Chapter 13

Medium

Ring Homomorphisms

00 · Symbol Glossary

$\varphi: R \to S$Ring homomorphism from R to S

A function respecting both ring operations: $\varphi(a+b)=\varphi(a)+\varphi(b)$ and $\varphi(ab)=\varphi(a)\varphi(b)$ . Twice the structure to preserve compared to a group homomorphism.

$\ker\varphi$Kernel of a ring homomorphism

$\ker\varphi = \{a\in R : \varphi(a)=0_S\}$ — everything collapsing to zero. Turns out to always be an ideal, not merely a subring.

$\mathrm{im}\,\varphi$Image of a ring homomorphism

$\mathrm{im}\,\varphi = \{\varphi(a):a\in R\}\subseteq S$ — everything actually hit. A subring of $S$ .

$R\cong S$Ring isomorphism

A bijective ring homomorphism exists between $R$ and $S$ — the rings are structurally identical.

01 · Ring Homomorphisms

Definition — Ring Homomorphism

A function $\varphi: R\to S$ between rings is a ring homomorphism if for all $a,b\in R$ :

\varphi(a+b)=\varphi(a)+\varphi(b), \qquad \varphi(ab)=\varphi(a)\varphi(b)

If both rings have unity, $\varphi$ is additionally required to satisfy $\varphi(1_R)=1_S$ (this course's convention).

Already half a group homomorphism

The additive condition $\varphi(a+b)=\varphi(a)+\varphi(b)$ says exactly that $\varphi$ is a group homomorphism from $(R,+)$ to $(S,+)$ . Every fact from Chapter 06 about additive behavior — preserving $0$ , preserving additive inverses — transfers immediately. The multiplicative condition is the genuinely new ingredient.

02 · Basic Properties

Theorem 13.1 — Ring Homomorphisms Preserve Additive Structure

For a ring homomorphism $\varphi: R\to S$ : (a) $\varphi(0_R)=0_S$ ; (b) $\varphi(-a)=-\varphi(a)$ for every $a\in R$ .

Proof

Both are Theorem 6.1(a) and (b) applied to $\varphi$ as a group homomorphism of $(R,+)\to(S,+)$ — no new argument is needed, since these facts only ever used the additive structure.

Theorem 13.2 — Ring Homomorphisms Preserve Units

If $\varphi: R\to S$ is a (unital) ring homomorphism and $a\in R^\times$ , then $\varphi(a)\in S^\times$ , with $\varphi(a^{-1})=\varphi(a)^{-1}$ .

Proof

Since $a\in R^\times$ , $aa^{-1}=a^{-1}a=1_R$ . Applying $\varphi$ : $\varphi(a)\varphi(a^{-1}) = \varphi(aa^{-1}) = \varphi(1_R) = 1_S$ , and similarly $\varphi(a^{-1})\varphi(a)=1_S$ . So $\varphi(a^{-1})$ satisfies the defining property of $\varphi(a)^{-1}$ in $S$ ; by uniqueness of inverses (Theorem 1.2, applied inside the monoid $(S,\times)$ ), $\varphi(a^{-1})=\varphi(a)^{-1}$ , and $\varphi(a)\in S^\times$ .

03 · Kernel and Image

Theorem 13.3 — The Kernel Is an Ideal

$\ker\varphi = \{a\in R : \varphi(a)=0_S\}$ is an ideal of $R$ .

Proof

By Theorem 6.3 (applied to the additive group homomorphism), $\ker\varphi$ is an additive subgroup of $R$ : condition I1 holds. For absorption (I2): take $r\in R$ , $a\in\ker\varphi$ . Then $\varphi(ra) = \varphi(r)\varphi(a) = \varphi(r)\cdot0_S = 0_S$ (using $\varphi(a)=0_S$ and Theorem 11.1 applied in $S$ ). So $ra\in\ker\varphi$ . Both ideal conditions hold.

Ideals are exactly ring-homomorphism kernels

This is the ring-theoretic echo of Theorem 6.3: just as every normal subgroup is exactly the kernel of some group homomorphism (namely the quotient map), every ideal is exactly the kernel of some ring homomorphism (the quotient map $R\to R/I$ , Chapter 12). Absorption — the extra condition ideals need beyond subring closure — is precisely what's forced by wanting $\ker\varphi$ to swallow multiplication by the whole ring, not just by itself.

