---
title: "Ideals, Quotient Rings, and Homomorphisms"
module: Ring Theory
moduleNumber: 6
lessonNumber: 2
order: 602
summary: >
  Ring homomorphisms have kernels that absorb multiplication; such subsets are
  ideals, and every ideal is the kernel of the projection onto a quotient ring.
  The quotient construction yields the ring isomorphism theorems and classifies
  ideals by their quotients: R/I is a field exactly when I is maximal, an integral
  domain exactly when I is prime.
topics: [Ring Theory]
sources:
  - book: Dummit & Foote
    ref: "Ch. 7 — Introduction to Rings; §7.3 Ring Homomorphisms and Quotient Rings"
  - book: Dummit & Foote
    ref: "§7.4 Properties of Ideals"
  - book: Judson
    ref: "Ch. 16 — Rings; §16.3 Ring Homomorphisms and Ideals, §16.4 Maximal and Prime Ideals"
draft: false
---

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

> **Definition (Ring homomorphism).** A map $\varphi : R \to S$ is a **ring
> homomorphism** if for all $a, b \in R$
> $$
> \varphi(a + b) = \varphi(a) + \varphi(b) \qquad\text{and}\qquad \varphi(ab) = \varphi(a)\varphi(b).
> $$
> Its **kernel** is $\ker\varphi = \{\, a \in R \mid \varphi(a) = 0 \,\}$. A
> bijective ring homomorphism is an **isomorphism**, written $R \cong S$.

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

The kernel behaves differently from the image.

> **Proposition (Image and kernel).** Let $\varphi : R \to S$ be a homomorphism.
>
> - The image $\varphi(R)$ is a subring of $S$.
> - The kernel is a subring of $R$, and moreover $ra, ar \in \ker\varphi$ for
>   every $r \in R$ and $a \in \ker\varphi$.

The absorption clause is what distinguishes a kernel. If $a \in \ker\varphi$ then $\varphi(ra) =
\varphi(r)\varphi(a) = \varphi(r) \cdot 0 = 0$, so $ra \in \ker\varphi$, and
symmetrically for $ar$.[^df-hom] 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

> **Definition (Ideal).** Let $R$ be a ring and $I \subseteq R$. Then $I$ is a
> **left ideal** if it is a subring and $rI \subseteq I$ for all $r \in R$; a
> **right ideal** if instead $Ir \subseteq I$ for all $r$; and a **(two-sided)
> ideal** if both. For commutative rings the three notions coincide.

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 $\tfrac12 \cdot 1 \notin
\mathbb{Z}$.

$$
% caption: An ideal I absorbs multiplication from the whole ring: multiplying any
% ambient element r by any a in I lands back inside I.
\begin{tikzpicture}[font=\footnotesize, >=stealth]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black, fill=black!4] (0,0) ellipse (4.6 and 2.9);
\node[black, anchor=north west] at (-4.5,2.8) {ring $R$};
\draw[acc, thick, fill=acc!10] (-1.2,-0.2) ellipse (2.1 and 1.5);
\node[text=acc, anchor=south] at (-1.2,1.35) {ideal $I$};
\fill[acc] (-1.7,-0.3) circle (1.6pt); \node[acc, anchor=north, font=\scriptsize] at (-1.7,-0.45) {$a$};
\fill[black!70] (3.3,1.2) circle (1.6pt); \node[black!70, anchor=south, font=\scriptsize] at (3.3,1.35) {$r$};
\fill[acc] (-0.3,0.55) circle (1.6pt); \node[acc, anchor=south, font=\scriptsize] at (-0.15,0.62) {$ra$};
\draw[->, black!70, thick] (3.15,1.05) to[bend right=18] (-0.35,0.55);
\node[black, font=\scriptsize] at (1.9,-0.1) {product stays in $I$};
\end{tikzpicture}
$$

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.

> **Theorem (Quotient ring).** If $I$ is an ideal of $R$, the cosets $R/I$ form a
> ring under
> $$
> (r + I) + (s + I) = (r + s) + I, \qquad (r + I)(s + I) = rs + I.
> $$
> Conversely, if these operations on the cosets of a subgroup $I$ are well
> defined, then $I$ is an ideal.

