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Suppose $A$ is a $k$-algebra, with $k$ a field, and let $\ell$ be a field extension of $k$. Is there an easy way to see/recover $\mathrm{Spec}(A)$ in/from $\mathrm{Spec}(A \otimes_k \ell)$, using the action of $\mathrm{Aut}(\ell/k)$ ? Important special case: $\ell$ is an algebraic closure of $k$.

To illustrate my question, let $\ell$ be an algebraic closure of $k$. Then a maximal ideal $\frak{m}$ of $A$ corresponds to an $\mathrm{Aut}(\ell/k)$-orbit of maximal ideals in $A \otimes_k \ell$ (the maximal ideals in $A \otimes_k \ell$ ''over $\frak{m}$''). Is such a thing true for the other prime ideals (and for any field extension $\ell$, not ''just'' algebraic closures) ? Is there a similar way (so using the $\mathrm{Aut}(\ell/k)$-action as above) to describe the closed sets of $\mathrm{Spec}(A)$ ?

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    $\begingroup$ Recover as a space or as a scheme? As a scheme it's just classical Galois descent, I believe. $\endgroup$ Feb 16, 2018 at 13:15
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    $\begingroup$ EGA IV$_2$, 2.3.12 (applies with $\ell/k$ an arbitrary extension of fields as in the initial part of the question, whereas "${\rm{Gal}}(\ell/k)$" doesn't make sense in general). $\endgroup$
    – nfdc23
    Feb 16, 2018 at 13:43
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    $\begingroup$ @THC. A subset of $\text{Spec}(A\otimes_k \ell)$ is the inverse image of a subset of $\text{Spec}(A)$ if and only it has equal inverse images under the two projections $\text{Spec}(A\otimes_k \ell\otimes_k \ell) \to \text{Spec}(A\otimes_k \ell)$. For a finite, Galois extension, equality of these two inverse images is equivalent to be fixed (as a subset) by the Galois group. Anyway, this is a different part of descent: it is not about "openness" or "closedness" of subsets, it is about which subsets are inverse images. $\endgroup$ Feb 16, 2018 at 17:08
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    $\begingroup$ You are very focused on the action of ${\rm{Aut}}(\ell/k)$, but that is useless when the finite extension is non-trivial but so far from Galois that it has no nontrivial automorphisms at all (e.g., $\mathbf{Q}(m^{1/3})/\mathbf{Q}$ for a non-cube integer $m$). What is your motivation for asking about general extensions and yet trying to shoehorn the (possibly trivial) automorphism group into that case? $\endgroup$
    – nfdc23
    Feb 16, 2018 at 17:39
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    $\begingroup$ For the record, there is an easy elementary argument that shows that the map $spec(A\otimes_k K)\to spec(A)$, $K$ the algebraic closure of $k$, is a quotient map (on topological spaces). No need to appeal to EGA IV, 2.3.12,... $\endgroup$
    – anon
    Feb 16, 2018 at 18:51

1 Answer 1

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Let $A$ be a finitely generated $k$-algebra, and let $K$ be the algebraic closure of $k$. I explain how to recover $spec(A)$ from $spec(A\otimes K)$. Equivalently, I explain how to recover $spm(A)$ from $spm(A\otimes K)$ (max specs). Consider the map $spm(A\otimes K)\rightarrow spm(A)$.

(a) The map is surjective. Every maximal ideal $\mathfrak{m}$ of $A$ is the kernel of a $k$-algebra homomorphism $A\rightarrow K$, which extends to a $K$-algebra homomorphism $A_{K}\rightarrow K$, whose kernel is a maximal ideal lying over $\mathfrak{m}$.

(b) The map is continuous. Let $S=Z(f_{1},\ldots,f_{s})$ be closed in $spm(A)$. Then $\pi^{-1}(S)=Z(f_{1},\ldots,f_{s})$ in $spm(A_{K})$.

(c) The map is closed. Let $T=Z(f_{1},\ldots ,f_{s})$ be a closed subset of $spm(A_{K})$. Choose a basis $(e_{j})_{j\in J}$ for $K$ as a $k$-vector space, and write $f_{i}=\sum_{j}e_{j}f_{ij}$ (finite sum) with $f_{ij}\in A$. Every maximal ideal of $A_{K}$ is the kernel of a $K$-algebra homomorphism $A_{K}\rightarrow K$, and the $f_{i}$ map to zero under such a homomorphism if and only if every $f_{ij}$ does. Therefore $T=Z(\ldots,f_{ij},\ldots)$ in $spm(A_{K})$, and it follows that $\pi(T)=Z(\ldots,f_{ij},\ldots)$ in $spm(A)$.

Thus, the map is a quotient map of topological spaces, and the fibres are the orbits of $Aut(K/k)$.

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  • $\begingroup$ The OP does not assume that the $k$-algebra $A$ is finitely generated. $\endgroup$ Feb 16, 2018 at 19:38
  • $\begingroup$ @Jason Starr. This is the key case. $\endgroup$
    – anon
    Feb 16, 2018 at 19:43
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    $\begingroup$ "This is the key case." I do not understand this comment. $\endgroup$ Feb 16, 2018 at 20:06
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    $\begingroup$ @anon: when $A$ is a general $k$-algebra, closed subsets of ${\rm{Spec}}(A)$ don't arise from ${\rm{Spec}}(A_0)$ for a finitely generated $k$-subalgebra $A_0 \subset A$, but rather are just an intersection of such pullbacks. Thus, there is no evident argument to reduce the problem to the case of finitely generated $A$; i.e., in what precise sense is this "the key case"? (This is just an elaboration on Jason Starr's comment.) $\endgroup$
    – nfdc23
    Feb 18, 2018 at 21:58

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