Ideals, Quotient Rings and the Isomorphism Theorems for Rings

Closure under $x-y$ is exactly the additive subgroup criterion, and the two absorption conditions then give the ideal criterion.

An intersection has subtraction closure and both absorption properties because each member ideal has them; for the empty family the intersection is $R$.

The criterion and intersection assertion follow.

Let $J$ be the finite sums $\sum r_is_i$; it contains $S$ and is closed under subtraction and multiplication by arbitrary ring elements.

Thus $J$ is an ideal containing $S$, while every ideal containing $S$ contains each such finite sum.

Hence $J=(S)$; taking $S=\{a\}$ gives $(a)=Ra$.

Subtraction closure of $I+J$ and of $IJ$ follows by subtracting representatives and concatenating finite sums.

Multiplying a representative on either side keeps it in $I+J$, and distributivity keeps each summand of a product in $IJ$.

The closure established in step 2.1 proves both claims.

The additive-group kernel of $f$ is an additive subgroup of $R$.

If $x\in\ker f$ and $r\in R$, then $f(rx)=f(r)f(x)=0$ and $f(xr)=0$.

Thus the additive and absorption properties make $\ker f$ a two-sided ideal.

If $J$ is an ideal and $r'=r+i$, $s'=s+j$, then $r's'-rs=rj+is+ij\in J$, so the product is well defined.

Conversely, compare $(r+J)(0+J)$ with $(r+J)(j+J)$ and the reversed product; [L4] gives $rj,jr\in J$ for all $r,j$.

Thus the rule is well defined exactly when $J$ is a two-sided ideal.

By [L3], coset addition has an abelian-group structure, and [L2] makes coset multiplication a defined operation.

Associativity and both distributive laws follow by applying the corresponding ring law in $R$ to representatives; $1+I$ is a multiplicative identity.

Hence $R/I$ is a ring with identity $1+I$.

The identities $\pi(r+s)=\pi(r)+\pi(s)$, $\pi(rs)=\pi(r)\pi(s)$, and $\pi(1)=1+I$ show that $\pi$ is a ring homomorphism.

By [L3] it is surjective; by [L4] and [L5], its kernel is exactly $I$.

Thus $\ker\pi=I$.

Map $a+n\mathbb Z$ to $[a]_n$; [L2] makes this an equality of additive quotient sets.

The quotient product maps to $[ab]_n$, exactly the modular product in [L4], and the identities agree.

Therefore the congruence-class ring is literally the quotient ring, including at $n=0,1$.

By [L1]--[L3], $I$ is normal in the additive group of $R$; applying [L5] to the additive homomorphism underlying $f$ defines $\bar f(r+I)=f(r)$ and proves representative independence.

Since [L4] gives $(r+I)(s+I)=rs+I$, one has $\bar f((r+I)(s+I))=f(rs)=f(r)f(s)$, and $\bar f(1+I)=f(1)=1$; thus $\bar f$ is a ring homomorphism.

The factor identity $f=\bar f\circ\pi$ holds by step 1.1, and any ring-homomorphic factor is additive, so the uniqueness in [L5] proves its uniqueness as a ring factor.

By [L2] and [L1], $f$ induces a ring homomorphism $\bar f:R/\ker f\to\operatorname{im}f$ with $\bar f(r+\ker f)=f(r)$.

This map is surjective by the definition of image, and its additive kernel is trivial, hence it is injective by [L4].

Thus $\bar f$ is a ring isomorphism.

$S+I$ contains $1_R$ and is subtraction-closed; expanding $(s+i)(t+j)=st+sj+it+ij$ proves multiplication closure.

For $s+i\in S+I$ and $j\in I$, both $(s+i)j$ and $j(s+i)$ lie in $I$, while $S\cap I$ is subtraction-closed and absorbed by $S$.

Hence $S+I$ is a unital subring, $I\mathrel{\trianglelefteq}S+I$, and $S\cap I\mathrel{\trianglelefteq}S$.

