----
(Everything here is [[commutative ring|commutative]], and we consider finitely many elements at a time.)
> [!definition] Definition. ([[algebraically independent]])
> Let $A$ be an [[algebra]] over a [[field]] $k$. Fix a finite indexed set $I=\{x_{1},\dots,x_{n} \}\subset A$. The [[inclusion map|inclusion]] $\iota:I \subset A$ [[free commutative algebra|induces]] a unique $k$-[[algebra]] [[algebra homomorphism|homomorphism]] $\overline{\iota}:k [T_{1},\dots,T_{n}]$ satisfying $\overline{\iota}(T_{i})=x_{i}$ for all $i \in [n]$.
>
> ```tikz
> \usepackage{tikz-cd}
> \usepackage{amsmath}
> \begin{document}
> % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAGtkAVAfQEZSAAgA6wqBBxwhvMBRABfUuky58hFAIBMVWoxZsAkgF5RwAB78ho8ZKEWwo+QqUgM2PASIC+O+s1aIIKIAxkxozsrual7kvnoBIACCCjowUADm8ESgAGYAThAAtkgCIDgQSADM1ABGMGBQSAC0lWS6-myi+Dh0INQMdHUMAAoqHuogeVjpABY4ESD5RSXU5UjEirkFxYhta4jV7fqBojBmWHCSggCEIsIQNDB5DFhgMMBdEnRO8hTyQA
> \begin{tikzcd}
> {k[T_1, \dots, T_n]} \arrow[r, "\exists ! \overline{\iota}"] & A \\
> & \cup \\
> & {I=\{x_1, \dots, x_n\}} \arrow[uu, "\iota"', bend right] \arrow[luu]
> \end{tikzcd}
> \end{document}
> ```
>
> We call $I$ **$k$-algebraically independent** if $\overline{\iota}$ is [[injection|injective]], that is, if having $p(x_{1},\dots,x_{n})=0$ for some $p \in k[T_{1},\dots,T_{n}]$ implies $x_{1}=\dots=x_{n}=0$.
>
> By definition, the $k$-[[subalgebra]] $k[x_{1},\dots,x_{n}]$ [[subalgebra generated by a subset|generated by]] $x_{1},\dots,x_{n}$ is given by the image of the embedding $k[T_{1},\dots,T_{n}] \xhookrightarrow{\overline{\iota}}A$: $k[T_{1},\dots,T_{n}] \cong_{\overline{\iota}} k[x_{1},\dots,x_{n}]$And so one has the mantra: '$x_{1},\dots,x_{n}$ are $k$-algebraically independent iff they generate a copy of $k[T_{1},\dots,T_{n}]$ inside $A