I'm self-studying some algebraic geometry from Cox, Little, and O'Shea, and I wish to prove that when $k$ is algebraically closed and $\mathbf{V}(I)$ is finite, the $k$-vector space $k[x_1,\dots,x_n]/I$ is finite-dimensional.
The proof I cooked up is different than the one in the book, and it would be nice to have a reality check. I have already proved that: "if for each $i$, there is some $m_i$ such that $x_i^{m_i}$ is in $\langle \{\text{leading monomials of $I$}\}\rangle$, then $k[x_1,\dots,x_n]/I$ is finite-dimensional," so we will deduce that instead from the hypothesis.
Proof: Suppose $k$ is algebraically closed and $I$ is an ideal such that $\mathbf{V}(I)$ is a finite collection of points $\{\mathbf{p}_i\}_{i=1}^s$ (if $\mathbf{V}(I) = \varnothing$, the result follows from the Weak Nullstellensatz). Since: $\mathbf{V}(I) = \bigcup_{i=1}^s \mathbf{V}(I_i)$, where $I_i = \langle x_1 - p_{i,1},\dots,x_n-p_{i,n}\rangle$, it follows that $\mathbf{V}(I) = \mathbf{V}\left(\prod I_i\right)$. Taking ideals, we see that $\mathbf{I}(\mathbf{V}(I)) = \mathbf{I}\left(\mathbf{V}\left(\prod I_i\right)\right) = \sqrt{I} = \sqrt{\prod I_i}$ by the Nullstellensatz.
Since $\prod I_i$ is generated by polynomials which are products of exactly one generator from each $I_i$, it follows that: $$p_i(x) = \prod_{j=1}^s (x_i - p_{j,i}) = x_i^s + (\dots)$$ is in $\prod I_i$, hence is also in $\sqrt{\prod I_i} = \sqrt{I}$. It follows that there is some $N_i$ so that $p_i^{N_i} \in I$, which has leading monomial $x_i^{sN_i}$, so picking $m_i = sN_i$ satisfies our goal. $\square$
Is this proof correct? Thank you.
Looks good to me.
A couple of simplifications: you don't actually need to consider $I_i$, it's easy to directly see that $p_i(x) \in \mathbf{I}(\{\mathbf{p}_j\})$. And you don't need to split out the $\mathbf{V}(I) = \varnothing$ case if you define $p_i(x) = 1$ (which is what the empty product should be anyway).