Chapter 12

Hard

Ideals & Quotient Rings

00 · Symbol Glossary

$I \trianglelefteq R$I is an ideal of R

Borrowing the normal-subgroup symbol deliberately: ideals play exactly the role in ring theory that normal subgroups played in group theory — the substructures that make quotients possible.

$(a)$Principal ideal generated by a

$(a) = aR = \{ar : r\in R\}$ — the smallest ideal containing $a$ , built by absorbing every possible multiple of $a$ .

$I+J$Sum of two ideals

$I+J = \{i+j : i\in I, j\in J\}$ — the smallest ideal containing both $I$ and $J$ .

$R/I$Quotient ring

The ring of cosets $a+I$ , with $(a+I)+(b+I) := (a+b)+I$ and $(a+I)(b+I) := ab+I$ . Well-defined precisely when $I$ is an ideal.

01 · Ideals

A subring only needs to be closed under $R$ 's operations. An ideal demands something much stronger from multiplication: it must absorb multiplication by anything in the ambient ring, not merely multiply things that are already inside it.

Definition — Ideal

A subset $I\subseteq R$ (commutative ring with unity) is an ideal if: (I1) $(I,+)$ is a subgroup of $(R,+)$ ; (I2) for every $r\in R$ and $a\in I$ , $ra\in I$ (absorption).

Absorption is strictly stronger than closure

A subring only requires $ab\in S$ when both $a,b\in S$ . An ideal requires $ra\in I$ even when $r$ comes from all of $R$ , not just from $I$ . Every ideal is automatically a subring's additive part, but ideals are rarely subrings with unity — see the FailBlock below.

02 · The Ideal Test

Theorem 12.1 — Ideal Test

A nonempty subset $I\subseteq R$ is an ideal if and only if: (a) $a,b\in I \implies a-b\in I$ ; (b) $r\in R,\,a\in I \implies ra\in I$ .

Proof

( $\Rightarrow$ ) I1 gives subtraction-closure (identical to the subgroup-test argument, Theorem 3.1) and I2 is exactly (b).

( $\Leftarrow$ ) Condition (a) is precisely the subgroup test (Theorem 3.1) for $(I,+)\le(R,+)$ , giving I1. Condition (b) is I2 directly.

03 · Examples of Ideals

Example — nZ is an ideal of Z

$n\mathbb{Z} = \{nk : k\in\mathbb{Z}\}$ : subtraction-closed (already a subgroup, Chapter 03). Absorption: for $r\in\mathbb{Z}$ and $nk\in n\mathbb{Z}$ , $r(nk) = n(rk) \in n\mathbb{Z}$ . $n\mathbb{Z}\trianglelefteq\mathbb{Z}$ .

Example — The trivial ideals

$\{0\}$ : trivially satisfies both conditions. $R$ itself: trivially satisfies both. Every ring has at least these two ideals, exactly mirroring $\{e\}$ and $G$ always being normal subgroups.

Definition — Principal Ideal

For $a\in R$ , $(a) := aR = \{ar : r\in R\}$ is the principal ideal generated by $a$ — verified to be an ideal by the same style of check as $n\mathbb{Z}$ above (with $a$ playing the role of $n$ ).

Common mistake — A proper ideal is never a subring with unity

Is $2\mathbb{Z}$ (an ideal of $\mathbb{Z}$ , since $2\mathbb{Z}=(2)$ ) a subring with unity? It would need some $u\in2\mathbb{Z}$ with $ua=a$ for every $a\in2\mathbb{Z}$ — forcing $u=1$ , but $1\notin2\mathbb{Z}$ . More generally: if a proper ideal $I\subsetneq R$ contained $1$ , absorption (I2) would give $r\cdot1=r\in I$ for every $r\in R$ , forcing $I=R$ — contradicting properness. A proper ideal never contains $1$ , and hence is never a subring with unity, even when it happens to be a perfectly good subring otherwise.

Theorem 12.2 — A Field's Only Ideals Are Trivial

If $F$ is a field (Chapter 17 defines this formally; for now, a commutative ring with unity in which every nonzero element is a unit), then its only ideals are $\{0\}$ and $F$ .

