \in R$ , so $uv \in I$ by left absorption again — alternatively, $uv \in I$ because $v \in I$ and $u \in R$ . Since $I$ is closed under addition, the sum $r v + u s + uv$ lies in $I$ , that is $r^{'} s^{'} - rs \in I$ , that is $r^{'} s^{'} + I = rs + I$ .

Note precisely which axiom did which job: closure under addition handled the recombination, left absorption handled $r v$ , and right absorption handled $u s$ . A one-sided ideal in a non-commutative ring is therefore not enough; both sides are needed, which is why quotient rings are formed only by two-sided ideals.

Failure without absorption. Take $R = Q$ and $I = Z$ , an additive subgroup which by the example A subring that is not an ideal is not an ideal. Then $\frac{1}{2} + Z = \frac{3}{2} + Z$ and $\frac{1}{2} + Z = \frac{1}{2} + Z$ , but the products are $\frac{1}{4} + Z$ and $\frac{3}{4} + Z$ , and $\frac{3}{4} - \frac{1}{4} = \frac{1}{2} \in / Z$ . The two cosets are different, so the formula assigns two different values to the same pair of inputs and defines no operation at all.∎

Pitfall. Students often check only the additive half and declare the quotient constructed. The additive half is automatic for any subgroup; it is the multiplicative half that fails, and it fails silently — the formula still produces an output, just not a consistent one. Whenever you see a quotient by a subset, the question to ask is not "is it closed?" but "does it absorb?".

Theorem 5.22 ( $R / I$ is a ring and $π$ is a surjective homomorphism). With the operations above, $R / I$ is a ring with zero element $0 + I = I$ and identity $1 + I$ . The map $π : R \to R / I$ , $π (r) = r + I$ , is a surjective ring homomorphism with $ker π = I$ . If $R$ is commutative, so is $R / I$ .

Proof. Associativity, commutativity of addition, distributivity and the behaviour of the identity are all inherited one line at a time from $R$ through the representative formulas; for instance

((r + I) (s + I)) (t + I) = (rs + I) (t + I) = (rs) t + I = r (s t) + I = (r + I) ((s + I) (t + I)),

and each such computation is legitimate because the theorem The quotient operations are well defined guarantees the intermediate cosets do not depend on the representatives chosen. The additive group $R / I$ is a group by the corresponding fact for quotient groups. The coset $0 + I = I$ is additively neutral and $1 + I$ satisfies $(1 + I) (r + I) = r + I$ .

The map $π$ is a homomorphism because the operations were defined to make it one: $π (r) + π (s) = (r + I) + (s + I) = (r + s) + I = π (r + s)$ , likewise for products, and $π (1) = 1 + I$ . It is surjective since every coset is $π$ of any of its representatives. Finally $π (r) = 0 + I$ means $r - 0 \in I$ , i.e. $r \in I$ , so $ker π = I$ . Commutativity passes to the quotient because $rs + I = sr + I$ when $rs = sr$ .∎

Corollary 5.23 (Every ideal is a kernel). A subset $I \subseteq R$ is a two-sided ideal if and only if $I = ker ϕ$ for some ring homomorphism $ϕ$ out of $R$ .

Proof. If $I$ is an ideal, take $ϕ = π : R \to R / I$ ; the theorem $R / I$ is a ring and $π$ is a surjective homomorphism gives $ker π = I$ . Conversely a kernel is an ideal by the proposition Kernels are ideals below.∎

Figure 5.24. The four cosets of

(4)

partition

Z

into residue classes. Points in a common row are identified in

Z /4 Z

; the ideal

(4)

is the bottom row.

Example 5.25 (The quotient $Z /12 Z$ and its zero divisors). Describe $Z / (12)$ , find its units and its zero divisors, and decide whether it is a field or an integral domain.

Solution.

The cosets are $\overset{ˉ}{0}, \overset{ˉ}{1}, \dots, \overline{11}$ , one for each remainder on division by $12$ , so $∣ Z / (12) ∣ = 12$ .
A class $\overset{a}{ˉ}$ is a unit exactly when $g cd (a, 12) = 1$ , since $a x \equiv 1 (mod 12)$ is solvable precisely then by Bézout from the ideal picture. That gives the four units $\overset{ˉ}{1}, \overset{ˉ}{5}, \overset{ˉ}{7}, \overline{11}$ .
The classes $\overset{a}{ˉ}$ with $1 < g cd (a, 12) < 12$ are zero divisors: for instance $\overset{ˉ}{3} \cdot \overset{ˉ}{4} = \overline{12} = \overset{ˉ}{0}$ with neither factor zero.
Since zero divisors exist, $Z / (12)$ is not an integral domain, hence not a field. Equivalently, $12$ is not prime, and the theorem Maximal ideals and fields will make that equivalence a theorem rather than a coincidence.

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Example 5.26 (Computing in $R [x] / (x^{2} + 1)$ ). Show that $R [x] / (x^{2} + 1) ≅ C$ , and use the isomorphism to invert the class of $1 + x$ .

Solution.

Normal forms. Divide any $f \in R [x]$ by the monic polynomial $x^{2} + 1$ : $f = q (x^{2} + 1) + (a + b x)$ with $a, b \in R$ . Since $q (x^{2} + 1) \in (x^{2} + 1)$ , every coset contains a representative $a + b x$ of degree $< 2$ , and it is unique, since a difference of two such lying in $(x^{2} + 1)$ would be a nonzero multiple of $x^{2} + 1$ of degree $< 2$ .
The defining relation. Writing $α = \overset{x}{ˉ}$ , the coset of $x$ , we have $α^{2} + 1 = \overline{x^{2} + 1} = \overset{ˉ}{0}$ , so $α^{2} = - 1$ .
Matching multiplication. $(a + b α) (c + d α) = a c + (a d + b c) α + b d α^{2} = (a c - b d) + (a d + b c) α$ , which is precisely the multiplication rule of $C$ under $a + b α \mapsto a + bi$ . The map is additive and bijective by step 1, so it is a ring isomorphism.
Inverting $1 + x$ . Under the isomorphism $\overline{1 + x} \mapsto 1 + i$ , whose inverse is $\frac{1 - i}{2}$ . Translating back, $(\overline{1 + x})^{- 1} = \frac{1}{2} - \frac{1}{2} \overset{x}{ˉ}$ . Checking directly: $(1 + x) (\frac{1}{2} - \frac{1}{2} x) = \frac{1}{2} (1 - x^{2}) = \frac{1}{2} (2 - (x^{2} + 1)) \equiv 1 (mod (x^{2} + 1))$ .

