# evaluation homomorphism

Let $R$ be a commutative ring and let $R[X]$ be the ring of polynomials with coefficients in $R$.

###### 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 $\psi(r)$.

This amounts to saying that polynomial rings are free objects in the category of $R$-algebras; the theorem then states that they are projective. This is true in much greater generality; in fact, the property of being projective is intended to extract the essential property of being free.

###### 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, $\gamma(a_{i}X^{i})=\psi(a_{i})s^{i}$ and $\gamma(f)=\sum_{i}\psi(a_{i})s^{i}=\phi(f)$. ∎

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