A ring homomorphism is a group homomorphism for addition that also respects multiplication. Its kernel turns out to be an ideal, just as the kernel of a group homomorphism is a normal subgroup.
Definition. $\varphi: R \to S$ is a ring homomorphism if $$\varphi(a + b) = \varphi(a) + \varphi(b) \quad \text{and} \quad \varphi(ab) = \varphi(a)\varphi(b)$$ for all $a, b$. Its kernel is $\ker\varphi = \{a : \varphi(a) = 0\}$.
Why it works
- Reduction $\mathbb{Z} \to \mathbb{Z}_{n}$, $a \mapsto a \bmod n$: kernel $(n)$. Also $\mathbb{Z}_{12} \to \mathbb{Z}_{4}$, since $4 \mid 12$.
- Evaluation $\mathbb{R}[x] \to \mathbb{R}$, $f \mapsto f(c)$: kernel $\{f : f(c) = 0\} = (x - c)$, by the factor theorem.
- Complex conjugation $\mathbb{C} \to \mathbb{C}$: one-to-one and onto.
- Not one: $x \mapsto 2x$ on $\mathbb{Z}$ keeps sums but $\varphi(1 \cdot 1) = 2 \ne 4 = \varphi(1)\varphi(1)$.
Worked example: Evaluation
- $f(-2) = -8 + 6 + 1 = -1$.
- The kernel is the polynomials with root $-2$: the ideal $(x + 2)$.
Answer: $-1$; kernel $(x + 2)$.
Worked example: Two moduli at once
- $\varphi(a) = (0, 0)$ when $4 \mid a$ and $6 \mid a$, that is, $12 \mid a$.
- So $\ker\varphi = (12)$. The image has $12$ elements, fewer than $24$, so $\varphi$ is not onto (the moduli share a factor).
Answer: $(12)$.
Worked example: Is it a ring homomorphism?
- Both preserve sums. For products, $\varphi(xy) = kxy$ must equal $\varphi(x)\varphi(y) = k^{2}xy$, so we need $k^{2} \equiv k \pmod 6$.
- $3^{2} = 9 \equiv 3$: yes. $4^{2} = 16 \equiv 4$: yes. (But $2^{2} = 4 \not\equiv 2$, so $x \mapsto 2x$ is not.)
Answer: Both are, since $3$ and $4$ satisfy $k^{2} = k$ in $\mathbb{Z}_{6}$.
A ring homomorphism out of $\mathbb{Z}$ or $\mathbb{Z}_{n}$ is decided by $e = \varphi(1)$, because $\varphi(k) = \varphi(1 + \cdots + 1) = ke$. The product rule then forces $e^{2} = \varphi(1 \cdot 1) = e$: the image of $1$ must be an idempotent. In a field or an integral domain the only idempotents are $0$ and $1$ (from $e(e - 1) = 0$), but rings like $\mathbb{Z}_{6}$ have more, which is why the last example found extra homomorphisms.
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