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Theorem 5.27 (Structure of $F [x] / (f)$ ). Let $F$ be a field and $f \in F [x]$ with $de g f = n \geq 1$ . Then every element of $F [x] / (f)$ is represented by a unique polynomial of degree $< n$ , so $F [x] / (f)$ is an $F$ -vector space of dimension $n$ with basis $\overset{ˉ}{1}, \overset{x}{ˉ}, \dots, \overline{x^{n - 1}}$ . If $F$ is finite with $∣ F ∣ = q$ , then $∣ F [x] / (f) ∣ = q^{n}$ .

Proof. Existence. Given $g \in F [x]$ , division by the nonzero $f$ gives $g = q f + r$ with $r = 0$ or $de g r < n$ , and $g - r = q f \in (f)$ , so $\overset{g}{ˉ} = \overset{r}{ˉ}$ .

Uniqueness. If $\overset{r}{ˉ} = \overset{s}{ˉ}$ with both of degree $< n$ , then $r - s \in (f)$ , so $f ∣ (r - s)$ . But $de g (r - s) < n = de g f$ unless $r - s = 0$ , and a nonzero multiple of $f$ has degree at least $n$ because $F [x]$ is an integral domain. Hence $r = s$ .

Consequently the classes $\overset{ˉ}{1}, \dots, \overline{x^{n - 1}}$ span and are linearly independent over $F$ , giving dimension $n$ . A vector space of dimension $n$ over a field with $q$ elements has exactly $q^{n}$ elements, one for each choice of $n$ coordinates.∎

Example 5.28 (A field with eight elements). Let $f (x) = x^{3} + x + 1 \in F_{2} [x]$ . Determine $∣ F_{2} [x] / (f) ∣$ and compute $\overset{x}{ˉ}^{4}$ in normal form.

Solution.

$f$ has degree $3$ and $∣ F_{2} ∣ = 2$ , so by the theorem Structure of $F [x] / (f)$ the quotient has $2^{3} = 8$ elements, namely $a_{0} + a_{1} \overset{x}{ˉ} + a_{2} \overset{x}{ˉ}^{2}$ with $a_{i} \in {0, 1}$ .
In the quotient $\overset{x}{ˉ}^{3} = - \overset{x}{ˉ} - \overset{ˉ}{1} = \overset{x}{ˉ} + \overset{ˉ}{1}$ , since $- 1 = 1$ in $F_{2}$ .
Therefore $\overset{x}{ˉ}^{4} = \overset{x}{ˉ} \cdot \overset{x}{ˉ}^{3} = \overset{x}{ˉ} (\overset{x}{ˉ} + \overset{ˉ}{1}) = \overset{x}{ˉ}^{2} + \overset{x}{ˉ}$ , which is already in normal form.
Since $f$ has no root in $F_{2}$ ( $f (0) = f (1) = 1$ ) and has degree $3$ , it is irreducible, so by the theorem Maximal ideals and fields together with Irreducibles generate maximal ideals in a PID this quotient is the field $F_{8}$ .

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Exercises for §5.3 →

5.4Prime and maximal ideals

Two properties of an ideal are singled out because each is exactly equivalent to a property of the quotient. The definitions are stated inside $R$ ; the theorems translate them outside, into $R / I$ , and it is the translation that makes them usable.

Definition 5.29 (Prime ideal). A proper ideal $P$ of a commutative ring $R$ is prime if for all $a, b \in R$ ,

ab \in P ⟹ a \in P or b \in P .

Definition 5.30 (Maximal ideal). A proper ideal $M$ of a ring $R$ is maximal if the only ideals $J$ with $M \subseteq J \subseteq R$ are $J = M$ and $J = R$ ; that is, $M$ is maximal among proper ideals under inclusion.

The prime condition is a direct abstraction of Euclid's lemma from Prime numbers and factorisation: $p ∣ ab \Rightarrow p ∣ a$ or $p ∣ b$ , which in ideal language reads $ab \in (p) \Rightarrow a \in (p)$ or $b \in (p)$ . The maximal condition has no arithmetic ancestor; it is purely order-theoretic and is why the correspondence theorem is needed to work with it.

Theorem 5.31 (Prime ideals and integral domains). Let $R$ be commutative with $1 \neq = 0$ and let $P$ be a proper ideal. Then

P is prime ⟺ R / P is an integral domain .

Proof. Throughout, $R / P$ is commutative with identity $\overset{ˉ}{1}$ , and $\overset{ˉ}{1} \neq = \overset{ˉ}{0}$ precisely because $P$ is proper (if $\overset{ˉ}{1} = \overset{ˉ}{0}$ then $1 \in P$ and $P = R$ ). So the only thing at issue is the absence of zero divisors.

( $\Rightarrow$ ) Suppose $P$ is prime and $\overset{a}{ˉ} \overset{ˉ}{b} = \overset{ˉ}{0}$ in $R / P$ . By definition of the quotient multiplication, $\overset{a}{ˉ} \overset{ˉ}{b} = \overline{ab}$ , so $\overline{ab} = \overset{ˉ}{0}$ , which means $ab \in P$ . Primeness gives $a \in P$ or $b \in P$ , that is $\overset{a}{ˉ} = \overset{ˉ}{0}$ or $\overset{ˉ}{b} = \overset{ˉ}{0}$ . So $R / P$ has no zero divisors and is an integral domain.

