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Category Multiplication Obvious Polynomial Ring Addition Ring Commutative Ring Property Basic Polynomial

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1	Obvious
1	Commutative Ring
1	Ring
1	Multiplication
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Evaluation Homomorphism

Let be a commutative ring and let be the ring of polynomials with coefficients in .

Theorem 1 Let 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 to and taking every $r\in R$ to $\psi(r)$ .

This amounts to saying that polynomial rings are free objects in the category of -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

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 $\gamma(f) = \sum_i \psi(a_i) s^i = \phi(f)$ . $\qedsymbol$