Restrict the quotient map $S+I\to(S+I)/I$ to $\phi:S\to(S+I)/I$, $\phi(s)=s+I$.

Its kernel is $S\cap I$, and every $(s+i)+I$ equals $s+I$, so its image is all of $(S+I)/I$.

The kernel and image computation gives $S/(S\cap I)\cong(S+I)/I$.

$J/I=\{j+I:j\in J\}$ is an additive subgroup of $R/I$ because $J$ is an additive subgroup.

For $r+I\in R/I$ and $j+I\in J/I$, both $(r+I)(j+I)=rj+I$ and $(j+I)(r+I)=jr+I$ lie in $J/I$.

Hence $J/I$ is a two-sided ideal of $R/I$.

Define $\phi:R/I\to R/J$ by $\phi(r+I)=r+J$; it is a well-defined surjective ring homomorphism because $I\subseteq J$.

The equality $\phi(r+I)=0+J$ holds exactly when $r\in J$, so $\ker\phi=J/I$.

The kernel and image computation gives $(R/I)/(J/I)\cong R/J$.

If $J$ contains $I$, then $J/I=\pi[J]$ is an ideal of $R/I$; if $K$ is an ideal of $R/I$, then $\pi^{-1}(K)$ is an ideal containing $I$.

Surjectivity gives $\pi[\pi^{-1}(K)]=K$, and $I\subseteq J$ gives $\pi^{-1}(\pi[J])=J$; inclusion is preserved.

Therefore the two assignments give the claimed correspondence.

If $P$ is prime, then $P$ is proper, so $1+P\ne0+P$ by [L4]; and $(a+P)(b+P)=0+P$ gives $ab\in P$, hence $a+P=0+P$ or $b+P=0+P$, while commutativity of $R$ makes the quotient commutative.

If $R/P$ is a domain and $ab\in P$, then $(a+P)(b+P)=0+P$, so [L3] and [L4] give $a\in P$ or $b\in P$; its nonzero identity gives $1\notin P$, hence $P\subsetneq R$.

These implications prove the equivalence.

If $M$ is maximal and $a\notin M$, the set $J=\{m+ra:m\in M,r\in R\}$ is an ideal by the subtraction-and-absorption criterion, properly contains $M$, and hence is $R$; thus $m+ra=1$ for some $m\in M,r\in R$, giving $(a+M)(r+M)=1+M$.

If $R/M$ is a field and $M\subsetneq J\mathrel{\trianglelefteq}R$, choose $a\in J\setminus M$; an inverse $r+M$ of $a+M$ gives $ar-1\in M\subseteq J$, while $ar\in J$, so $1\in J$ and $J=R$.

Hence $R/M$ is a field exactly when $M$ is maximal.

By [L1], $R/M$ is a field.

By [L3], that quotient is an integral domain.

The domain conclusion of step 2.1 yields that $M$ is prime.

Let $\mathcal P$ be the poset of proper ideals of $R$ containing $I$, ordered by inclusion.

$\mathcal P$ is nonempty, because $I$ is a proper ideal containing $I$.

The empty chain of $\mathcal P$ has an upper bound in $\mathcal P$: every member of $\mathcal P$ is vacuously above all of its members, and $\mathcal P$ is nonempty, so $I$ is such an upper bound.

A nonempty chain $\mathcal C\subseteq\mathcal P$ has an upper bound in $\mathcal P$: $\bigcup\mathcal C$ is an ideal containing $I$, and it is proper, since $1\in\bigcup\mathcal C$ would place $1$ in some member of $\mathcal C$, forcing that member to equal $R$ and contradicting its properness.

Every chain of $\mathcal P$ has an upper bound in $\mathcal P$, and $\mathcal P$ is a nonempty poset, so Zorn's lemma yields a maximal element $M$ of $\mathcal P$.

$M$ is a proper ideal containing $I$ that is maximal among proper ideals of $R$, so $M$ is a maximal ideal containing $I$.