Proof

Let $I\trianglelefteq F$ with $I\neq\{0\}$ . Pick nonzero $a\in I$ . Since $F$ is a field, $a$ has an inverse $a^{-1}\in F$ . By absorption (I2), $a^{-1}\cdot a \in I$ , i.e. $1\in I$ . Then for any $r\in F$ , absorption gives $r\cdot1=r\in I$ . So $I=F$ .

A preview of what fields will mean structurally

Theorem 12.2 already hints at why fields will turn out to be exactly the commutative rings whose only quotients are trivial ( $R/\{0\}\cong R$ or $R/R\cong\{0\}$ ) — there's no "in-between" ideal to quotient by, unlike $\mathbb{Z}$ 's rich supply of ideals $n\mathbb{Z}$ . This becomes precise in Section 06.

04 · Sums and Intersections of Ideals

Theorem 12.3 — Sums and Intersections of Ideals Are Ideals

If $I,J\trianglelefteq R$ , then $I+J = \{i+j:i\in I,j\in J\}$ and $I\cap J$ are both ideals of $R$ .

Proof

$I+J$ : nonempty ( $0+0=0\in I+J$ ). Subtraction: $(i_1+j_1)-(i_2+j_2) = (i_1-i_2)+(j_1-j_2) \in I+J$ , using subtraction-closure inside $I$ and $J$ separately. Absorption: $r(i+j) = ri+rj \in I+J$ , using absorption inside $I$ and inside $J$ separately.

$I\cap J$ : nonempty ( $0\in I\cap J$ ). Subtraction: if $a,b\in I\cap J$ , then $a-b\in I$ (subtraction-closure in $I$ ) and $a-b\in J$ (subtraction-closure in $J$ ), so $a-b\in I\cap J$ . Absorption: if $a\in I\cap J$ and $r\in R$ , then $ra\in I$ and $ra\in J$ , so $ra\in I\cap J$ .

Example — Sums and intersections in Z

$4\mathbb{Z}+6\mathbb{Z}$ : contains $4$ and $6$ , hence (being an ideal, closed under subtraction) contains $6-4=2$ , hence contains $(2)=2\mathbb{Z}$ ; conversely every element of $4\mathbb{Z}+6\mathbb{Z}$ is a multiple of $\gcd(4,6)=2$ . So $4\mathbb{Z}+6\mathbb{Z}=2\mathbb{Z}$ — sum of ideals corresponds to gcd.

$4\mathbb{Z}\cap6\mathbb{Z}$ : multiples of both $4$ and $6$ are exactly multiples of $\mathrm{lcm}(4,6)=12$ . So $4\mathbb{Z}\cap6\mathbb{Z}=12\mathbb{Z}$ — intersection corresponds to lcm, a connection that resurfaces directly in Chapter 16's Chinese Remainder Theorem.

05 · Quotient Rings

Theorem 12.4 — R/I Is a Ring When I Is an Ideal

Let $I\trianglelefteq R$ . Then $(a+I) + (b+I) := (a+b)+I$ and $(a+I)(b+I) := ab+I$ are well-defined operations on the set of cosets $R/I$ , and $R/I$ is a commutative ring with unity $1+I$ .

Proof

Since $(R,+)$ is abelian, $(I,+)$ is automatically normal (Chapter 05, every subgroup of an abelian group is normal), so addition on $R/I$ is well-defined and $(R/I,+)$ is an abelian group by Theorem 5.3 — this part is pure group theory, already proven.

Multiplication well-defined. Suppose $a+I=a'+I$ and $b+I=b'+I$ , so $a'=a+i_1$ , $b'=b+i_2$ for $i_1,i_2\in I$ (Theorem 4.1, additive version). Then:

a'b' = (a+i_1)(b+i_2) = ab + ai_2 + i_1b + i_1i_2

Each of $ai_2$ , $i_1b$ , $i_1i_2$ lies in $I$ : $ai_2\in I$ and $i_1b\in I$ by absorption (I2), and $i_1i_2\in I$ by absorption applied to $i_1\in I$ multiplied by the ring element $i_2$ (or simply because $I$ is closed under multiplication by anything in $R$ , in particular by elements of $I\subseteq R$ ). So $a'b' - ab = ai_2+i_1b+i_1i_2 \in I$ (sum of elements of $I$ , using subtraction-closure repeatedly), giving $a'b'+I = ab+I$ .

