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

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 $V(I)$.