( $\Leftarrow$ ) Suppose $R / P$ is an integral domain and let $ab \in P$ . Then $\overline{ab} = \overset{ˉ}{0}$ , i.e. $\overset{a}{ˉ} \overset{ˉ}{b} = \overset{ˉ}{0}$ in $R / P$ . Having no zero divisors, $R / P$ forces $\overset{a}{ˉ} = \overset{ˉ}{0}$ or $\overset{ˉ}{b} = \overset{ˉ}{0}$ , i.e. $a \in P$ or $b \in P$ . Since $P$ is proper by hypothesis, $P$ is prime.∎

Theorem 5.32 (Maximal ideals and fields). Let $R$ be commutative with $1 \neq = 0$ and let $M$ be a proper ideal. Then

M is maximal ⟺ R / M is a field .

Proof. The bridge in both directions is the theorem Trivial ideals characterise fields, applied to $R / M$ , together with the following observation: the ideals of $R / M$ are exactly the sets $J / M$ for ideals $J$ of $R$ containing $M$ . We prove the two directions directly, so that the argument does not presuppose the correspondence theorem.

( $\Rightarrow$ ) Let $M$ be maximal. Since $M$ is proper, $\overset{ˉ}{1} \neq = \overset{ˉ}{0}$ in $R / M$ . Let $\overset{a}{ˉ} \neq = \overset{ˉ}{0}$ , so $a \in / M$ . Consider $J = M + (a)$ , an ideal by the proposition Sums and products are ideals. It contains $M$ , and it contains $a \in / M$ , so $J ⊋ M$ . Maximality forces $J = R$ , hence $1 \in M + (a)$ : there are $m \in M$ and $r \in R$ with $1 = m + r a$ . Passing to the quotient, $\overset{ˉ}{1} = \overset{m}{ˉ} + \overset{r}{ˉ} \overset{a}{ˉ} = \overset{r}{ˉ} \overset{a}{ˉ}$ , since $\overset{m}{ˉ} = \overset{ˉ}{0}$ . So $\overset{a}{ˉ}$ is invertible with inverse $\overset{r}{ˉ}$ . Every nonzero class is a unit and $\overset{ˉ}{1} \neq = \overset{ˉ}{0}$ , so $R / M$ is a field.

( $\Leftarrow$ ) Let $R / M$ be a field, and suppose $J$ is an ideal with $M ⊊ J \subseteq R$ . Pick $a \in J ∖ M$ . Then $\overset{a}{ˉ} \neq = \overset{ˉ}{0}$ in $R / M$ , so there is $\overset{r}{ˉ}$ with $\overset{r}{ˉ} \overset{a}{ˉ} = \overset{ˉ}{1}$ , i.e. $r a - 1 \in M \subseteq J$ . Also $r a \in J$ by absorption, since $a \in J$ . Therefore $1 = r a - (r a - 1) \in J$ , and $J = R$ by the corollary A proper ideal contains no unit. So no ideal lies strictly between $M$ and $R$ , and $M$ is maximal. Note that $\overset{ˉ}{1} \neq = \overset{ˉ}{0}$ in the field $R / M$ gives $1 \in / M$ , so $M$ is proper.∎

These two theorems are the reason the chapter exists. Each converts a statement quantified over all elements or all ideals of $R$ into a statement about a single, usually much smaller, ring. The immediate payoff is a relation between the two conditions that would be awkward to see inside $R$ .

Corollary 5.33 (Maximal implies prime). In a commutative ring with $1 \neq = 0$ , every maximal ideal is prime.

Proof. Let $M$ be maximal. By the theorem Maximal ideals and fields, $R / M$ is a field. Every field is an integral domain: if $ab = 0$ with $a \neq = 0$ , multiply by $a^{- 1}$ to get $b = 0$ . So $R / M$ is an integral domain, and by the theorem Prime ideals and integral domains, $M$ is prime.∎

Proposition 5.34 (Prime does not imply maximal). The ideal $(0)$ is prime but not maximal in $Z$ , and the ideal $(x)$ is prime but not maximal in $Z [x]$ .

Proof. For $(0) ⊴ Z$ : the quotient $Z / (0) ≅ Z$ is an integral domain but not a field, since $2$ has no integer inverse. By the two translation theorems, $(0)$ is prime and not maximal. Concretely, $(0) ⊊ (2) ⊊ Z$ exhibits an ideal strictly between.

For $(x) ⊴ Z [x]$ : evaluation at $0$ gives a surjection $Z [x] \to Z$ with kernel $(x)$ , so $Z [x] / (x) ≅ Z$ by the theorem First isomorphism theorem for rings. Again this is a domain and not a field, so $(x)$ is prime and not maximal; indeed $(x) ⊊ (2, x) ⊊ Z [x]$ .∎

Figure 5.35. The ideals of

Z

above

(0)

, ordered by inclusion. Maximal ideals sit one step below the whole ring;

(6)

is neither prime nor maximal, and

(0)

is prime but sits at the bottom.

Theorem 5.36 (Primes and maximals of $Z$ ). The prime ideals of $Z$ are $(0)$ and $(p)$ for $p$ a prime number; the maximal ideals are exactly the $(p)$ .

Proof. Every ideal of $Z$ is $(n)$ with $n \geq 0$ by the theorem Every ideal of $Z$ is principal. We have $Z / (1) = {0}$ , so $(1)$ is not proper. For $n \geq 2$ , $Z / (n)$ is a domain if and only if $n$ is prime: if $n = ab$ with $1 < a, b < n$ then $\overset{a}{ˉ} \overset{ˉ}{b} = \overset{ˉ}{0}$ with both factors nonzero, so $n$ composite fails; and if $n = p$ is prime then $p ∣ ab$ implies $p ∣ a$ or $p ∣ b$ by Euclid's lemma, so there are no zero divisors. For $n = 0$ , $Z / (0) ≅ Z$ is a domain. The theorem Prime ideals and integral domains now gives the list of primes.