Theorem 13.4 — The Image Is a Subring

$\mathrm{im}\,\varphi \le S$ (as a subring).

Proof

Nonempty: $0_S=\varphi(0_R)\in\mathrm{im}\,\varphi$ . Subtraction-closed: $\varphi(a)-\varphi(b) = \varphi(a)+(-\varphi(b)) = \varphi(a) + \varphi(-b) = \varphi(a-b) \in \mathrm{im}\,\varphi$ (using Theorem 13.1(b) and additivity). Multiplication-closed: $\varphi(a)\varphi(b)=\varphi(ab)\in\mathrm{im}\,\varphi$ . By Theorem 11.6, $\mathrm{im}\,\varphi$ is a subring.

Theorem 13.5 — Injective iff Trivial Kernel

A ring homomorphism $\varphi: R\to S$ is injective if and only if $\ker\varphi=\{0_R\}$ .

Proof

Identical to Theorem 6.4's proof, applied to the additive group homomorphism — injectivity is purely an additive-structure question, unaffected by multiplication.

04 · The First Isomorphism Theorem for Rings

Theorem 13.6 — First Isomorphism Theorem (Rings)

For any ring homomorphism $\varphi: R\to S$ :

R/\ker\varphi \;\cong\; \mathrm{im}\,\varphi

Proof

Write $I=\ker\varphi$ (an ideal by Theorem 13.3, so $R/I$ is a ring by Theorem 12.4). Define $\bar\varphi: R/I\to\mathrm{im}\,\varphi$ by $\bar\varphi(a+I)=\varphi(a)$ .

By Theorem 6.5 (applied to the additive group homomorphism), $\bar\varphi$ is well-defined, additive, injective, and surjective onto $\mathrm{im}\,\varphi$ — every part of that proof used only the additive structure, so it transfers unchanged.

It remains to check $\bar\varphi$ respects multiplication: $\bar\varphi\big((a+I)(b+I)\big) = \bar\varphi(ab+I) = \varphi(ab) = \varphi(a)\varphi(b) = \bar\varphi(a+I)\bar\varphi(b+I)$ . And $\bar\varphi(1+I)=\varphi(1_R)=1_S$ . So $\bar\varphi$ is a bijective ring homomorphism: a ring isomorphism.

Example — Z/nZ ≅ Z_n, now as rings

$\pi:\mathbb{Z}\to\mathbb{Z}_n$ , $\pi(a)=[a]$ , is a ring homomorphism (additivity from Chapter 06; multiplicativity is exactly Theorem 1.5's well-definedness of $[a][b]=[ab]$ ; $\pi(1)=[1]$ ). $\ker\pi = n\mathbb{Z}$ (Chapter 06), and $\pi$ is surjective. By Theorem 13.6:

\mathbb{Z}/n\mathbb{Z} \;\cong\; \mathbb{Z}_n

now as an isomorphism of rings, not merely of additive groups — confirming Chapter 12's quotient-ring construction reproduces $\mathbb{Z}_n$ 's full ring structure, multiplication included.

Example — Evaluation homomorphisms

Let $R$ be the ring of all functions $f:[0,1]\to\mathbb{R}$ under pointwise addition and multiplication: $(f+g)(x):=f(x)+g(x)$ , $(fg)(x):=f(x)g(x)$ . Fix $c\in[0,1]$ and define $\mathrm{ev}_c: R\to\mathbb{R}$ by $\mathrm{ev}_c(f) = f(c)$ .

Checking the axioms: $\mathrm{ev}_c(f+g) = (f+g)(c) = f(c)+g(c) = \mathrm{ev}_c(f)+\mathrm{ev}_c(g)$ . $\mathrm{ev}_c(fg) = (fg)(c) = f(c)g(c) = \mathrm{ev}_c(f)\mathrm{ev}_c(g)$ . $\mathrm{ev}_c(\mathbf{1}) = \mathbf{1}(c) = 1$ (where $\mathbf{1}$ is the constant function 1). A genuine ring homomorphism.

$\ker(\mathrm{ev}_c) = \{f\in R : f(c)=0\}$ — every function vanishing at the single point $c$ — is, by Theorem 13.3, an ideal of $R$ , with no direct computation needed.