Ring axioms. Associativity, distributivity, and commutativity of the coset multiplication are inherited directly from the corresponding properties in $R$ (the same style of verification as Theorem 5.3's group axioms for $G/N$ ). $1+I$ is the multiplicative identity: $(1+I)(a+I) = a+I$ .

Example — $\mathbb{Z}/n\mathbb{Z}$ as a ring, confirmed

The quotient ring construction applied to $I=n\mathbb{Z}\trianglelefteq\mathbb{Z}$ gives exactly $\mathbb{Z}_n$ with its ring structure from Chapter 11 — Theorem 12.4 confirms that the well-definedness proof carried out by hand in Theorem 1.5 was a special case of this general machinery.

06 · Prime and Maximal Ideals

Definition — Prime and Maximal Ideals

A proper ideal $I\subsetneq R$ is prime if $ab\in I \implies a\in I$ or $b\in I$ . It is maximal if there is no ideal $J$ with $I\subsetneq J\subsetneq R$ (no proper ideal strictly between $I$ and $R$ ).

Theorem 12.5 — R/I Is a Domain iff I Is Prime

$R/I$ is an integral domain if and only if $I$ is a prime ideal.

Proof

$R/I$ has $1+I\neq0+I$ exactly when $1\notin I$ , i.e. $I\neq R$ : proper. $R/I$ has no zero divisors exactly when: $(a+I)(b+I) = 0+I \implies a+I=0+I$ or $b+I=0+I$ , i.e. $ab\in I \implies a\in I$ or $b\in I$ . This is precisely the definition of $I$ being prime. So " $R/I$ is a domain" and " $I$ is prime" are, unpacked, the identical statement.

Theorem 12.6 — R/I Is a Field iff I Is Maximal

$R/I$ is a field if and only if $I$ is a maximal ideal.

Proof

( $\Leftarrow$ ) Suppose $I$ is maximal. $R/I$ is a nonzero commutative ring with unity (since $I$ is proper). Take any nonzero $a+I$ , i.e. $a\notin I$ . Consider $J=I+(a)$ (an ideal, by Theorem 12.3, since $(a)$ is an ideal). Since $a=a\cdot1\in(a)\subseteq J$ but $a\notin I$ , $J\supsetneq I$ (strictly larger). By maximality of $I$ , the only ideal strictly containing $I$ is $R$ itself, so $J=R$ . In particular $1\in J=I+(a)$ , so $1=i+ar$ for some $i\in I$ , $r\in R$ . Then $ar-1 = -i\in I$ , so $(a+I)(r+I) = ar+I = 1+I$ : $a+I$ has inverse $r+I$ . Since $a+I$ was an arbitrary nonzero element, $R/I$ is a field.

( $\Rightarrow$ ) Suppose $R/I$ is a field; in particular $I\neq R$ (proper, since $1+I\neq0+I$ ). Let $J$ be an ideal with $I\subsetneq J\subseteq R$ ; pick $a\in J\setminus I$ . Since $a\notin I$ , $a+I\neq0+I$ in $R/I$ ; as $R/I$ is a field, $a+I$ has an inverse $r+I$ : $(a+I)(r+I)=1+I$ , i.e. $ar-1\in I \subseteq J$ . Since $a\in J$ and $J$ absorbs multiplication by $r\in R$ , $ar\in J$ . Then $1 = ar - (ar-1)$ is a difference of two elements of $J$ ( $ar\in J$ and $ar-1\in J$ ), so $1\in J$ (subtraction-closure). By absorption, $J\ni r\cdot1=r$ for every $r\in R$ , so $J=R$ . No ideal sits strictly between $I$ and $R$ : $I$ is maximal.

Example — Maximal ideals of Z are exactly (p) for prime p

$\mathbb{Z}/(n) = \mathbb{Z}_n$ is a field exactly when every nonzero element is a unit — by Theorem 11.4, that happens iff $\gcd(a,n)=1$ for every $a\in\{1,\ldots,n-1\}$ , which holds iff $n$ is prime. By Theorem 12.6, $(n)$ is maximal in $\mathbb{Z}$ exactly when $n$ is prime — matching the elementary number-theoretic fact that primes have no nontrivial factorizations, now derived as a ring-theoretic statement.

