dimension of a coordinate ring Let $I$ be an ideal of $\mathbb{C}[x,y]$ such that its zero set in $\mathbb{C}^2$ has cardinality $n$. Is it true that $\mathbb{C}[x,y]/I$ is an $n$-dimensional $\mathbb{C}$-vector space (and why)?

As Jim points out, this is false in general. The correct statement is that $\mathbb C[x,y]/\sqrt{I}$ is an $n$-dimensional vector space.
Proof: Write $\sqrt{I} = P_1\cap \cdots \cap P_m$, where $P_i$ are the primes minimal over $I$. Since $V(P_i)\subseteq V(I)$, we see that each $V(P_i)$ is a single point. Thus since each $V(P_i)$ is distinct, we get $m=n$. Furthermore, each $P_i$ is maximal, so the $P_i$ are comaximal hence by the Chinese Remainder Theorem we get
$$\frac{\mathbb C[x,y]}{P_1\cap\cdots\cap P_n}\cong\frac{\mathbb C[x,y]}{P_1}\times\cdots\times\frac{\mathbb C[x,y]}{P_n}\cong \mathbb C\times \cdots \times \mathbb C$$
with $n$ copies of $\mathbb C$.