For maximality, $Z / (p)$ is a finite integral domain, hence a field: for $\overset{a}{ˉ} \neq = \overset{ˉ}{0}$ the map $\overset{x}{ˉ} \mapsto \overset{a}{ˉ} \overset{x}{ˉ}$ is injective on a finite set, hence surjective, so $\overset{a}{ˉ} \overset{x}{ˉ} = \overset{ˉ}{1}$ has a solution. By the theorem Maximal ideals and fields, $(p)$ is maximal. And $(0)$ is not maximal since $(0) ⊊ (2) ⊊ Z$ .∎

Corollary 5.37 (Irreducibles generate maximal ideals in a PID). Let $R$ be a PID and $p \in R$ a nonzero non-unit. Then $(p)$ is maximal if and only if $p$ is irreducible. In particular, for a field $F$ and $f \in F [x]$ of degree $\geq 1$ , $F [x] / (f)$ is a field exactly when $f$ is irreducible.

Proof. ( $\Leftarrow$ ) Let $p$ be irreducible and let $J \supseteq (p)$ be an ideal. Since $R$ is a PID, $J = (d)$ for some $d$ , and $p \in (d)$ gives $p = d c$ . Irreducibility forces $d$ or $c$ to be a unit. If $d$ is a unit then $J = R$ ; if $c$ is a unit then $d = p c^{- 1}$ is an associate of $p$ and $J = (p)$ . Also $(p) \neq = R$ because $p$ is not a unit. So $(p)$ is maximal.

( $\Rightarrow$ ) Let $(p)$ be maximal and suppose $p = ab$ . Then $(p) \subseteq (a)$ , so by maximality $(a) = (p)$ or $(a) = R$ . In the second case $a$ is a unit. In the first, $a = p u$ for some $u$ , whence $p = ab = p u b$ ; cancelling $p \neq = 0$ in the domain $R$ gives $u b = 1$ , so $b$ is a unit. Hence $p$ is irreducible.

The polynomial statement is this applied to the PID $F [x]$ of the theorem Polynomials over a field form a PID, combined with Maximal ideals and fields.∎

Example 5.38 (Classifying small quotients of $Z [x]$ ). Decide for each of $(x)$ , $(x, 5)$ and $(x^{2})$ in $Z [x]$ whether it is prime, maximal, or neither.

Solution.

$Z [x] / (x) ≅ Z$ , a domain but not a field, so $(x)$ is prime and not maximal.
$Z [x] / (x, 5)$ : setting $x = 0$ and $5 = 0$ leaves $Z / (5) = F_{5}$ , a field. So $(x, 5)$ is maximal, and prime by the corollary Maximal implies prime.
$Z [x] / (x^{2})$ contains $\overset{x}{ˉ} \neq = \overset{ˉ}{0}$ with $\overset{x}{ˉ}^{2} = \overset{ˉ}{0}$ , a nonzero nilpotent, hence a zero divisor. So the quotient is not a domain and $(x^{2})$ is neither prime nor maximal.
The pattern generalises: $(x^{n})$ for $n \geq 2$ is never prime, because $x \cdot x^{n - 1} \in (x^{n})$ while neither factor is.

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Example 5.39 (When is $Z / (n)$ a field?). For which $n \geq 2$ is $Z / (n)$ a field, and how many ideals does $Z / (n)$ have?

Solution.

By the theorem Maximal ideals and fields, $Z / (n)$ is a field exactly when $(n)$ is maximal, which by Primes and maximals of $Z$ happens exactly when $n$ is prime.
By the correspondence theorem below, the ideals of $Z / (n)$ correspond to ideals of $Z$ containing $(n)$ , i.e. to ideals $(d)$ with $d ∣ n$ and $d \geq 1$ .
So the number of ideals of $Z / (n)$ is the number of positive divisors of $n$ , namely $τ (n)$ .
For $n = 12 = 2^{2} \cdot 3$ this is $(2 + 1) (1 + 1) = 6$ ideals, matching the six nodes of the lattice figure below.

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Figure 5.40. The six ideals of

Z /12 Z

, one for each divisor of

12

, drawn with an edge whenever one contains the other with nothing in between. The lattice is the divisor lattice of

12

turned upside down.

Intuition. Read the lattice from the top down as "how much information is thrown away". At the top, $(1)$ is the whole ring and the quotient is the zero ring: everything is forgotten. At the bottom, $(0)$ forgets nothing and the quotient is $Z /12 Z$ itself. A maximal ideal is a node one step below the top — the largest amount you can forget while still having something nonzero left, which is why what remains is a field, the simplest possible nonzero ring.

Exercises for §5.4 →

5.5Homomorphisms, kernels and the isomorphism theorems

Ideals were introduced as the subsets one may quotient by. The corollary Every ideal is a kernel already showed they are exactly the kernels of ring maps. This section makes that correspondence quantitative: it shows that a surjective homomorphism carries no information beyond its kernel, and that the ideal structure above the kernel is faithfully reproduced in the image.

Definition 5.41 (Ring homomorphism). A map $ϕ : R \to S$ between rings is a ring homomorphism if for all $a, b \in R$

ϕ (a + b) = ϕ (a) + ϕ (b), ϕ (ab) = ϕ (a) ϕ (b), ϕ (1_{R}) = 1_{S} .

Its kernel is $ker ϕ = {r \in R : ϕ (r) = 0}$ and its image is $im ϕ = ϕ (R)$ .

Proposition 5.42 (Kernels are ideals, images are subrings). For a ring homomorphism $ϕ : R \to S$ , $ker ϕ$ is a two-sided ideal of $R$ and $im ϕ$ is a subring of $S$ . Moreover $ϕ$ is injective if and only if $ker ϕ = {0}$ .