$\mathrm{ev}_c$ is surjective onto $\mathbb{R}$ (every real number is $f(c)$ for the constant function $f\equiv$ that number). By Theorem 13.6, $R/\ker(\mathrm{ev}_c) \cong \mathbb{R}$ — a field. By Theorem 12.6 (Chapter 12), this means $\ker(\mathrm{ev}_c)$ is a maximal ideal of $R$ , purely as a consequence of the First Isomorphism Theorem, without any direct argument about ideals strictly between it and $R$ .

05 · Exercises

EXERCISE 13.1

Check both the additive and multiplicative homomorphism conditions directly using properties of complex conjugation.

Let $\varphi:\mathbb{C}\to\mathbb{C}$ , $\varphi(z)=\bar z$ (complex conjugation). Additivity: $\overline{z+w} = \bar z+\bar w$ (standard fact about conjugation). Multiplicativity: $\overline{zw}=\bar z\bar w$ (also standard). Unity: $\bar1=1$ . All conditions hold: complex conjugation is a ring homomorphism $\mathbb{C}\to\mathbb{C}$ — in fact bijective (its own inverse), so a ring automorphism ( $\mathbb{C}\cong\mathbb{C}$ via a nontrivial map).

Show that complex conjugation $\varphi(z)=\bar z$ is a ring homomorphism $\mathbb{C}\to\mathbb{C}$ .

EXERCISE 13.2

Apply Theorem 13.5 directly: compute the kernel of the reduction-mod-6 map on $\mathbb{Z}_{12}$ .

Let $\varphi:\mathbb{Z}_{12}\to\mathbb{Z}_6$ , $\varphi([a]_{12}) = [a]_6$ (well-defined by a Theorem 1.5-style argument, since $12=2\times6$ ). $\ker\varphi = \{[a]_{12} : a\equiv0\pmod6\} = \{[0],[6]\}$ — not just $\{[0]\}$ .

By Theorem 13.5, since $\ker\varphi\neq\{0\}$ , $\varphi$ is not injective.

Let $\varphi:\mathbb{Z}_{12}\to\mathbb{Z}_6$ send $[a]_{12}\mapsto[a]_6$ . Find $\ker\varphi$ and determine injectivity using Theorem 13.5.

EXERCISE 13.3

Apply Theorem 13.6 to the map from Exercise 13.2, using that it's surjective onto $\mathbb{Z}_6$ .

By Theorem 13.6, $\mathbb{Z}_{12}/\{[0],[6]\} \cong \mathbb{Z}_6$ (as rings). Order check: $\lvert\mathbb{Z}_{12}\rvert/\lvert\{[0],[6]\}\rvert = 12/2=6=\lvert\mathbb{Z}_6\rvert$ . ✓

State the conclusion of Theorem 13.6 applied to the homomorphism from Exercise 13.2, and check the orders match.

EXERCISE 13.4

Compute $\varphi(0)$ two ways: via Theorem 13.1, and by testing whether the candidate map even respects addition.

Consider $\varphi:\mathbb{Z}\to\mathbb{Z}$ , $\varphi(n)=n+1$ (the same non-example from Chapter 06). $\varphi(0)=1\neq0$ — violating Theorem 13.1(a) directly. Since every genuine ring homomorphism must send $0_R\mapsto0_S$ , $\varphi$ cannot be a ring homomorphism (nor even a group homomorphism, as Chapter 06 already showed).

Explain, using Theorem 13.1, why $\varphi(n)=n+1$ cannot be a ring homomorphism $\mathbb{Z}\to\mathbb{Z}$ .

06 · Chapter Summary

Concept	Statement
Ring homomorphism	$\varphi(a+b)=\varphi(a)+\varphi(b)$ , $\varphi(ab)=\varphi(a)\varphi(b)$ , $\varphi(1_R)=1_S$
Preserves $0$ , additive inverses	Thm 13.1 (inherited from group homomorphism facts)
Preserves units	$a\in R^\times \implies \varphi(a)\in S^\times$ (Thm 13.2)
Kernel	Always an ideal of $R$ (Thm 13.3)
Image	Always a subring of $S$ (Thm 13.4)
Injective test	$\varphi$ injective $\iff \ker\varphi=\{0\}$ (Thm 13.5)
First Isomorphism Theorem	$R/\ker\varphi \cong \mathrm{im}\,\varphi$ (Thm 13.6)

Next: Chapter 14 — Polynomial Rings builds $R[x]$ , the ring of polynomials over a ring $R$ , setting up the central examples — degree, roots, and factorization — needed for Euclidean domains, PIDs, and eventually field extensions.