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.[^df-hom] The construction closes the loop between ideals
and homomorphisms.

> **Theorem (First Isomorphism Theorem for Rings).** If $\varphi : R \to S$ is a
> ring homomorphism, then $\ker\varphi$ is an ideal, $\varphi(R)$ is a subring of
> $S$, and
> $$
> R / \ker\varphi \;\cong\; \varphi(R).
> $$
> Conversely, for any ideal $I$ the natural projection $\pi : R \to R/I$, $r
> \mapsto r + I$, is a surjective homomorphism with kernel $I$. Every ideal is a
> kernel and every kernel is an ideal.

$$
% caption: The first isomorphism theorem: a homomorphism f factors through its
% quotient. The projection q collapses R onto the cosets of the kernel I, and the
% induced map matches them bijectively with the image f(R).
\begin{tikzpicture}[font=\footnotesize, >=stealth,
  nd/.style={draw=black, minimum width=17mm, minimum height=9mm, align=center}]
\definecolor{acc}{HTML}{4A6FA5}
\node[nd] (R) at (0,0) {$R$};
\node[nd] (S) at (6,0) {$S$};
\node[nd, draw=acc, thick, fill=acc!10] (Q) at (3,-2.4) {$\frac{R}{I}$};
\draw[->, thick] (R) -- (S) node[midway, above] {$f$};
\draw[->, thick] (R) -- (Q) node[midway, left, xshift=-1mm] {$q$};
\draw[->, acc, thick] (Q) -- (S) node[midway, right, xshift=1mm] {onto image $f(R)$};
\node[black, anchor=north, font=\scriptsize] at (3,-3.15) {$I = $ kernel; nodes are cosets};
\end{tikzpicture}
$$

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-$\leq 1$ part, and
there $\bar x \cdot \bar x = \overline{x^2} = 0$, so the quotient acquires zero
divisors the base ring lacked.[^df-hom]

### The remaining isomorphism theorems

The higher isomorphism theorems transfer verbatim from
[the group versions](/abstract-algebra/subgroups-and-quotients/isomorphism-theorems):
prove the additive-group statement, then check the bijection respects
multiplication.

> **Theorem (Second, Third, Fourth).** Let $R$ be a ring.
>
> - **(Second)** If $A$ is a subring and $B$ an ideal, then $A + B$ is a subring,
>   $A \cap B$ an ideal of $A$, and $(A + B)/B \cong A/(A \cap B)$.
> - **(Third)** If $I \subseteq J$ are ideals, then $J/I$ is an ideal of $R/I$ and
>   $(R/I)/(J/I) \cong R/J$.
> - **(Fourth / Correspondence)** For an ideal $I$, the map $A \mapsto A/I$ is an
>   inclusion-preserving bijection between subrings of $R$ containing $I$ and
>   subrings of $R/I$, and $A$ is an ideal of $R$ iff $A/I$ is an ideal of $R/I$.

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 $d \mid 12$.

$$
% caption: The correspondence theorem for I = 12Z. Ideals of Z containing 12Z
% (the divisor lattice of 12) match the ideals of Z/12Z, order preserved.
\begin{tikzpicture}[font=\footnotesize, >=stealth,
  nd/.style={draw=black, minimum size=6mm, inner sep=2pt}]
\definecolor{acc}{HTML}{4A6FA5}
% left lattice: ideals of Z containing 12Z
\node[nd] (Z)  at (0,3.3) {$Z$};
\node[nd] (2Z) at (-1.1,2.1) {$2Z$};
\node[nd] (3Z) at (1.1,2.1) {$3Z$};
\node[nd] (4Z) at (-1.1,0.9) {$4Z$};
\node[nd] (6Z) at (1.1,0.9) {$6Z$};
\node[nd] (12Z) at (0,-0.3) {$12Z$};
\draw[black] (12Z)--(4Z) (12Z)--(6Z) (4Z)--(2Z) (6Z)--(2Z) (6Z)--(3Z) (2Z)--(Z) (3Z)--(Z);
\node[black, font=\scriptsize] at (0,-1.1) {ideals of $Z$ over $12Z$};
% right lattice: ideals of Z/12Z
\begin{scope}[xshift=6.4cm]
\node[nd, draw=acc] (R)  at (0,3.3) {$R$};
\node[nd, draw=acc] (b2) at (-1.1,2.1) {$(2)$};
\node[nd, draw=acc] (b3) at (1.1,2.1) {$(3)$};
\node[nd, draw=acc] (b4) at (-1.1,0.9) {$(4)$};
\node[nd, draw=acc] (b6) at (1.1,0.9) {$(6)$};
\node[nd, draw=acc] (b0) at (0,-0.3) {$(0)$};
\draw[acc] (b0)--(b4) (b0)--(b6) (b4)--(b2) (b6)--(b2) (b6)--(b3) (b2)--(R) (b3)--(R);
\node[text=acc, font=\scriptsize] at (0,-1.1) {ideals of the quotient};
\end{scope}
\draw[->, black, thick, dashed] (2.1,1.5) -- (4.3,1.5) node[midway, above, font=\scriptsize, black] {reduce mod $I$};
\end{tikzpicture}
$$

