Universal Property of Polynomial Ring/Free Monoid on Set

Theorem
Let $$R,S$$ be a commutative and unitary rings, and $$s\in S$$.

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

Let $$R[X]$$ be the Ring of Polynomial Forms with coefficients in $$R$$.

Then there exists a unique homomorphism $$\phi:R[X]\to S$$ extending $$\psi$$ such that $$\phi(X)=s$$.

Proof
For $$f=a_0+a_1X+\cdots+a_nX^n\in R[X]$$, let $$\phi(f)=\psi(a_0)+\psi(a_1)s+\cdots+\psi(a_n)s^n$$.

It is immediate that this map has the required properties and that it is unique.

Remarks

 * The homomorphism $$\phi$$ is often called "evaluation at s".


 * 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.