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)?
 A: No, the zero set of $I = (x, y^k)$ has cardinality $1$, it's just the point $(0, 0)$.  But $\mathbb C[x, y]/(x, y^k)$ has dimension $k$.
A: The answer is no, but your very interesting question leads to about the most  elementary motivation for the introduction of scheme theory in elementary algebraic geometry.  
You see, if the common zero set $X_{\mathrm{classical}}=V_{\mathrm{classical}}(I)$ consists  set-theoretically (I would even say physically) in $n$ points, then $N:=\dim_\mathbb C \mathbb{C}[x,y]/I\geq n$.
If $N\gt n$, this is an indication that some interesting geometry is present: $X_{\mathrm{classical}}$ is described by equations that are not transversal enough, so that morally they describe a variety bigger than the naked physical set.
 The most elementary example is given by $I=\langle y,y-x^2\rangle$: we have $V_{\mathrm{classical}}(I)=\{(0,0)\}=\{O\}$
Everybody feels that it is a bad idea do describe the origin as $V(I)$, i.e. as the  intersection of a parabola and one of its tangents: a better description would be to describe it by the ideal $J=\langle x,y\rangle,$ in other words as the intersection of two transversal lines.
However the ideal $I$ describes an interesting  structure, richer than a naked point,  and this structure is called a scheme.
This is all reflected in the strict inequality $$\dim_\mathbb C \mathbb{C}[x,y]/I=2=N\gt \dim_\mathbb C \mathbb{C}[x,y]/J=1=n=\text { number of physical points}.$$ Scheme theory in its most elementary incarnation disambiguates  these two cases by adding the relevant algebra in the algebro-geometric structure, now defined as pairs consisting of a physical set plus an algebra: $$V_{\mathrm{scheme}}(J)=(\{O\},\mathbb{C}[x,y]/J )\subsetneq  V_{\mathrm{scheme}}(I)= (\{O\},\mathbb{C}[x,y]/I ).$$
Bibliography
Perrin's Algebraic Geometry is the most elementary introduction to this down-to-earth vision of schemes (cf. beginning of Chapter VI). 
A: 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 $\mathbb C$.
A: Here is a positive result (see Fulton, Algebraic curves, Corollary I.7.4 on page 23). Let $k$ be some algebraically closed field. If $I \subseteq k[x_1,\dotsc,x_n]$ is an ideal, then $V(I) \subseteq \mathbb{A}^n(k)$ is finite if and only if $k[x_1,\dotsc,x_n]/I$ is a finite-dimensional $k$-vector space. In this case, we have $|V(I)| \leq \dim_k(k[x_1,\dotsc,x_n]/I)$.
A more precise result is that that $k[x_1,\dotsc,x_n]/\sqrt{I} \cong \prod_{P \in V(I)} k$, in particular $\dim_k(k[x_1,\dotsc,x_n]/\sqrt{I})=|V(I)|$.
