Universal Property of Polynomial Ring/Free Monoid on Set

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

Let $\family {s_j}_{j \mathop \in J}$ be an indexed family of elements of $S$.

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

Let $R \sqbrk {\set {X_j: j \in J} }$ be a polynomial ring.

Then there exists a unique evaluation homomorphism $\phi: R \sqbrk {\set {X_j: j \in J} } \to S$ at $\family {s_j}_{j \mathop \in J}$ extending $\psi$.

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

Define:


 * $\ds \map \phi f = \sum_{k \mathop \in Z} \map \psi {a_k} \prod_{j \mathop \in J} s_j^{k_j}$

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

If $\ds g = \sum_{k \mathop \in Z} b_k \prod_{j \mathop \in J} X_j^{k_j}$, 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 $\map {\phi'} {X_j} = s_j$ and $\map {\phi'} r = \map \psi r$ 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 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.

Also see

 * Equivalence of Definitions of Polynomial Ring
 * Universal Property of Polynomial Algebra