Generic question about commutative algebra I don't know how to categorize better this question: I'm studing from the notes of a course of commutative algebra. Anyway, I read that for a finitely generated $k$-algebra $A$, with $k=\bar{k}$ an algebraically closed field, we can view $A$ as an algebra of functions from the set $X:=\operatorname{Max}(A)$ to $k$. In fact any element $f\in A$ defines a function on $X$ such that $f(x)=\bar{f}\in A/x$, with $x\in X$, and $A/x$ must be $k$ because it is algebraically closed. In this context we studied also the Hilbert Nullstellensatz, that here basically says that taken a finitely generated $k$-algebra $A:=k[x_1,\dots ,x_n]/I$, with $I$ an ideal of $k[x_1,\dots ,x_n]$, then $\sqrt{I}$ can be recovered from the polynomials $f$ in $k[x_1,\dots ,x_n]$ such that, viewed as an element of $A$, one has $f(x)=0$ for all $x\in\operatorname{Max}(A)$. (Obviously $f(x)$ is defined as above).
A few pages later, these notes define the so-called geometric points of $A$, that are $k$-homomorphisms from $A$ (a $k$-algebra) to a field extension $K\supset k$. It turns out that there is a bijection between certain equivalence classes of geometric points of $A$ and the prime spectrum of $A$. Again, the elements of $A$ can be regarded as functions on the elements of $Y:=\operatorname{Spec}(A)$. (In the same  way as before, $f(y)=\bar{f}\in A/y$, if $y\in Y$).
I don't have a precise question, because I'm not very practised yet in this commutative algebra \ algebraic geometry sector, I'd just like to know if these two "constructions" are linked one to the other or they are simply independent; I feel like there is something that I am missing overall. Thanks in advance
 A: If $K = k$ then the second construction reduces to the first one; the Nullstellensatz implies that maximal ideals correspond exactly to $k$-algebra homomorphisms $A \to k$.
In general the second construction is more general (as it has to be to give all prime ideals instead of just maximal ones). The simplest example to keep in mind is $A = k[x], K = k(x)$ and $A \to K$ the usual inclusion, which defines what is called the generic point of the affine line $\mathbb{A}^1$ and corresponds to the prime-and-not-maximal ideal $(0)$. The second construction also does not require $k$ to be algebraically closed, and can be used to recover maximal ideals by taking $K$ to be a finite extension of $k$ (this follows from a more general version of the Nullstellensatz).
A: For $A$ a finitely generated $k$-algebra, let $X = \operatorname{m-spec} A$, and let $X(k)$ be the set of $k$-algebra homomorphisms from $A$ to $k$ (we call $X(k)$ the set of $k$-rational points of $X$).  There is a natural injective map
$X(k) \rightarrow X$ given by sending a $k$-algebra homomorphism to its kernel.  One way of stating the Nullstellensatz is that for $k$ algebraically closed, this is a bijection.
Now assume that $k$ is perfect but not necessarily algebraically closed, and let $Y = \operatorname{m-spec} A \otimes_k \overline{k}$.  The natural map $\mathfrak m \mapsto \mathfrak m \cap A$ can be shown to define a surjection $Y \rightarrow X$.
The geometric points of $A$ as you call them are the same as $k$-algebra homomorphisms from $A$ into $\overline{k}$, and these are the same as $\overline{k}$-algebra homomorphisms from $A \otimes_k \overline{k}$ into $\overline{k}$.  In other words, a geometric point of $A$ is just an element of $Y(\overline{k})$.  Now we have a diagram
$$\begin{matrix} X(k) & \subset & Y(\overline{k}) \\ \cap & &|| \\X & \leftarrow & Y\end{matrix}$$
where $Y = Y(\overline{k})$ because of the Nullstellensatz.  How do we interpret this diagram?  The Galois group $\operatorname{Gal}(\overline{k}/k)$ acts on $Y$ because it acts on $\overline{k}$.  It can be shown that $Y \rightarrow X$ is actually the quotient map under this action (this is even a topological quotient if $X$ and $Y$ are taken in the Zariski topologies).  Therefore, a maximal ideal of $A$ corresponds to an equivalence class of maximal ideals of $A \otimes_k \overline{k}$ (or geometric points of $A$) under the action of the Galois group, and the $k$-rational points of $X$ are exactly the fixed points of this action.
