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I'm not sure at all which theorem(s) could be applied in order to get started on the following problem. Any suggestion is very much appreciated.

Suppose $k$ is a field and $A$ is a finitely generated $k$-algebra. Let $K$ be a fixed algebraic closure of $k$. Construct a bijection from the set of maximal ideals of $A$ to the set of orbits of the absolute Galois group Gal$(K/k)$ on the set of $k$-algebra homomorphisms from $A$ to $K$.

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2 Answers 2

up vote 2 down vote accepted

Consider a $k$-algebra morphism $\chi:A\to K$.
Its image $\chi(A)$ is a finitely generated algebraic domain over the field $k$, hence is a finite extension of $k$.
The kernel of $\chi$ is consequently a maximal ideal of $A$ and this yields a map $$\text{ker}: \text{Hom}_{k-\text{alg}} (A,K) \to \text{Specmax} (A): \chi\mapsto \text{ker}( \chi ) $$ The absolute Galois group $Gal(K/k)$ acts on $ \text{Hom}_{k-\text{alg}} (A,K)$ by composition on the left : $g*\chi=g\circ \chi$.
Since obviously $\chi$ and $g\circ \chi$ have the same kernel, we obtain the required map $$\text{Ker}:( \text{Hom}_{k-\text{alg}} (A,K) )/ Gal(K/k)\to \text{Specmax} (A): \overline {\chi}\mapsto \text{ker}( \chi ) $$ One must then check that this map is bijective, which isn't too difficult.

Note carefully:
Despite the unfortunate terminology "absolute Galois group" and the notation $Gal(K/k)$, the extension $k\subset K$ is not Galois in general since it is not separable if $k$ is not perfect.
Ideally, one should write $Aut(K/k)$ and speak of the absolute automorphism group.

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Thanks very much Georges! –  Sean Feb 28 at 4:22
1  
You are welcome, Sean. –  Georges Elencwajg Feb 28 at 4:23

Here's a hint: Think about the kernel of any k-algebra homomorphism from A to K.

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