Theorem 1   Let $S$ be a commutative ring, and let $\psi\colon R\to S$ be a homomorphism. Further, let $s\in S$ . Then there is a unique homomorphism $\phi\colon R[X]\to S$ taking $X$ to $s$ and taking every $r\in R$ to $\psi(r)$ .

Proof. We first prove existence. Let $f\in R[X]$ . Then by definition there is some finite list of $a_i$ such that $f = \sum_i a_i X^i$ . Then define $\phi(f)$ to be $\sum_i \psi(a_i) s^i$ . It is clear from the definition of addition and multiplication on polynomials that $\phi$ is a homomorphism; the definition makes it clear that $\phi(X)=s$ and $\phi(r)=\psi(r)$ .

Now, to show uniqueness, suppose $\gamma$ is any homomorphism satisfying the conditions of the theorem, and let $f\in R[X]$ . Write $f = \sum_i a_i X^i$ as before. Then $\gamma(a_i) = \psi(a_i)$ and $\gamma(s)$ by assumption. But then since $\gamma$ is a homomorphism, $\gamma(a_iX^i) = \psi(a_i)s^i$ and $\gamma(f) = \sum_i \psi(a_i) s^i = \phi(f)$ .