Every time we met a new algebraic object in the group chapter, the next questions were the same: what are the maps that respect it, what do their kernels look like, and how do you divide one out? This page runs that whole story again for rings. Most of it will feel familiar. The one genuinely new thing is the ideal, the ring version of a normal subgroup, and it turns out to be a stronger condition than you might guess.
A ring has two operations, so a map that "respects the structure" has to respect both.
Suppose $R, S$ are rings. A ring homomorphism is a map $\phi : R \rightarrow S$ satisfying:
Condition (i) says $\phi$ is a group homomorphism $(R, +) \rightarrow (S, +)$, so everything from Homomorphisms applies for free: $\phi(0) = 0$, $\phi(-a) = -\phi(a)$, and $\phi$ is injective exactly when its kernel is $\{0\}$.
The kernel is defined using the additive identity, exactly as for groups, since $(R, +)$ is the group in the room.
The kernel of a ring homomorphism $\phi : R \rightarrow S$ is $\ker(\phi) = \{a \in R : \phi(a) = 0_S\}$.
And a homomorphism with an inverse is a relabeling, just like before.
If a ring homomorphism $\phi$ is bijective it is a ring isomorphism.
Now some examples. The first is the one you've been using since the Modular Arithmetic page.
$\phi : \mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z}$, $a \mapsto \overline{a}$.
This is a ring homomorphism because addition and multiplication of residue classes modulo $n$ are well-defined: $\overline{a + b} = \overline{a} + \overline{b}$ and $\overline{ab} = \overline{a}\,\overline{b}$ are the definitions. Its kernel is $n\mathbb{Z}$.
Polynomial rings give a pair of examples that look almost identical, where one works and one doesn't. Write $f(x) = \sum_{i=0}^n a_i x^i$ and $g(x) = \sum_{j=0}^m b_j x^j$.
$\phi : R[x] \rightarrow R$, $f(x) \mapsto f(0)$.
$(f+g)(x) = (a_0 + b_0) + (a_1 + b_1)x + \dots \implies \phi(f+g) = a_0 + b_0 = \phi(f) + \phi(g)$.
$(fg)(x) = a_0 b_0 + (a_0 b_1 + a_1 b_0)x + \dots \implies \phi(fg) = a_0 b_0 = \phi(f)\phi(g)$.
So this is a ring homomorphism.
The constant term of a product only depends on the constant terms. The $x$ coefficient isn't so lucky.
$\phi : R[x] \rightarrow R$, $\phi(a_n x^n + \dots + a_0) = a_1$.
$(f+g)(x) = (a_0 + b_0) + (a_1 + b_1)x + \dots \implies \phi(f+g) = a_1 + b_1 = \phi(f) + \phi(g)$.
$(fg)(x) = a_0 b_0 + (a_0 b_1 + a_1 b_0)x + \dots \implies \phi(fg) = a_0 b_1 + a_1 b_0 \neq a_1 b_1$ (in general).
So $\phi$ is not a ring homomorphism. (Concretely, in $\mathbb{Z}[x]$: $\phi(x \cdot x) = \phi(x^2) = 0$, but $\phi(x)\phi(x) = 1$.)
A map can respect addition perfectly and still fail on multiplication. Here's the smallest example.
$\phi : \mathbb{Z} \rightarrow \mathbb{Z}$, $n \mapsto 2n$.
$\phi(m+n) = 2(m+n) = 2m + 2n = \phi(m) + \phi(n)$.
$\phi(1 \cdot 1) = 2 \neq \phi(1) \cdot \phi(1) = 4$.
This is not a ring homomorphism.
So "multiply by $k$" is always additive, but multiplicative only sometimes. Which $k$ work? In $\mathbb{Z}$, $\phi(n) = kn$ needs $k \cdot ab = k^2 \cdot ab$ for all $a, b$, so $k = k^2$, which leaves only $k = 0$ and $k = 1$. In $\mathbb{Z}/n\mathbb{Z}$ there can be more. Pick $n$ and $k$; each cell of the grid checks $\phi(ab) = \phi(a)\phi(b)$ for one pair.
Notice that $x \mapsto 3x$ on $\mathbb{Z}/6\mathbb{Z}$ is a ring homomorphism that sends $1$ to $3$, not to $1$. These notes don't require $\phi(1) = 1$, so that's allowed. Some books do require it (they call these "unital" homomorphisms). When you read other sources, check which convention they use.