Proof. $ker ϕ$ is nonempty since $ϕ (0) = 0$ , which follows from additivity. If $a, b \in ker ϕ$ then $ϕ (a - b) = ϕ (a) - ϕ (b) = 0$ , so $a - b \in ker ϕ$ . For absorption take $r \in R$ and $a \in ker ϕ$ : then $ϕ (r a) = ϕ (r) ϕ (a) = ϕ (r) \cdot 0 = 0$ and $ϕ (a r) = 0 \cdot ϕ (r) = 0$ , so $r a, a r \in ker ϕ$ . By the theorem The ideal test, $ker ϕ ⊴ R$ .

The image contains $1_{S} = ϕ (1_{R})$ , and is closed under differences and products because $ϕ (a) - ϕ (b) = ϕ (a - b)$ and $ϕ (a) ϕ (b) = ϕ (ab)$ ; so it is a subring.

For the last claim, injectivity clearly forces $ker ϕ = {0}$ . Conversely if $ker ϕ = {0}$ and $ϕ (a) = ϕ (b)$ , then $ϕ (a - b) = 0$ , so $a - b = 0$ .∎

Pitfall. The image is a subring but not in general an ideal of $S$ . The inclusion $Z ↪ Q$ is a ring homomorphism whose image $Z$ is not an ideal of $Q$ . Kernels are ideals of the source; images are subrings of the target. Confusing the two is the most common error in this chapter.

Theorem 5.43 (First isomorphism theorem for rings). Let $ϕ : R \to S$ be a ring homomorphism and $K = ker ϕ$ . Then there is a unique ring isomorphism

\overset{ˉ}{ϕ} : R / K ⟶ im ϕ, \overset{ˉ}{ϕ} (r + K) = ϕ (r),

and $ϕ = \overset{ˉ}{ϕ} \circ π$ where $π : R \to R / K$ is the quotient map. In particular $R / ker ϕ ≅ im ϕ$ .

Proof. Well defined. Suppose $r + K = r^{'} + K$ , so $r - r^{'} \in K$ . Then $ϕ (r) - ϕ (r^{'}) = ϕ (r - r^{'}) = 0$ , so $ϕ (r) = ϕ (r^{'})$ . The value assigned to a coset therefore does not depend on the representative, and $\overset{ˉ}{ϕ}$ is a genuine function on $R / K$ .

Homomorphism. Using the quotient operations and then the fact that $ϕ$ is a homomorphism,

\overset{ˉ}{ϕ} ((r + K) + (s + K)) = \overset{ˉ}{ϕ} ((r + s) + K) = ϕ (r + s) = ϕ (r) + ϕ (s) = \overset{ˉ}{ϕ} (r + K) + \overset{ˉ}{ϕ} (s + K),

and identically for products, $\overset{ˉ}{ϕ} ((r + K) (s + K)) = \overset{ˉ}{ϕ} (rs + K) = ϕ (rs) = ϕ (r) ϕ (s)$ . Also $\overset{ˉ}{ϕ} (1 + K) = ϕ (1) = 1_{S}$ .

Injective. If $\overset{ˉ}{ϕ} (r + K) = 0$ then $ϕ (r) = 0$ , so $r \in K$ , so $r + K = 0 + K$ is the zero of $R / K$ . Hence $ker \overset{ˉ}{ϕ}$ is trivial and $\overset{ˉ}{ϕ}$ is injective by the proposition Kernels are ideals, images are subrings.

Surjective onto the image. Every element of $im ϕ$ is $ϕ (r) = \overset{ˉ}{ϕ} (r + K)$ for some $r$ .

Factorisation and uniqueness. By construction $\overset{ˉ}{ϕ} (π (r)) = \overset{ˉ}{ϕ} (r + K) = ϕ (r)$ , so $ϕ = \overset{ˉ}{ϕ} \circ π$ . If $ψ$ is any map with $ψ \circ π = ϕ$ , then $ψ (r + K) = ψ (π (r)) = ϕ (r) = \overset{ˉ}{ϕ} (r + K)$ for every $r$ , so $ψ = \overset{ˉ}{ϕ}$ .∎

The theorem is a factory for isomorphisms. To identify a quotient $R / I$ , one does not manipulate cosets; one finds a homomorphism out of $R$ whose kernel is $I$ and reads off the image. The following examples are the standard ones, and the technique is the same every time.

Example 5.44 (Evaluation and the quotient by $(x - c)$ ). Let $R$ be commutative with $1$ and $c \in R$ . Identify $R [x] / (x - c)$ .

Solution.

Define $ϕ : R [x] \to R$ by $ϕ (f) = f (c)$ . Evaluation at a fixed element of a commutative ring respects sums and products, and $ϕ (1) = 1$ , so $ϕ$ is a ring homomorphism.
$ϕ$ is surjective: the constant polynomial $b$ maps to $b$ .
$ker ϕ = (x - c)$ . Indeed if $f (c) = 0$ , divide by the monic $x - c$ — legitimate over any commutative ring — to get $f = q \cdot (x - c) + r$ with $r$ constant; evaluating at $c$ gives $r = f (c) = 0$ , so $f \in (x - c)$ . Conversely every multiple of $x - c$ vanishes at $c$ .
By the theorem First isomorphism theorem for rings, $R [x] / (x - c) ≅ R$ .
Sanity check with $R = Z$ , $c = 0$ : $Z [x] / (x) ≅ Z$ , exactly as used in the proposition Prime does not imply maximal.

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Example 5.45 (Reduction of coefficients modulo a prime). Identify $Z [x] / (p)$ for a prime $p$ , and deduce that $(p)$ is prime but not maximal in $Z [x]$ .

Solution.

