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This question is about Galois theory. So let $K / k$ be a Galois extension of fields. Let us assume that $K / k$ is finite dimensional, though everything can be made to work in the profinite case by adding some continuity conditions here and there. I should perhaps remark that I do not (yet) know Grothendieck's Galois theory. If this theory answers my question, I will be happy with some precise statement. $\newcommand{\Mod}{\mathrm{Mod}}\newcommand{\Gal}{\mathrm{Gal}}$

Let $\Mod(K/k)$ be the category of $K/k$-vector spaces: Objects are $K$-vector spaces $V$ together with an action of the Galois group $\Gal(K/k)$ on $V$ which is semilinear in the sense that $\sigma(a v) = \sigma(a) \sigma(v)$ for $\sigma \in \Gal(K/k)$, $a \in K$ and $v \in V$. Morphisms are $K$-linear maps which commute with the action.

Now Galois descent is the statement that the functors $$\Mod(k) \to \Mod(K/k), \qquad V \mapsto K \otimes_k V$$ and $$\Mod(K/k) \to \Mod(k), \qquad W \mapsto W^{\Gal(K/k)}$$ form an equivalence of ($k$-linear, symmetric monoidal, …) categories.

This induces equivalences of the respective categories of monoids (i.e. $k$-algebras on the left hand side), which is related to Brauer groups, or of “objects with a blinear form” (i.e. objects $V$ equipped with a map from $V \otimes V$ to the monoidal unit, satisfying some axioms), which is related to the Galois cohomology of the orthogonal group.

Is there some equivalence between categories of non-linear objects, which induces the above equivalence between $k$-vector spaces and $K/k$-vector spaces? For example, is there some ringed topos $(X, \mathcal{O}_X)$ such that $\Mod(K/k)$ is more or less by definition the category of modules in this ringed topos and such that $(X, \mathcal{O}_X)$ is equivalent to $(\mathrm{Set},k)$?

I have to admit that I do not really expect the answer to the above question to be “yes”. For example one way to prove Galois descent is to realise that $\Mod(K/k)$ is basically the category of modules over the twisted group algebra $K[\Gal(K/k)]$ and that by linear indpendence of characters, this algebra is isomorphic to $\mathrm{End}_k(K)$, which by Morita theory has the same module category as $k$. So one (though not commutative) ringed topos giving rise to $\Mod(K/k)$ would be $(\mathrm{Set},K[\Gal(K/k)])$, which is, of course, not equivalent to $(\mathrm{Set}, K)$. But maybe there is some category of sets with a “semi-linear“ Galois action (whatever that would mean) which could work? Or maybe there is a better suited topos than $(\mathrm{Set},k)$, which gives rise to $\Mod(k)$?

But in fact, every proof I can think of to prove Galois descent is based on the linear independence of characters and hence something very linear-ish. Another thing making me sceptical is that the functor $V \mapsto K \otimes_k V$ does not seem to have a non-linear equivalent. Therefore I would like to know:

Is there any high-level explenation (more convincing than the one given in the above paragraph) why the answer to question 1 has to be “no”?

This question is about Galois theory. So let $K / k$ be a Galois extension of fields. Let us assume that $K / k$ is finite dimensional, though everything can be made to work in the profinite case by adding some continuity conditions here and there. I should perhaps remark that I do not (yet) know Grothendieck's Galois theory. If this theory answers my question, I will be happy with some precise statement. $\newcommand{\Mod}{\mathrm{Mod}}\newcommand{\Gal}{\mathrm{Gal}}$

Let $\Mod(K/k)$ be the category of $K/k$-vector spaces: Objects are $K$-vector spaces $V$ together with an action of the Galois group $\Gal(K/k)$ on $V$ which is semilinear in the sense that $\sigma(a v) = \sigma(a) \sigma(v)$ for $\sigma \in \Gal(K/k)$, $a \in K$ and $v \in V$. Morphisms are $K$-linear maps which commute with the action.

Now Galois descent is the statement that the functors $$\Mod(k) \to \Mod(K/k), \qquad V \mapsto K \otimes_k V$$ and $$\Mod(K/k) \to \Mod(k), \qquad W \mapsto W^{\Gal(K/k)}$$ form an equivalence of ($k$-linear, symmetric monoidal, …) categories.

