Universal Property of Polynomial Ring/Free Monoid on Set

Theorem
Let $R, S$ be commutative and unitary rings.

Let $\left\{{s_j: j \in J}\right\} \subseteq S$ 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 $\phi \left({X_j}\right) = s_j$ for all $j \in J$.

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

Let $k_j$ be the $j$th component of a multiindex $k$.

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

Define:


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

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

If $\displaystyle g = \sum_{k \mathop \in Z} b_k \prod_{j \mathop \in J} X_j^{k_j}$, 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 $\phi' \left({r}\right) = \psi \left({r}\right)$ 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 $\left\{{s_j: j \in J}\right\}$.


 * 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.