As with groups, the image and the kernel are substructures.
If $\phi : R \rightarrow S$ is a ring homomorphism then
Proof. Both are additive subgroups, by the group version. (i) $\phi(a)\phi(b) = \phi(ab) \in \operatorname{im}(\phi)$. (ii) If $a, b \in \ker\phi$ then $\phi(ab) = \phi(a)\phi(b) = 0 \cdot 0 = 0$, so $ab \in \ker\phi$. $\blacksquare$
But the kernel is much more than a subring. Look at the proof of (ii) again: we only needed one of $a, b$ to be in the kernel. If $a \in \ker\phi$ and $r$ is anything in $R$, then $\phi(ra) = \phi(r) \cdot 0 = 0$. The kernel swallows products with arbitrary elements. That property is the star of the rest of this chapter.
Let's see why that swallowing property matters, by trying to build a quotient ring.
Suppose $\phi : R \rightarrow S$ is a ring homomorphism, let $K = \ker \phi$ and let $R/K$ be the collection of additive cosets $a + K$.
$(R/K, +)$ is a group, because $(R, +)$ is Abelian, & $(K, +) \le (R, +)$. (Every subgroup of an Abelian group is normal.)
Multiplication of cosets by the rule $(a+K)(b+K) = ab+K$ is also well-defined, and $(R/K, +, \cdot)$ is a ring.
Why is multiplication of cosets well-defined? Suppose we used different representatives: $(a+x)+K = a+K$ and $(b+y)+K = b+K$, for some $x, y \in K$. Then
$$(a+x)(b+y) = ab + ay + xb + xy,$$
and $ay + xb + xy \in K$ (each term has a factor in $K$) $\implies (ab + ay + xb + xy)+K = ab+K$.
On Quotient Groups, the coset product $(aH)(bH) = abH$ was well-defined exactly when $H$ was normal. For rings, addition is always fine (everything is Abelian), and the whole question moves to multiplication.
The proof used exactly one fact about $K$: a product $ay$ or $xb$, with one factor in $K$ and the other anywhere in $R$, lands back in $K$. That's the definition we want. In a non-commutative ring, "multiply on the left" and "multiply on the right" are different conditions, so there are three versions.
A subring $I \subseteq R$ is a left ideal if $\forall a \in R$, $aI \subseteq I$.
The mirror image:
A subring $I \subseteq R$ is a right ideal if $\forall a \in R$, $Ia \subseteq I$.
And both at once, which is the one that makes quotients work:
A subring $I \subseteq R$ is an ideal if $\forall a \in R$, $aI \subseteq I$ and $Ia \subseteq I$.
In a commutative ring the three notions coincide. The point of the definition is this theorem, which says ideals are exactly the subsets you can divide out.
If $I \subseteq R$ is an ideal then the additive group $R/I$ forms a ring with $(a+I)(b+I) = ab+I$.
Conversely if $I$ is an additive subgroup and multiplication as above is well-defined, then $I$ is an ideal.
Proof. Well-definedness is the computation in the Motivation box, with $K$ replaced by $I$: $ay \in I$ and $xb, xy \in I$ because $I$ is an ideal. The ring axioms for $R/I$ are inherited from $R$, e.g. $(a+I)\big((b+I)(c+I)\big) = a(bc) + I = (ab)c + I = \big((a+I)(b+I)\big)(c+I)$.
Conversely, suppose the rule is well-defined, and let $r \in R$, $x \in I$. Since $x + I = 0 + I$, well-definedness gives $(r + I)(x + I) = (r + I)(0 + I)$, i.e. $rx + I = 0 + I$, so $rx \in I$. Similarly $(x + I)(r + I) = (0 + I)(r + I)$ gives $xr \in I$. $\blacksquare$
When $I$ is an ideal of $R$, the ring $R/I$ with $(a+I) + (b+I) = (a+b) + I$ and $(a+I)(b+I) = ab+I$ is the quotient ring of $R$ by $I$.
The converse is the interesting half, and it's easiest to see in a ring with two coordinates. In $\mathbb{Z} \times \mathbb{Z}$ (add and multiply coordinatewise), compare the ideal $\mathbb{Z} \times \{0\}$ with the diagonal $\Delta = \{(n, n) : n \in \mathbb{Z}\}$. The diagonal is a perfectly good subring (it even contains $(1, 1)$), but it's not an ideal: $(1, 0) \cdot (5, 5) = (5, 0) \notin \Delta$. Drag $a$ and $b$ along their cosets and watch which coset the product lands in.