This induces equivalences of the respective categories of monoids (i.e. $k$-algebras on the left hand side), which is related to Brauer groups, or of “objects with a bilinear form” (i.e. objects $V$ equipped with a map from $V \otimes V$ to the monoidal unit, satisfying some axioms), which is related to the Galois cohomology of the orthogonal group.

Is there some equivalence between categories of non-linear objects, which induces the above equivalence between $k$-vector spaces and $K/k$-vector spaces? For example, is there some ringed topos $(X, \mathcal{O}_X)$ such that $\Mod(K/k)$ is more or less by definition the category of modules in this ringed topos and such that $(X, \mathcal{O}_X)$ is equivalent to $(\mathrm{Set},k)$?

I have to admit that I do not really expect the answer to the above question to be “yes”. For example one way to prove Galois descent is to realise that $\Mod(K/k)$ is basically the category of modules over the twisted group algebra $K[\Gal(K/k)]$ and that by linear independence of characters, this algebra is isomorphic to $\mathrm{End}_k(K)$, which by Morita theory has the same module category as $k$. So one (though not commutative) ringed topos giving rise to $\Mod(K/k)$ would be $(\mathrm{Set},K[\Gal(K/k)])$, which is, of course, not equivalent to $(\mathrm{Set}, k)$. But maybe there is some category of sets with a “semi-linear“ Galois action (whatever that would mean) which could work? Or maybe there is a better suited topos than $(\mathrm{Set},k)$, which gives rise to $\Mod(k)$?

But in fact, every proof I can think of to prove Galois descent is based on the linear independence of characters and hence something very linear-ish. Another thing making me sceptical is that the functor $V \mapsto K \otimes_k V$ does not seem to have a non-linear equivalent. Therefore I would like to know:

Is there any high-level explenation (more convincing than the one given in the above paragraph) why the answer to question 1 has to be “no”?

This question is about Galois theory. So let $K / k$ be a Galois extension of fields. Let us assume that $K / k$ is finite dimensional, though everything can be made to work in the profinite case by adding some continuity conditions here and there. I should perhaps remark that I do not (yet) know Grothendieck's Galois theory. If this theory answers my question, I will be happy with some precise statement. $\newcommand{\Mod}{\mathrm{Mod}}\newcommand{\Gal}{\mathrm{Gal}}$

Let $\Mod(K/k)$ be the category of $K/k$-vector spaces: Objects are $K$-vector spaces $V$ together with an action of the Galois group $\Gal(K/k)$ on $V$ which is semilinear in the sense that $\sigma(a v) = \sigma(a) \sigma(v)$ for $\sigma \in \Gal(K/k)$, $a \in K$ and $v \in V$. Morphisms are $K$-linear maps which commute with the action.

Now Galois descent is the statement that the functors $$\Mod(k) \to \Mod(K/k), \qquad V \mapsto K \otimes_k V$$ and $$\Mod(K/k) \to \Mod(k), \qquad W \mapsto W^{\Gal(K/k)}$$ form an equivalence of ($k$-linear, symmetric monoidal, …) categories.

This induces equivalences of the respective categories of monoids (i.e. $k$-algebras on the left hand side), which is related to Brauer groups, or of “objects with a blinear form” (i.e. objects $V$ equipped with a map from $V \otimes V$ to the monoidal unit, satisfying some axioms), which is related to the Galois cohomology of the orthogonal group.

Is there some equivalence between categories of non-linear objects, which induces the above equivalence between $k$-vector spaces and $K/k$-vector spaces? For example, is there some ringed topos $(X, \mathcal{O}_X)$ such that $\Mod(K/k)$ is more or less by definition the category of modules in this ringed topos and such that $(X, \mathcal{O}_X)$ is equivalent to $(\mathrm{Set},k)$?

