Universal Property of Polynomial Ring/Free Monoid on Set

Theorem
Let $R, S$ be commutative and unitary rings, and $\left\{s_j:j\in J\right\}\subseteq S$ be a subset of $S$ indexed by $J$.

Let $\psi: R \to S$ be a ring homomorphism.

Let $R \left[{ \left\{X_j:j\in J\right\} } \right]$ be the Ring of Polynomial Forms with coefficients in $R$.

Then there exists a unique homomorphism $\phi: R \left[ { \left\{ X_j: j \in J \right \} } \right] \to S$ extending $\psi$ such that $\phi(X_j) = s_j$ for all $j \in J$.

Proof
Let $Z$ be the set of all multiindices indexed by $J$.

Let $k_j$ be the $j^\text{th}$ component of a multiindex $k$.

Let $\displaystyle f=\sum_{k\in Z}a_k\prod_{j\in J}X_j^{k_j}$ be a polynomial over $R$, and define:


 * $\displaystyle \phi(f)=\sum_{k\in Z}\psi(a_k)\prod_{j\in J}s_j^{k_j}$

It is clear that $\phi$ extends $\psi$.

If $\displaystyle g=\sum_{k\in Z}b_k\prod_{j\in J}X_j^{k_j}$, then:

Therefore $\phi$ preserves addition.

Also:

This shows that $\phi$ is a homomorphism.

Now suppose that $\phi'$ is another such homomorphism.

For each $j\in J$, $\phi'$ must satisfy $\phi'(X_j)=s_j$ and $\phi'(r)=\psi(r)$ for all $r\in R$.

In addition $\phi'$ must be a homomorphism, so we compute:

and therefore $\phi'=\phi$.

This concludes the proof.

Remarks

 * The homomorphism $\phi$ is often called evaluation at $\{s_j:j\in J\}$.


 * The requirement that the rings be commutative is vital. A fundamental difference for polynomials over non-commutative rings is additional difficulty identifying polynomial forms and functions using this method.