I am working (slowly and with much labor) through Vakil's Algebraic Geometry and came upon this problem.
Suppose $k$ is an algebraically closed field, and $A = k[x_1,... ,x_n]/I$ is a finitely generated $k$-algebra with $\mathfrak{N}(A) = {0}$ (so the discussion of §3.2.13 applies). Consider the set $X = SpecA$ as a subset of $\mathbb{A}_n^k$. Show that functions on $X$ are determined by their values on the closed points (by the weak Nullstellensatz 3.2.4, the “classical” points $k^n$. The space $\mathbb{A}_n^k \cap SpecA$ of $Spec A$). Hint: if $f$ and $g$ are different functions on $X$, then $f − g$ is nowhere zero on an open subset of $X$. Use Exercise 3.6.J(a).
The reference exercise 3.6.J(a) shows that for a field $k$ and a finitely-generated $k$-algebra $A$, the set of closed points in $SpecA$ (previously established as corresponding to maximal ideals of $A$) is dense in $SpecA$.
Okay I stumbled through exercise 3.6.J(a) on my own, and I think I see where the starting line is for this race. I imagine that I should look at $f(\mathfrak{m}) - g(\mathfrak{m})$ for every $\mathfrak{m}$. If $f$ and $g$ are equal at every point $\mathfrak{m}$, then $f(\mathfrak{m}) - g(\mathfrak{m}) = 0$ for every $\mathfrak{m}$ so that $f-g$ has a zero on every open subset of $X$ (using the fact that the set of closed points is dense in $X$) and therefore $f = g$. But why is the comment about $f-g$ being nowhere zero on an open subset of $X$ true?
Moreover, given the (not entirely rigorous) proof above, wouldn't the comment in question imply that $f = g$ even if $f-g=0$ on just one open subset of $X$?
Thanks for any help anyone can offer. I imagine I might be asking a trivial question and just forgetting some topology basics or some fact about $SpecA$. (Anyway it's good to practice some LaTeX ;)
Here's a hint:
Observe that the open set $D(f-g)$ contains no closed points. It is therefore an open set disjoint from a dense set. What does this imply about $D(f-g)$? What does that implication tell you about $f-g$?