Reducing every coefficient mod $p$ gives a map $ϕ : Z [x] \to F_{p} [x]$ , which is a surjective ring homomorphism because reduction $Z \to F_{p}$ is one and the polynomial operations are computed coefficientwise.
$ϕ (f) = 0$ exactly when every coefficient of $f$ is divisible by $p$ , that is when $f \in (p)$ . So $ker ϕ = (p)$ .
Hence $Z [x] / (p) ≅ F_{p} [x]$ by the theorem First isomorphism theorem for rings.
$F_{p} [x]$ is an integral domain but not a field, since $x$ has no inverse. By Prime ideals and integral domains and Maximal ideals and fields, $(p)$ is prime but not maximal in $Z [x]$ . Indeed $(p) ⊊ (p, x) ⊊ Z [x]$ .

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Example 5.46 (The characteristic of a ring as a quotient of $Z$ ). Show that every ring $R$ receives a unique homomorphism from $Z$ , and that its image is $Z / (n)$ where $n$ is the characteristic of $R$ .

Solution.

A homomorphism $χ : Z \to R$ must send $1 \mapsto 1_{R}$ , and additivity then forces $χ (k) = k \cdot 1_{R}$ for all $k$ ; this formula is a homomorphism, so $χ$ exists and is unique.
$ker χ$ is an ideal of $Z$ , hence $(n)$ for a unique $n \geq 0$ by the theorem Every ideal of $Z$ is principal. By definition $n$ is the characteristic of $R$ .
By the theorem First isomorphism theorem for rings, $Z / (n) ≅ im χ \subseteq R$ , the prime subring of $R$ .
If $R$ is an integral domain then $im χ$ is a domain too, so $(n)$ is prime and $n$ is $0$ or a prime number. This is why characteristics of domains are $0$ or prime and never $6$ .

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Theorem 5.47 (Correspondence theorem for ideals). Let $I ⊴ R$ and $π : R \to R / I$ the quotient map. The assignments

J ⟼ J / I = π (J), \overset{ˉ}{J} ⟼ π^{- 1} (\overset{ˉ}{J})

are mutually inverse, inclusion-preserving bijections between the set of ideals of $R$ containing $I$ and the set of ideals of $R / I$ . Under this bijection, $J$ is prime (respectively maximal) in $R$ if and only if $J / I$ is prime (respectively maximal) in $R / I$ .

Proof. The maps land where claimed. If $J \supseteq I$ is an ideal of $R$ , then $π (J)$ is an additive subgroup as the image of one, and for $\overset{r}{ˉ} \in R / I$ and $π (a) \in π (J)$ we have $\overset{r}{ˉ} π (a) = π (r a) \in π (J)$ by surjectivity of $π$ ; so $π (J)$ is an ideal. If $\overset{ˉ}{J}$ is an ideal of $R / I$ , then $π^{- 1} (\overset{ˉ}{J})$ is an ideal by the same computation run backwards, and it contains $π^{- 1} (\overset{ˉ}{0}) = I$ .

Inverse on one side. Let $\overset{ˉ}{J} ⊴ R / I$ and $J = π^{- 1} (\overset{ˉ}{J})$ . Then $π (J) = \overset{ˉ}{J}$ because $π$ is surjective.

Inverse on the other side. Let $J \supseteq I$ be an ideal and put $J^{'} = π^{- 1} (π (J))$ . Clearly $J \subseteq J^{'}$ . Conversely let $r \in J^{'}$ , so $π (r) = π (a)$ for some $a \in J$ . Then $r - a \in ker π = I \subseteq J$ , and since $a \in J$ we get $r = (r - a) + a \in J$ . Hence $J^{'} = J$ . This step is exactly where the hypothesis $J \supseteq I$ is used, and it is why ideals not containing $I$ are not in the correspondence.

Inclusions. Both maps are monotone for inclusion, since images and preimages are.

Prime and maximal. Composing $R \to R / I \to (R / I) / (J / I)$ gives a surjection with kernel $J$ , so by the theorem First isomorphism theorem for rings,

(R / I) / (J / I) ≅ R / J .

The quotients agree, so by the theorems Prime ideals and integral domains and Maximal ideals and fields the primeness or maximality of $J$ in $R$ and of $J / I$ in $R / I$ are the same condition.∎

Corollary 5.48 (Third isomorphism theorem for rings). If $I \subseteq J$ are ideals of $R$ , then $J / I$ is an ideal of $R / I$ and

(R / I) / (J / I) ≅ R / J .

Proof. This was established in the last paragraph of the proof of the theorem Correspondence theorem for ideals: the composite $R \to R / I \to (R / I) / (J / I)$ is surjective with kernel ${r : r + I \in J / I} = J$ , and the first isomorphism theorem does the rest.∎

Example 5.49 (Counting the ideals of $Z /36 Z$ ). How many ideals does $Z /36 Z$ have, and how many of them are maximal?

Solution.

By the theorem Correspondence theorem for ideals, ideals of $Z / (36)$ correspond to ideals of $Z$ containing $(36)$ .
By Every ideal of $Z$ is principal these are the $(d)$ with $(36) \subseteq (d)$ , that is $d ∣ 36$ , with $d \geq 1$ .
$36 = 2^{2} 3^{2}$ has $(2 + 1) (2 + 1) = 9$ positive divisors, so there are $9$ ideals.
The maximal ones correspond to maximal ideals of $Z$ containing $(36)$ , i.e. $(p)$ with $p ∣ 36$ prime: $p = 2$ and $p = 3$ . So there are $2$ maximal ideals, namely $(2) / (36)$ and $(3) / (36)$ .

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Example 5.50 (Using the third isomorphism theorem). Identify $(Z / (36)) / ((4) / (36))$ .

Solution.

Take $R = Z$ , $I = (36)$ , $J = (4)$ ; note $I \subseteq J$ since $4 ∣ 36$ .
By the corollary Third isomorphism theorem for rings, the displayed ring is isomorphic to $Z / (4)$ .
A count confirms it: $∣ Z / (36) ∣ = 36$ and $∣ (4) / (36) ∣ = 36/4 = 9$ , so the quotient has $36/9 = 4$ elements.
Since $4$ is not prime, this is not a field; it has the zero divisor $\overset{ˉ}{2}$ .

