Ideals, Quotient Rings, and Homomorphisms

For groups, a homomorphism has a normal-subgroup kernel, and every normal subgroup is a kernel; quotienting by it produces a new group. Rings add one operation, and the same correspondence holds. A ring homomorphism has a kernel, the kernel is closed under multiplication by every element of the ring, and that closure property defines the substructure — an ideal — by which a ring may be quotiented.

Ring homomorphisms and their kernels

Both operations must be respected. The additive-scaling map ${\phi }_n(x)=nx$ on $\mathbb{Z}$ is a group homomorphism but a ring homomorphism only for $n\in \{0,1\}$, since ${\phi }_n(xy)=nxy$ while ${\phi }_n(x){\phi }_n(y)=n^{2}xy$. Checking both laws is not optional.¹

The kernel behaves differently from the image.

The absorption clause is what distinguishes a kernel. If $a\in \ker \phi$ then $\phi (ra)=\phi (r)\phi (a)=\phi (r)\cdot 0=0$, so $ra\in \ker \phi$, and symmetrically for $ar$.¹ The kernel is not merely closed under its own multiplication; it absorbs multiplication by the entire ring. A subring with that absorption property is an ideal.

Ideals

To verify $I$ is an ideal, check that it is nonempty, closed under subtraction, and closed under multiplication by every element of $R$ (not just by elements of $I$). The absorption condition is strictly stronger than being a subring: $\mathbb{Z}$ is a subring of $\mathbb{Q}$ but not an ideal, because $\frac{1}{2}\cdot 1\notin \mathbb{Z}$.

The last clause of the image/kernel proposition says that every kernel is an ideal. The converse also holds, and it comes from the quotient construction.

The quotient ring

Because $(I,+)$ is a subgroup of the abelian group $(R,+)$, the additive cosets $r+I$ already form a quotient group. The question is whether coset multiplication $(r+I)(s+I):=rs+I$ is well defined. It is well defined exactly when $I$ is an ideal, and then the distributive laws pass to the cosets.

Writing $\bar{r}=r+I$, addition and multiplication in $R/I$ are just $\bar{r}+\bar{s}=\overline{r+s}$ and $\bar{r}\bar{s}=\overline{rs}$: compute with any representatives and reduce.¹ The construction closes the loop between ideals and homomorphisms.

For $R=\mathbb{Z}$ the ideals coincide with the subgroups $n\mathbb{Z}$, and the quotient is $\mathbb{Z}/n\mathbb{Z}$ — the construction that named modular arithmetic. In $\mathbb{Z}[x]$, quotienting by the ideal of polynomials with no constant or linear term collapses each polynomial to its degree-$\le 1$ part, and there $\bar{x}\cdot \bar{x}=\overline{x^2}=0$, so the quotient acquires zero divisors the base ring lacked.¹

The remaining isomorphism theorems

The higher isomorphism theorems transfer verbatim from the group versions: prove the additive-group statement, then check the bijection respects multiplication.

The correspondence theorem is the one used most in practice. The ideals of $\mathbb{Z}/12\mathbb{Z}$ are the images of the ideals of $\mathbb{Z}$ that contain $12\mathbb{Z}$, namely the $d\mathbb{Z}$ for $d\mid 12$.

Generated ideals

Given a subset $A\subseteq R$, the ideal generated by $A$, written $(A)$, is the smallest ideal containing $A$ — equivalently the intersection of all ideals that contain $A$. In a commutative ring with $1$ it is the set of finite sums $\sum r_i{a}_i$ with $r_i\in R$ and $a_i\in A$. An ideal generated by a single element is principal; one generated by a finite set is finitely generated.²

• In $\mathbb{Z}$, every ideal is principal: $n\mathbb{Z}=(n)$, and $(n,m)=(d)$ where $d=\gcd (n,m)$. Divisibility becomes containment, $(b)\subseteq (a)\leftrightarrow a\mid b$, so the ideal lattice mirrors the divisor lattice.
• In $\mathbb{Z}[x]$, the ideal $(2,x)$ of polynomials with even constant term is not principal. If $(2,x)=(a(x))$, then $a(x)\mid 2$ forces $a(x)=\pm 1$ or $\pm 2$; a unit would make the ideal all of $\mathbb{Z}[x]$, and $\pm 2$ fails to divide $x$. Whether ideals are principal turns out to be a deep structural property, studied under principal ideal domains.

Containing a unit is the same as being everything.

If $u\in I$ is a unit with inverse $v$, then $1=vu\in I$, so $r=r\cdot 1\in I$ for all $r$.² The second part is why fields have a trivial ideal structure, and it forces every nonzero homomorphism out of a field to be injective: the kernel is a proper ideal, hence $0$.

Maximal and prime ideals

Two classes of proper ideal are singled out by their quotients. They generalize, respectively, as large as possible and prime number from the integers.

Existence is guaranteed in any ring with identity: every proper ideal is contained in a maximal ideal, proved by applying Zorn's lemma to the poset of proper ideals containing it, whose chains have their union as an upper bound.² The value of both definitions is that each translates into a property of the quotient ring.

The maximal case is the correspondence theorem plus the field iff only trivial ideals proposition: $M$ is maximal iff $R/M$ has no ideals but $0$ and itself, iff $R/M$ is a field. The prime case is a direct translation: $\bar{a}\bar{b}=0$ in $R/P$ means $ab\in P$, and no zero divisors is what the prime condition asks. Because every field is a domain, maximal implies prime.²³

The integers show that the converse fails. In $\mathbb{Z}$ the prime ideals are the $p\mathbb{Z}$ for $p$ prime together with the zero ideal $0=(0)$; only the $p\mathbb{Z}$ are maximal. So $0$ is prime but not maximal, matching $\mathbb{Z}/0=\mathbb{Z}$ being a domain but not a field. In $\mathbb{Z}[x]$ the chain $(x)\subset (2,x)\subset \mathbb{Z}[x]$ shows $(x)$ is prime — the quotient $\mathbb{Z}[x]/(x)\cong \mathbb{Z}$ is a domain — but not maximal.²

This is how fields are built: quotient a familiar ring by a maximal ideal. Every finite field arises as $\mathbb{Z}[x]/M$ for a suitable maximal $M$, the route taken in field theory.

Sums, products, and reduction as a tool

Two ideals combine into new ideals. The sum $I+J=\{a+b:a\in I,b\in J\}$ is the smallest ideal containing both; the product $IJ$ is the set of finite sums of products $ab$ with $a\in I$, $b\in J$. Always $IJ\subseteq I\cap J$. In $\mathbb{Z}$, $m\mathbb{Z}+n\mathbb{Z}=\gcd (m,n)\mathbb{Z}$ and $(m\mathbb{Z})(n\mathbb{Z})=mn\mathbb{Z}$, so ideal sum is gcd and ideal product is the product of the integers.¹ These operations underlie the Chinese Remainder Theorem.

Reduction modulo an ideal also serves as a proof technique. Because $\pi :R\to R/I$ is a homomorphism, any polynomial relation among ring elements survives reduction. To show $x^2+y^2=3z^2$ has no nonzero integer solution with $x,y,z$ coprime, reduce mod $4$: squares are $0$ or $1$ there, and checking the few combinations forces $x,y,z$ all even, contradicting coprimality.¹

The absorption loop ranges over the whole ring $R$, not over $I$: that is what separates an ideal from a mere subring, and what makes $R/I$ a ring.