Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2005 / 3

An ideal is an additive subgroup of a ring closed under multiplication by arbitrary ring elements; in a potentially noncommutative ring, a two-sided ideal must absorb multiplication on both sides. For a ring homomorphism $\varphi$, its kernel of a ring homomorphism $K=\{r:\varphi (r)=0\}$ contains zero, is closed under subtraction, and satisfies $\varphi (rk)=\varphi (r)\varphi (k)=0$ and $\varphi (kr)=0$. Thus $K$ is an ideal.

Define $\tilde{\varphi }:R/K\to S$ by $r+K\mapsto \varphi (r)$. It is well-defined because two representatives differ by an element of $K$, preserves addition and multiplication, is surjective by hypothesis, and is injective because its zero class is precisely $K$. Therefore $\boxed{R/\ker \varphi \cong S}$, the first isomorphism theorem for rings.

Apply this to the evaluation homomorphism $\mathbb{C}[z]\to \mathbb{C}$, $p\mapsto p(1)$. Constants prove surjectivity, and division by $z-1$ shows that its kernel of a ring homomorphism is $(z-1)$. Hence $\boxed{\mathbb{C}[z]/(z-1)\cong \mathbb{C}}$.