Is algebraic closure necessary? (3.6.K Ravi Vakil's notes)

Question

I've just done exercise 3.6.K in Ravi Vakil's notes and noticed that my solution does not seem to rely on algebraic closure, so I'd like a sanity check. I understand it's important to make the "classical" points the only closed points, but it seems like an extraneous assumption for the problem at hand? The problem is:

Suppose $k$ is an algebraically closed field, and $A = k[x_1,...,x_n]/I$ is a finitely generated $k$-algebra with $\mathfrak{N}(A) = \{0\}$. Show that functions on $\operatorname{Spec}A$ are determined by their values on the closed points.

My attempt is that for two different $f,g\in A$, $\operatorname{D}(f-g)$ is nonempty since there are no nilpotents, so exercise 3.6.J(a) should apply immediately...

Answer 1

I think the exact phrasing of the question tries to push you to view the elements of $k^n$ as being classical, and then to note that those correspond to the closed points of $\mathbb{A}^n_k$ when $k = \bar{k}$ and that evaluation is then the obvious map into $k$.

I guess I've never thought of closed points in general as being all that classical -- I don't see how to easily interpret $(x^2 - 2)$ in $\mathbb{Q}[x]$ as something having just a single set of coordinates in some affine space. There's some sort of gluing going on.

On the other hand, the hypothesis is a little misleading and I agree that your argument works for any $k$ as long as "value" is interpreted suitably. You might write to Ravi suggesting that the question begin in a more general setting and then proceed to a discussion of the "classical" case. I do think the general fact is useful at a few other places in the notes.

Answer 2

1) For an arbitrary field $k$, classes in $A = k[x_1,...,x_n]/I$ are certainly not determined by their associated functions on the classical points $V(I)\subset k^n$.
For example if $k=\mathbb F_2, n=1, I=(0)$, the polynomials $x$ and $x^2$ give rise to the same function on $k^1=\mathbb F_2$.

2) However the associated functions on $\operatorname {Specmax(A)}$ do determine the elements in $A$.
The point (as noted by Zhen) is that an element $\phi=\bar f\in A=k[x_1,...,x_n]/I$ which is zero at all points $[\mathfrak m]\in \operatorname {Specmax(A)}$ corresponding to maximal ideals $\mathfrak m\subset A$ is an element in the Jacobson radical $J=\cap_{\mathfrak m\in \operatorname {Specmax(A)}} \mathfrak m$, which is known to be zero for reduced finitely generated algebras over a field.

3) In particular if $k$ is algebraically closed, the (weak) Nullstellensatz states that the maximal ideals of $A$ correspond to the points of $V(I)\subset k^n$, so that the elements of $A$ are completely determined by their associated functions on $V(I)$.