Proof of the first isomorphism theorem for rings

Proof

1. Let $f \colon R \to S$ be a ring homomorphism, and write $K = \ker f$ for its kernel.
2. As for any additive map, $f(0) = 0$ and $f(a - b) = f(a) - f(b)$ for $a, b \in R$. Indeed, $f(0) = f(0) + f(0)$ gives $f(0) = 0$, and then $f(a - b) + f(b) = f(a)$.
3. The set $K$ is an additive subgroup of $R$. Indeed, $0 \in K$, and if $a, b \in K$, then $f(a - b) = 0 - 0 = 0$.
4. For $r \in R$ and $a \in K$, $f(r a) = f(r) \cdot 0 = 0$ and $f(a r) = 0 \cdot f(r) = 0$. So $r a, a r \in K$, and $K$ is a two-sided ideal of $R$.
5. The image $f(R)$ is a subring of $S$. Indeed, $1 = f(1) \in f(R)$, $f(a) - f(b) = f(a - b)$ and $f(a) f(b) = f(a b)$.
6. Since $K \subseteq \ker f$, the universal property of the quotient ring gives a ring homomorphism $\bar{f} \colon R/K \to S$ with $\bar{f}(a + K) = f(a)$. Here $R/K$ is the quotient ring.
7. The values of $\bar{f}$ are the values $f(a)$, so $\bar{f}$ takes values in $f(R)$. As a map $R/K \to f(R)$, it is still a ring homomorphism, because $f(R)$ is a subring with the operations of $S$.
8. This map is surjective onto $f(R)$, because each $f(a)$ equals $\bar{f}(a + K)$.
9. Its kernel is $\set{0 + K}$. Indeed, $\bar{f}(a + K) = 0$ means $f(a) = 0$, that is, $a \in K$. Then $a - 0 \in K$, so $a + K = 0 + K$.
10. A ring homomorphism with zero kernel is injective. So $\bar{f} \colon R/K \to f(R)$ is bijective, and it is a ring isomorphism.
11. If $f$ is surjective, then $f(R) = S$. So $\bar{f}$ is an isomorphism from $R/\ker f$ to $S$, and $R/\ker f \cong S$.