Universal property of the polynomial ring

Statement

Let $R$ and $S$ be commutative rings, let $\varphi \colon R \to S$ be a ring homomorphism and let $s \in S$. Then there is exactly one ring homomorphism $\Phi \colon R[x] \to S$ such that $\Phi(r) = \varphi(r)$ for all $r \in R$ and $\Phi(x) = s$. It is given by $\Phi(a_0 + a_1 x + \dots + a_n x^n) = \varphi(a_0) + \varphi(a_1) s + \dots + \varphi(a_n) s^n$.

Remarks

With $S = R$, $\varphi$ the identity and $s = c$, the map $\Phi$ is the evaluation homomorphism at $c$. In the language of algebras: for each commutative $R$-algebra $A$ and each $a \in A$, there is exactly one $R$-algebra homomorphism $R[x] \to A$ that sends $x$ to $a$. This property characterizes $R[x]$ up to a unique isomorphism.

Explanations

L1 human

A ring homomorphism from $R[x]$ is fixed by two choices: what it does on the coefficients, and where it sends $x$. Every pair of choices works, and it gives exactly one homomorphism. Evaluation at a number is the main example.