Equivalent conditions for an ideal to be zero-dimensional. 
For an ideal $I \subset \mathbb{C} [x_1, ... , x_n]$ show that dim$_{\mathbb{C}}R/I$ is finite iff $I$ is contained in only finitely many maximal ideals.

Thoughts so far: I'm not sure how to get started, so a hint to get things going would be appreciated! 

Set $R = \mathbb{C}[x_1, \ldots, x_n]$.  
If $I \subset R$ is an ideal such that $\dim_{\mathbb{C}}(R/I) < \infty$ then $R/I$ is Artinian, and consider a descending sequence of ideals $\mathfrak{m}_1 \supset \mathfrak{m}_1 \cap \mathfrak{m}_2 \supset \cdots$ where $\mathfrak{m_i} \subset R$ is maximal.  
If $I \subset R$ is an ideal that is contained in only a finite number of maximal ideals, then $\sqrt{I} = \mathfrak{m}_1 \cap \cdots \mathfrak{m}_r$ for some max ideals $\mathfrak{m}_i, \ldots$ of $R$, and since $\mathfrak{m}_i$'s are coprime to each other $\sqrt{I} = \mathfrak{m}_1 \cdots \mathfrak{m}_r$, so $\mathfrak{m}_1^{e} \cdots \mathfrak{m}_r^{e} \subseteq I$ for some positive integer $e \geq 1$.  But each $R/\mathfrak{m}_i^{e_i}$ is finite dimensional over $\mathbb{C}$ because each factor $\mathfrak{m_i}^v / \mathfrak{m_i}^{v+1}$ is finite dimensional over $\mathbb{C}$ ($\mathfrak{m}_i^v / \mathfrak{m}_i^{v+1}$ is a f.g. module over $R/\mathfrak{m}_i = \mathbb{C}$).
