First isomorphism theorem for rings

Statement

Let $f \colon R \to S$ be a ring homomorphism. Then the kernel $\ker f$ is a two-sided ideal of $R$, and the image $f(R)$ is a subring of $S$. The map $R/\ker f \to f(R)$, $a + \ker f \mapsto f(a)$, is a ring isomorphism. In particular, if $f$ is surjective, then $R/\ker f \cong S$.

Remarks

The reduction map $\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$ has kernel $n\mathbb{Z}$. The evaluation map $\mathbb{R}[x] \to \mathbb{C}$, $p \mapsto p(i)$, is surjective with kernel $(x^2 + 1)$. So $\mathbb{R}[x]/(x^2 + 1) \cong \mathbb{C}$. The theorem has the same form for groups and for modules.

Explanations

L1 human

A homomorphism forgets exactly the information in its kernel. If you divide out the kernel, the homomorphism becomes a bijection onto its image. So every image of a ring is a quotient of that ring.