$k[x_1,...,x_n]/I$ is an Artin ring if and only if $k[x_1,...,x_n]/I$ is of finite dimension as $k$ vector space. ($k$ is algebraically closed)
My attempt:
If $k[x_1,...,x_n]/I$ is Artin and $k[x_1,...,x_n]/I$ is of infinite dimension as k-vector space, let's take $\{b_1,b_2,...\}$ a base for $k[x_1,...,x_n]/I$, then it would be fulfilled that $\left \langle b_1, b_2,... \right \rangle\supset \left \langle b_2,... \right \rangle\supset\left \langle b_3,... \right \rangle $ is a descending chain that does not stabilize which contradicts that $k[x_1,...,x_n]/I$ is Artin. (This is OK?)
On the other hand, if $k[x_1,...,x_n]/I$ is of finite dimension as a k-vector space, then let $\{b_1,..., b_n\}$ take a basis for $k[x_1,...,x_n]/I$, are the ideals of $k[x_1,...,x_n]/I$ of finite dimension as k-vector space?
