----
> [!theorem] Theorem. ([[universal property of polynomial rings]])
> [[polynomial 4|Polynomial rings]] play a role in [[ring]] theory analogous to that of [[free group|free groups]] in [[group]] theory.
>
> Let $A=\{ a_{1},\dots,a_{n} \}$ be a finite set. Consider the [[category]] $\mathscr{R}^{A}$ whose objects are pairs $(j,R)$ with $R$ a [[commutative ring]] and $j:A \to R$ a set-function and whose morphisms $(j_{1},R_{1}) \to (j_{2}, R_{2})$
> are commutative diagrams of set-functions
>
> ```tikz
> \usepackage{tikz-cd}
> \usepackage{amsmath}
> \begin{document}
> % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBoBGAXVJADcBDAGwFcYkQBBEAX1PU1z5CKMsWp0mrdgCUA+uR58QGbHgJFypMTQYs2iEHIBMPcTCgBzeEVAAzAE4QAtkjIgcEJJol72AK3lFO0cXRDcPJCMdSX0QAJMaRnoAIxhGAAUBNWEQeywLAAscIJAHZy8aCMQonykDAB16hns0AqxTbiA
> \begin{tikzcd}
> R_1 \arrow[r, "\varphi"] & R_2 \\
> A \arrow[u, "j_1"] \arrow[ru, "j_2"'] &
> \end{tikzcd}
> \end{document}
> ```
>
> in which $\varphi$ is required to be a [[ring homomorphism]]. [^1] Then $(i, \mathbb{Z}[x_{1},\dots, x_{n}])$ is [[terminal object|initial]] in $\mathscr{R}^{A}$, where $i: A \to \mathbb{Z}[x_{1},\dots,x_{n}]$ sends $a_{k}$ to the monomial $x_{k}$. That is, given any [[commutative ring]] $R$ and set-function function $A \xrightarrow{j} R$ there exists a unique [[ring homomorphism]] $\varphi$ making the following diagram commute:
>
> ```tikz
> \usepackage{tikz-cd}
> \usepackage{amsmath}
> \usepackage{amssymb}
> \begin{document}
> % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBoBGAXVJADcBDAGwFcYkQBBEAX1PU1z5CKMsWp0mrdgB1pAW3o4AFgCMVwAFrdkADwD65UgAJZUCDjjH9YSjz4gM2PASKGxNBizaIQAJR7iMFAA5vBEoABmAE4QckhkIDgQSIYSXuxYdpExcYipSUgATB6S3iCyMDpYcBYAhCbSDFFoSpm82bHxNAWIxWlSPgBWIDSM9CowjAAKAs7CIIwwETgB3EA
> \begin{tikzcd}
> {\mathbb{Z}[x_1, \dots, x_n]} \arrow[r, "\exists! \varphi"] & R \\
> A \arrow[u, "i"] \arrow[ru, "j"'] &
> \end{tikzcd}
> \end{document}
> ```
>
^theorem
> [!proof]- Proof. ([[universal property of polynomial rings]])
> **Uniqueness.** To say $\varphi \circ i = j$ is to say $\varphi(x_{k})=j(a_{k})$, so commutativity has enforced $\varphi