## 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**.[^df-ideal]

- **In $\mathbb{Z}$**, every ideal is principal: $n\mathbb{Z} = (n)$, and $(n, m) =
  (d)$ where $d = \gcd(n, m)$. Divisibility becomes containment, $(b) \subseteq (a)
  \iff a \mid b$, 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](/abstract-algebra/factorization-and-polynomials/euclidean-domains-pids-ufds).

> **Worked example.** Compute the ideal $(4, 6)$ of $\mathbb{Z}$ and test whether
> $3 \in (4, 6)$.
>
> Every element of $(4, 6)$ has the form $4s + 6t$ with $s, t \in \mathbb{Z}$, so
> it is a multiple of $\gcd(4, 6) = 2$; conversely $2 = 6 - 4 \in (4, 6)$. Hence
> $$
> (4, 6) = (2) = 2\mathbb{Z}.
> $$
> Membership now reduces to divisibility by $2$: $8 = 4 \cdot 2 \in (4, 6)$, while
> $3$ is odd, so $3 \notin (4, 6)$. In $\mathbb{Z}$ a two-generator ideal always
> collapses to the single generator $\gcd(m, n)$.

Containing a unit is the same as being everything.

> **Proposition (Units and fields via ideals).** Let $I$ be an ideal of a ring
> with $1$.
>
> - $I = R$ if and only if $I$ contains a unit.
> - A commutative ring $R$ is a field if and only if its only ideals are $0$ and
>   $R$.

If $u \in I$ is a unit with inverse $v$, then $1 = vu \in I$, so $r = r \cdot 1 \in
I$ for all $r$.[^df-ideal] 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.

> **Definition (Maximal and prime ideal).** Let $R$ be commutative with $1 \neq
> 0$.
>
> - $M \neq R$ is **maximal** if the only ideals containing $M$ are $M$ and $R$.
> - $P \neq R$ is **prime** if $ab \in P$ implies $a \in P$ or $b \in P$.

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.[^df-ideal] The value
of both definitions is that each translates into a property of the quotient ring.

> **Theorem (Quotient characterizations).** Let $R$ be commutative with $1 \neq
> 0$, and let $M, P$ be ideals.
>
> - $M$ is maximal $\iff$ $R/M$ is a field.
> - $P$ is prime $\iff$ $R/P$ is an integral domain.
>
> Consequently every maximal ideal is prime.

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.[^df-ideal][^ju-max]

$$
% caption: The implication chain. Passing to the quotient turns the ideal property
% into a ring property; a field is a domain, so maximal ideals are prime.
\begin{tikzpicture}[font=\footnotesize, >=stealth,
  nd/.style={draw=black, minimum width=26mm, minimum height=11mm, align=center}]
\definecolor{acc}{HTML}{4A6FA5}
\node[nd, draw=acc, thick, fill=acc!10] (M) at (0,1.5) {$M$ maximal};
\node[nd] (P) at (0,-1.5) {$P$ prime};
\node[nd, draw=acc, fill=acc!8] (F) at (5.6,1.5) {$\frac{R}{M}$: all nonzero\\ invert};
\node[nd] (D) at (5.6,-1.5) {$\frac{R}{P}$: no zero\\ divisors};
\draw[<->, acc, thick] (M) -- (F) node[midway, above, font=\scriptsize] {exactly when};
\draw[<->, thick] (P) -- (D) node[midway, above, font=\scriptsize] {exactly when};
\draw[->, black, thick] (M) -- (P) node[midway, left, font=\scriptsize] {implies};
\draw[->, black, thick] (F) -- (D) node[midway, right, font=\scriptsize] {units cancel};
\end{tikzpicture}
$$