When you're asked "is $I$ an ideal?", checking that $I$ is a subring is only half the job, and it's the easy half. The real test is absorption: take an arbitrary $r \in R$ (not just $r \in I$!) and an arbitrary $x \in I$, and show $rx, xr \in I$. A subring like $\Delta$ or $\mathbb{Z} \subseteq \mathbb{Q}$ fails exactly there: $\frac{1}{2} \cdot 1 \notin \mathbb{Z}$.
An ideal absorbs everything it touches. Multiply it by anything in the ring and it just... keeps it. It's the black hole of subrings. The name "ideal" is a very generous way to describe that personality.
With ideals and quotient rings in hand, all four isomorphism theorems carry over from groups. The proofs are the group proofs plus one line checking multiplication, so we'll state them and focus on examples.
Compare each of these with its group version on Isomorphism Theorems. Every "normal subgroup" became "ideal," every "subgroup" became "subring," and every $\cong$ is now an isomorphism of rings.
If $\phi : R \rightarrow S$ is a ring homomorphism then $\ker \phi$ is an ideal of $R$ and $R/\ker \phi \cong \operatorname{im}(\phi)$ (isomorphic as rings).
Also: If $I \subseteq R$ is an ideal then $I$ is the kernel of the ring homomorphism $\pi : R \rightarrow R/I$, $a \mapsto a+I$.
So "ideal" and "kernel of a ring homomorphism" are the same thing. In practice the first isomorphism theorem is the fastest way to identify a quotient: find a surjective homomorphism whose kernel is your ideal. Here's the model example.
$R = C([0,1])$. Let $I = \{f \in R : f(1/2) = 0\}$. What can you say about $R/I$?
$I$ is an ideal:
The quotient: Let $\phi : R \rightarrow \mathbb{R}$ be defined by $\phi(f) = f(1/2)$. Then $\phi$ is a surjective ring homomorphism (the constant function $c$ maps to $c$) and $\ker \phi = I \implies R/I \cong \mathbb{R}$ (1st isomorphism theorem).
What does a coset $f + I$ actually look like? It's every continuous function $g$ with $g - f \in I$, i.e. $g(1/2) = f(1/2)$. All the functions through one point. The whole coset is pinned down by one real number, and that's the isomorphism $R/I \cong \mathbb{R}$. Press the button to draw new members of the same coset.
Nothing about $[0,1]$ or $\mathbb{R}$ was special there. The same argument works for any function ring, at any point.
Suppose $A$ is a ring, $X$ is a non-empty set, $R = \{f : X \rightarrow A\}$. Then $\forall c \in X$, $I_c = \{f \in R : f(c) = 0_A\}$ is an ideal of $R$, and $R/I_c \cong A$.
Evaluation at $c$, $\phi(f) = f(c)$, is a ring homomorphism $R \rightarrow A$ (addition and multiplication are pointwise). It's surjective: the constant function $a$ maps to $a$. Its kernel is $I_c$. Apply the first isomorphism theorem.
The second theorem mixes a subring with an ideal.
Suppose $A \le R$ (a subring of $R$) and $B \subseteq R$ is an ideal. Then $A+B = \{a+b : a \in A, b \in B\}$ is a subring of $R$, $A \cap B$ is an ideal of $A$, and
$$(A+B)/B \cong A/(A \cap B)$$
(ring isomorphism). (Since $B$ is an ideal of $R$, it's an ideal of $A + B$, so the left side makes sense.)
The third is "cancelling" in a quotient of a quotient.
Suppose $I, J \subseteq R$ are ideals, $I \subseteq J$. Then $J/I$ is an ideal of $R/I$ and
$$(R/I)/(J/I) \cong R/J$$
(ring isomorphism).
$(R/I)/(J/I) \cong R/J$. Cancel the $I$'s like fractions. Your middle-school teacher would be proud and horrified in exactly equal amounts.
And the fourth says the ideals of a quotient are just the ideals upstairs that contain what you divided out.
If $I \subseteq R$ is an ideal then the correspondence $A \leftrightarrow A/I$ is an inclusion preserving bijection between the set of all subrings of $R$ that contain $I$ and the set of subrings of $R/I$. Also, $A$ is an ideal of $R$ if and only if $A/I$ is an ideal of $R/I$.
