evaluation homomorphismEvaluationHomomorphism2004-03-08 23:08:392005-03-18 22:34:08Theorempolynomial ringevaluation homomorphism% this is the default PlanetMath preamble. as your knowledge
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\newtheorem{cor}{Corollary}Let $R$ be a commutative ring and let $R[X]$ be the ring of polynomials with coefficients in $R$.
\begin{theorem}
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 $\psi(r)$.
\end{theorem}
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.
\begin{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 $\gamma(f) = \sum_i \psi(a_i) s^i = \phi(f)$.
\end{proof}