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.[^df-ideal]

| Ideal | Quotient | Class of quotient | Example in $\mathbb{Z}$ |
| --- | --- | --- | --- |
| $I = R$ | $0$ | trivial | — |
| prime, not maximal | integral domain, not a field | domain | $0$ |
| maximal | field | field | $p\mathbb{Z}$ |

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](/abstract-algebra/field-theory/field-extensions-and-algebraic-elements).

> **Worked example.** Identify the quotient $\mathbb{R}[x]/(x^2 + 1)$.
>
> Since $x^2 + 1$ has degree $2$, division leaves each coset a unique
> representative $a + bx$ with $a, b \in \mathbb{R}$. Write $\bar x$ for the class
> of $x$; the relation $x^2 + 1 \equiv 0$ gives $\bar x^2 = -1$. A product of two
> representatives reduces by that relation:
> $$
> (a + b\bar x)(c + d\bar x) = ac + (ad + bc)\bar x + bd\,\bar x^2 = (ac - bd) + (ad + bc)\bar x,
> $$
> the multiplication rule of $\mathbb{C}$ under $\bar x \leftrightarrow i$. Thus
> $\mathbb{R}[x]/(x^2 + 1) \cong \mathbb{C}$. The quotient is a field, so
> $(x^2 + 1)$ is a maximal ideal of $\mathbb{R}[x]$.

> **Worked example.** Show $(2, x)$ is maximal in $\mathbb{Z}[x]$ by identifying
> its quotient.
>
> Evaluation at $0$ followed by reduction mod $2$ is a surjective homomorphism
> $$
> \varphi : \mathbb{Z}[x] \to \mathbb{Z}/2\mathbb{Z}, \qquad p(x) \mapsto p(0) \bmod 2.
> $$
> Its kernel is $(2, x)$: a polynomial equals $2a(x) + x\,b(x)$ exactly when its
> constant term is even, and those are the polynomials with $p(0)$ even.
> The first isomorphism theorem gives
> $$
> \mathbb{Z}[x]/(2, x) \cong \mathbb{Z}/2\mathbb{Z} = \mathbb{F}_2,
> $$
> a field, so $(2, x)$ is maximal. Contrast the chain $(x) \subset (2, x)$: the
> smaller ideal yields $\mathbb{Z}[x]/(x) \cong \mathbb{Z}$, a domain but not a
> field, so $(x)$ is prime but not maximal.

## 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.[^df-hom] 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.[^df-hom]

```algorithm
caption: $\textsc{IsIdeal}(I, R)$ — the ideal test in a ring with identity
if $I$ is empty then return false
for each $a, b \in I$ do
  if $a - b \notin I$ then return false // subgroup under subtraction
for each $r \in R$ and each $a \in I$ do
  if $ra \notin I$ or $ar \notin I$ then return false // absorption
return true
```

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.

[^df-hom]: **Dummit & Foote**, _Abstract Algebra_, §7.3 — Ring Homomorphisms and Quotient Rings: homomorphisms and kernels, ideals as kernels, the quotient ring, the four isomorphism theorems, sums and products of ideals, and reduction modulo $n$ as a Diophantine tool.
[^df-ideal]: **Dummit & Foote**, _Abstract Algebra_, §7.4 — Properties of Ideals: generated and principal ideals, $(2,x)$ not principal, the unit/field criterion, existence of maximal ideals via Zorn's lemma, and the quotient characterizations of maximal and prime ideals.
[^ju-max]: **Judson**, _Abstract Algebra: Theory and Applications_, §16.3–16.4 — Ring Homomorphisms and Ideals; Maximal and Prime Ideals: the factor ring, the isomorphism theorems, and $R/M$ a field iff $M$ maximal, $R/P$ a domain iff $P$ prime.