For $R = \mathbb{Z}$ and $I = 12\mathbb{Z}$, the ideals of $\mathbb{Z}$ containing $12\mathbb{Z}$ are the $d\mathbb{Z}$ with $d \mid 12$ (we'll see on the next page that every ideal of $\mathbb{Z}$ is some $d\mathbb{Z}$). The fourth theorem matches them, inclusion for inclusion, with the ideals of $\mathbb{Z}/12\mathbb{Z}$.
The correspondence for $I = 12\mathbb{Z} \subseteq \mathbb{Z}$. Lines go up from smaller ideals to bigger ones.
Each $d\mathbb{Z}$ on the left becomes $d\mathbb{Z}/12\mathbb{Z} = \{\overline{0}, \overline{d}, \overline{2d}, \dots\}$ on the right, the ideal of $\mathbb{Z}/12\mathbb{Z}$ generated by $\overline{d}$. The two pictures have the same shape. By the third theorem, the quotients match too: $(\mathbb{Z}/12\mathbb{Z})/(4\mathbb{Z}/12\mathbb{Z}) \cong \mathbb{Z}/4\mathbb{Z}$.
Two about homomorphisms, one about ideals.
Prove that if $\phi : R \rightarrow S$ is a surjective ring homomorphism and $R$ has an identity $1$, then $\phi(1)$ is an identity of $S$.
Proof. Let $s \in S$. Since $\phi$ is surjective, $s = \phi(r)$ for some $r \in R$. Then $\phi(1)s = \phi(1)\phi(r) = \phi(1r) = \phi(r) = s$, and similarly $s\phi(1) = s$. $\blacksquare$
(Without surjectivity this fails: $x \mapsto 3x$ on $\mathbb{Z}/6\mathbb{Z}$ sends $1$ to $3$, which is not an identity of $\mathbb{Z}/6\mathbb{Z}$. It is the identity of the image $\{0, 3\}$, though.)
Find all ring homomorphisms $\phi : \mathbb{Z} \rightarrow \mathbb{Z}$.
Let $k = \phi(1)$. Additivity gives $\phi(n) = nk$ for every $n \in \mathbb{Z}$ (it's a group homomorphism out of the cyclic group $\mathbb{Z} = \langle 1 \rangle$). Multiplicativity gives $k = \phi(1 \cdot 1) = \phi(1)^2 = k^2$, so $k(k - 1) = 0$, and since $\mathbb{Z}$ has no zero divisors, $k = 0$ or $k = 1$.
Both work, so there are exactly two: the zero map and the identity map.
Let $R = M_2(\mathbb{R})$ and let $L$ be the set of matrices whose second column is zero. Show that $L$ is a left ideal but not a right ideal.
$L$ is closed under subtraction and multiplication, so it's a subring. For any $A \in R$ and $X \in L$, the second column of $AX$ is $A$ times the second column of $X$, which is $A \cdot 0 = 0$. So $AX \in L$: $L$ is a left ideal.
But $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \in L$ and $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \notin L$. So $L$ is not a right ideal.
The second isomorphism theorem was stated without proof. Here's your chance, using the first.
Let $A$ be a subring of $R$ and $B$ an ideal of $R$. Prove that $A + B$ is a subring of $R$, and that $(A+B)/B \cong A/(A \cap B)$. (Hint: consider $\psi : A \rightarrow (A+B)/B$, $\psi(a) = a + B$.)
Proof. $A + B$ is a subring. It contains $0 = 0 + 0$. For $a, a' \in A$ and $b, b' \in B$: $(a + b) - (a' + b') = (a - a') + (b - b') \in A + B$, and $(a + b)(a' + b') = aa' + (ab' + ba' + bb')$, where $aa' \in A$ and the bracket is in $B$ because $B$ is an ideal.
$B$ is an ideal of $A + B$, since it absorbs products with every element of $R$. So $(A+B)/B$ is a ring.
The map. $\psi(a) = a + B$ is a ring homomorphism, since it's the restriction of $\pi : R \rightarrow R/B$. It's surjective: any coset $(a + b) + B$ equals $a + B = \psi(a)$, since $b \in B$. Its kernel is $\{a \in A : a \in B\} = A \cap B$.
By the first isomorphism theorem, $A \cap B$ is an ideal of $A$ and $A/(A \cap B) \cong (A + B)/B$. $\blacksquare$