Maximal ideals are always prime

Since fields are always integral domains (a field has no zero divisors: if $ab=0$ with $a\neq0$ , multiply by $a^{-1}$ to get $b=0$ ), Theorem 12.6 combined with Theorem 12.5 shows every maximal ideal is automatically prime — $R/I$ a field $\implies$ $R/I$ a domain $\implies$ $I$ prime. The converse can fail: $(0)$ is prime in $\mathbb{Z}$ (since $\mathbb{Z}$ has no zero divisors) but not maximal (since $\mathbb{Z}\subsetneq$ nothing is needed — rather, $(0)\subsetneq(2)\subsetneq\mathbb{Z}$ shows $(0)$ is not maximal).

07 · Exercises

EXERCISE 12.1

Apply Theorem 12.1 directly: check subtraction-closure and absorption for multiples of 5.

$I=5\mathbb{Z}$ . Subtraction: $5j-5k=5(j-k)\in5\mathbb{Z}$ . Absorption: for $r\in\mathbb{Z}$ , $r(5k)=5(rk)\in5\mathbb{Z}$ . Both conditions hold: $5\mathbb{Z}\trianglelefteq\mathbb{Z}$ .

Verify using Theorem 12.1 that $5\mathbb{Z}$ is an ideal of $\mathbb{Z}$ .

EXERCISE 12.2

Apply the gcd/lcm identification from Section 04's example.

$6\mathbb{Z}+10\mathbb{Z} = \gcd(6,10)\mathbb{Z} = 2\mathbb{Z}$ . $6\mathbb{Z}\cap10\mathbb{Z} = \mathrm{lcm}(6,10)\mathbb{Z} = 30\mathbb{Z}$ .

Find $6\mathbb{Z}+10\mathbb{Z}$ and $6\mathbb{Z}\cap10\mathbb{Z}$ .

EXERCISE 12.3

Apply Theorem 12.5: is $(4)$ a prime ideal of $\mathbb{Z}$ ? Find a specific product that breaks primality if not.

$2\times2=4\in(4)$ , but $2\notin(4)$ . So $(4)$ is not prime: the implication " $ab\in I\implies a\in I$ or $b\in I$ " fails with $a=b=2$ .

By Theorem 12.5, since $(4)$ is not prime, $\mathbb{Z}/(4)=\mathbb{Z}_4$ is not an integral domain — confirmed directly, since $[2][2]=[0]$ in $\mathbb{Z}_4$ with $[2]\neq[0]$ (Chapter 11, Exercise 11.3's sibling fact).

Is $(4)$ a prime ideal of $\mathbb{Z}$ ? Use Theorem 12.5 to conclude whether $\mathbb{Z}_4$ is an integral domain.

EXERCISE 12.4

Apply Theorem 12.6 directly, then recall which $\mathbb{Z}_n$ are fields.

$7$ is prime, so by Section 06's Example, $(7)$ is maximal in $\mathbb{Z}$ . By Theorem 12.6, $\mathbb{Z}/(7) = \mathbb{Z}_7$ is a field — every nonzero element has an inverse (confirmed directly for $\mathbb{Z}_5$ in Exercise 1.5, and the same argument works for any prime).

Using Theorem 12.6, determine whether $\mathbb{Z}_7$ is a field.

08 · Chapter Summary

Concept	Statement
Ideal	Additive subgroup that absorbs multiplication by all of $R$ (I1, I2)
Ideal test	Subtraction-closed + absorbs multiplication (Thm 12.1)
Principal ideal	$(a)=aR$
Field's ideals	Only $\{0\}$ and $F$ (Thm 12.2)
Sum, intersection of ideals	Both are ideals (Thm 12.3); correspond to gcd, lcm in $\mathbb{Z}$
Quotient ring $R/I$	Well-defined ring structure iff $I$ is an ideal (Thm 12.4)
Prime ideal	$ab\in I\implies a\in I$ or $b\in I$
Maximal ideal	No proper ideal strictly between $I$ and $R$
$R/I$ domain $\iff I$ prime	Thm 12.5
$R/I$ field $\iff I$ maximal	Thm 12.6

Next: Chapter 13 — Ring Homomorphisms builds the ring-theoretic analog of Chapter 06's homomorphism and isomorphism theorems, with ideals playing the role kernels of group homomorphisms played for normal subgroups.