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Theorem 5.51 (Second isomorphism theorem for rings). Let $A$ be a subring of $R$ and $I$ an ideal of $R$ . Then $A + I$ is a subring of $R$ , $I$ is an ideal of $A + I$ , $A \cap I$ is an ideal of $A$ , and

A / (A \cap I) ≅ (A + I) / I .

Proof. $A + I$ is closed under differences and, since $(a + u) (b + v) = ab + (a v + u b + uv)$ with the bracket in $I$ by absorption, under products; it contains $1 = 1 + 0$ . Clearly $I \subseteq A + I$ and $I$ absorbs multiplication by elements of $A + I \subseteq R$ , so $I ⊴ A + I$ .

Consider $ψ : A \to (A + I) / I$ , $ψ (a) = a + I$ . It is a ring homomorphism, being the restriction to $A$ of the quotient map of $R$ by $I$ . It is surjective: a general element of $(A + I) / I$ is $(a + u) + I = a + I = ψ (a)$ , since $u \in I$ . Its kernel is ${a \in A : a \in I} = A \cap I$ , which is therefore an ideal of $A$ by the proposition Kernels are ideals, images are subrings. The theorem First isomorphism theorem for rings gives $A / (A \cap I) ≅ (A + I) / I$ .∎

Exercises for §5.5 →

5.6The Chinese Remainder Theorem

The last theorem of the chapter says when a quotient factors as a product, and it is the statement that turns simultaneous congruences into a single one. The right hypothesis is not "the generators are coprime" but a statement about ideals that makes sense in any ring.

Definition 5.52 (Comaximal ideals). Ideals $I$ and $J$ of a commutative ring $R$ are comaximal (or coprime) if $I + J = R$ ; equivalently, if there exist $u \in I$ and $v \in J$ with $u + v = 1$ .

The equivalence is immediate: $I + J = R$ contains $1$ , and conversely an ideal containing $1$ is everything. In $Z$ , $(m) + (n) = (g cd (m, n))$ by the corollary Bézout from the ideal picture, so $(m)$ and $(n)$ are comaximal exactly when $g cd (m, n) = 1$ — the classical hypothesis, recovered.

Lemma 5.53 (Comaximal ideals have product equal to intersection). If $I, J$ are comaximal ideals of a commutative ring $R$ , then $I J = I \cap J$ .

Proof. The inclusion $I J \subseteq I \cap J$ holds always, by the proposition Sums and products are ideals. For the reverse, write $1 = u + v$ with $u \in I$ , $v \in J$ , and take $w \in I \cap J$ . Then

w = w \cdot 1 = w u + w v .

Now $w u$ has $w \in J$ and $u \in I$ , so $w u = u w \in I J$ ; and $w v$ has $w \in I$ and $v \in J$ , so $w v \in I J$ . Hence $w \in I J$ , giving $I \cap J \subseteq I J$ .∎

Theorem 5.54 (Chinese Remainder Theorem). Let $R$ be commutative with $1$ and let $I_{1}, \dots, I_{k}$ be pairwise comaximal ideals. Then the map

Φ : R ⟶ R / I_{1} \times \dots \times R / I_{k}, Φ (r) = (r + I_{1}, \dots, r + I_{k})

is a surjective ring homomorphism with kernel $I_{1} \cap \dots \cap I_{k} = I_{1} I_{2} \dots I_{k}$ , and therefore

R / (I_{1} \dots I_{k}) ≅ R / I_{1} \times \dots \times R / I_{k} .

Proof. $Φ$ is a homomorphism into the product ring because each component $r \mapsto r + I_{j}$ is one and the product ring's operations are componentwise; $Φ (1) = (\overset{ˉ}{1}, \dots, \overset{ˉ}{1})$ is the identity. Its kernel is the set of $r$ lying in every $I_{j}$ , that is $⋂_{j} I_{j}$ .

Surjectivity, case $k = 2$ . Write $1 = u + v$ with $u \in I_{1}$ , $v \in I_{2}$ . Given a target $(a + I_{1}, b + I_{2})$ , set $r = a v + b u$ . Since $u \in I_{1}$ , $r \equiv a v (mod I_{1})$ , and $v = 1 - u \equiv 1 (mod I_{1})$ , so $r \equiv a (mod I_{1})$ . Symmetrically $r \equiv b (mod I_{2})$ . So $Φ (r) = (a + I_{1}, b + I_{2})$ .

Surjectivity, general $k$ , by induction. Suppose the result for $k - 1$ factors. It suffices to show $I_{1}$ and $J = I_{2} \dots I_{k}$ are comaximal; then the case $k = 2$ combined with the inductive hypothesis applied to $R / J ≅ R / I_{2} \times \dots \times R / I_{k}$ finishes the argument. For each $j \geq 2$ choose $u_{j} \in I_{1}$ , $v_{j} \in I_{j}$ with $u_{j} + v_{j} = 1$ . Multiplying these equations together,

1 = j = 2 \prod k (u_{j} + v_{j}) = (\prod_{j \geq 2} v_{j}) + w,

where $w$ collects every term containing at least one factor $u_{j} \in I_{1}$ and hence lies in $I_{1}$ . The remaining product $\prod_{j \geq 2} v_{j}$ lies in $I_{2} \dots I_{k} = J$ . So $1 \in I_{1} + J$ and the two are comaximal.

Kernel as a product. By the lemma Comaximal ideals have product equal to intersection and induction, $⋂_{j} I_{j} = I_{1} \dots I_{k}$ : for $k = 2$ this is the lemma, and for larger $k$ apply the lemma to $I_{1}$ and $J = I_{2} \dots I_{k}$ , which were just shown comaximal, together with the inductive identity $⋂_{j \geq 2} I_{j} = J$ .

The final isomorphism is the theorem First isomorphism theorem for rings applied to the surjection $Φ$ .∎

Corollary 5.55 (The classical form). If $n = n_{1} n_{2} \dots n_{k}$ with the $n_{j}$ pairwise coprime, then

Z / (n) ≅ Z / (n_{1}) \times \dots \times Z / (n_{k})

as rings. Consequently a system $x \equiv a_{j} (mod n_{j})$ has a solution, unique modulo $n$ .

Proof. Pairwise coprimality gives $g cd (n_{i}, n_{j}) = 1$ , so $(n_{i}) + (n_{j}) = Z$ by Bézout from the ideal picture and the ideals are pairwise comaximal. Their product is $(n_{1} \dots n_{k}) = (n)$ , so the theorem Chinese Remainder Theorem applies verbatim. Surjectivity gives existence of a solution, and the kernel $(n)$ gives uniqueness modulo $n$ .∎

Example 5.56 (Solving a system of congruences). Find the unique $x$ with $0 \leq x < 105$ satisfying $x \equiv 2 (mod 3)$ , $x \equiv 3 (mod 5)$ , $x \equiv 2 (mod 7)$ .

Solution.

The moduli $3, 5, 7$ are pairwise coprime with product $105$ , so by the corollary The classical form a unique solution exists modulo $105$ .
Solve the first two: $x \equiv 2 (mod 3)$ and $x \equiv 3 (mod 5)$ . Writing $x = 3 t + 2$ gives $3 t + 2 \equiv 3 (mod 5)$ , i.e. $3 t \equiv 1 (mod 5)$ ; since $3 \cdot 2 = 6 \equiv 1$ , $t \equiv 2 (mod 5)$ . So $x = 3 (5 s + 2) + 2 = 15 s + 8$ , i.e. $x \equiv 8 (mod 15)$ .
Impose the third: $15 s + 8 \equiv 2 (mod 7)$ gives $s + 1 \equiv 2 (mod 7)$ (as $15 \equiv 1$ , $8 \equiv 1$ ), so $s \equiv 1 (mod 7)$ and $s = 7 m + 1$ .
Then $x = 15 (7 m + 1) + 8 = 105 m + 23$ , so $x = 23$ . Check: $23 = 3 \cdot 7 + 2$ , $23 = 5 \cdot 4 + 3$ , $23 = 7 \cdot 3 + 2$ . All three hold.

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Example 5.57 (Splitting a quotient ring). Decompose $Z / (60)$ as a product of quotients by prime powers, and count its units.

Solution.

$60 = 4 \cdot 3 \cdot 5$ with the three factors pairwise coprime.
By the corollary The classical form, $Z / (60) ≅ Z / (4) \times Z / (3) \times Z / (5)$ .
A tuple is a unit in a product exactly when each component is, so the unit group has order $φ (4) φ (3) φ (5) = 2 \cdot 2 \cdot 4 = 16$ .
Directly, $φ (60) = 60 (1 - \frac{1}{2}) (1 - \frac{1}{3}) (1 - \frac{1}{5}) = 16$ , which agrees. The multiplicativity of Euler's $φ$ , proved combinatorially in Congruences, is exactly this ring isomorphism.

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Summary. The chapter in five lines. An ideal is an additive subgroup that absorbs multiplication; absorption is precisely what makes coset multiplication well defined. Every ideal is the kernel of a homomorphism, namely the quotient map, and every kernel is an ideal. The first isomorphism theorem identifies $R / ker ϕ$ with $im ϕ$ , turning the problem of recognising a quotient into the problem of finding a surjection. Primeness of $I$ says $R / I$ is a domain; maximality says $R / I$ is a field; hence maximal implies prime, and the converse fails at $(0) ⊴ Z$ . Comaximal ideals split the quotient into a product, which is the Chinese Remainder Theorem.

Exercises for §5.6 →

Common Mistakes

Checking closure instead of absorption. A subset closed under its own multiplication need not be an ideal: $Z \subseteq Q$ is the standard example. The ideal axiom multiplies an element of $I$ by an arbitrary element of $R$ .
Forgetting that a proper ideal has no identity. If $1 \in I$ then $I = R$ . So an ideal is almost never a subring in the unital sense, and "ideal" and "subring" name genuinely different things.
Verifying only the additive half of well-definedness. Addition of cosets is well defined for any additive subgroup; it is multiplication that needs absorption, and it fails silently, returning inconsistent answers rather than an error.
Confusing $(2)$ and $(2, x)$ in $Z [x]$ . The first is the set of polynomials with every coefficient even; the second is the set with even constant term. Thus $x \in (2, x)$ but $x \in / (2)$ , and only the second is maximal.
Assuming every ideal is principal. True in $Z$ and $F [x]$ , false in $Z [x]$ and $F [x, y]$ . The proof that $Z$ is a PID uses the division algorithm and cannot be transplanted to a ring without one.
Assuming prime implies maximal. It does in a PID for nonzero ideals, and it is false in general: $(0)$ is prime in $Z$ , and $(x)$ is prime in $Z [x]$ , and neither is maximal. Only the direction maximal $\Rightarrow$ prime is a theorem.
Applying the correspondence theorem to ideals that do not contain $I$ . The bijection is between ideals of $R / I$ and ideals of $R$ containing $I$ . The ideal $(5) ⊴ Z$ has no counterpart in $Z / (12)$ because $(12) \neq \subseteq (5)$ .
Treating the image of a homomorphism as an ideal of the target. It is a subring. Only the kernel, in the source, is an ideal.
Using the Chinese Remainder Theorem without comaximality. $Z / (4)$ is not isomorphic to $Z / (2) \times Z / (2)$ : the first has an element of additive order $4$ and the second does not. The hypothesis $I + J = R$ is not decorative.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter

Introduction

5.1Ideals and the ideal test

Building new ideals from old

Ideals detect fields

5.2Generators, principal ideals and Z

5.3Quotient rings and the well-definedness crux

5.4Prime and maximal ideals

5.5Homomorphisms, kernels and the isomorphism theorems

5.6The Chinese Remainder Theorem

5.2Generators, principal ideals and $Z$