I have to admit that I do not really expect the answer to the above question to be “yes”. For example one way to prove Galois descent is to realise that $\Mod(K/k)$ is basically the category of modules over the twisted group algebra $K[\Gal(K/k)]$ and that by linear indpendence of characters, this algebra is isomorphic to $\mathrm{End}_k(K)$, which by Morita theory has the same module category as $k$. So one ringed topos giving rise to $\Mod(K/k)$ would be $(\mathrm{Set},K[\Gal(K/k)])$, which is, of course, not equivalent to $(\mathrm{Set}, K)$. But maybe there is some category of sets with a “semi-linear“ Galois action (whatever that would mean) which could work? Or maybe there is a better suited topos than $(\mathrm{Set},k)$, which gives rise to $\Mod(k)$?

But in fact, every proof I can think of to prove Galois descent is based on the linear independence of characters and hence something very linear-ish. Another thing making me sceptical is that the functor $V \mapsto K \otimes_k V$ does not seem to have a non-linear equivalent. Therefore I would like to know:

Is there any high-level explenation (more convincing than the one given in the above paragraph) why the answer to question 1 has to be “no”?

This question is about Galois theory. So let $K / k$ be a Galois extension of fields. Let us assume that $K / k$ is finite dimensional, though everything can be made to work in the profinite case by adding some continuity conditions here and there. I should perhaps remark that I do not (yet) know Grothendieck's Galois theory. If this theory answers my question, I will be happy with some precise statement. $\newcommand{\Mod}{\mathrm{Mod}}\newcommand{\Gal}{\mathrm{Gal}}$

Let $\Mod(K/k)$ be the category of $K/k$-vector spaces: Objects are $K$-vector spaces $V$ together with an action of the Galois group $\Gal(K/k)$ on $V$ which is semilinear in the sense that $\sigma(a v) = \sigma(a) \sigma(v)$ for $\sigma \in \Gal(K/k)$, $a \in K$ and $v \in V$. Morphisms are $K$-linear maps which commute with the action.

Now Galois descent is the statement that the functors $$\Mod(k) \to \Mod(K/k), \qquad V \mapsto K \otimes_k V$$ and $$\Mod(K/k) \to \Mod(k), \qquad W \mapsto W^{\Gal(K/k)}$$ form an equivalence of ($k$-linear, symmetric monoidal, …) categories.

This induces equivalences of the respective categories of monoids (i.e. $k$-algebras on the left hand side), which is related to Brauer groups, or of “objects with a blinear form” (i.e. objects $V$ equipped with a map from $V \otimes V$ to the monoidal unit, satisfying some axioms), which is related to the Galois cohomology of the orthogonal group.

Is there some equivalence between categories of non-linear objects, which induces the above equivalence between $k$-vector spaces and $K/k$-vector spaces? For example, is there some ringed topos $(X, \mathcal{O}_X)$ such that $\Mod(K/k)$ is more or less by definition the category of modules in this ringed topos and such that $(X, \mathcal{O}_X)$ is equivalent to $(\mathrm{Set},k)$?

I have to admit that I do not really expect the answer to the above question to be “yes”. For example one way to prove Galois descent is to realise that $\Mod(K/k)$ is basically the category of modules over the twisted group algebra $K[\Gal(K/k)]$ and that by linear indpendence of characters, this algebra is isomorphic to $\mathrm{End}_k(K)$, which by Morita theory has the same module category as $k$. So one (though not commutative) ringed topos giving rise to $\Mod(K/k)$ would be $(\mathrm{Set},K[\Gal(K/k)])$, which is, of course, not equivalent to $(\mathrm{Set}, K)$. But maybe there is some category of sets with a “semi-linear“ Galois action (whatever that would mean) which could work? Or maybe there is a better suited topos than $(\mathrm{Set},k)$, which gives rise to $\Mod(k)$?

But in fact, every proof I can think of to prove Galois descent is based on the linear independence of characters and hence something very linear-ish. Another thing making me sceptical is that the functor $V \mapsto K \otimes_k V$ does not seem to have a non-linear equivalent. Therefore I would like to know:

Is there any high-level explenation (more convincing than the one given in the above paragraph) why the answer to question 1 has to be “no”?

This question is about Galois theory. So let $K / k$ be a Galois extension of fields. Let us assume that $K / k$ is finite dimensional, though everything can be made to work in the profinite case by adding some continuity conditions here and there. I should perhaps remark that I do not (yet) know Grothendieck's Galois theory. If this theory answers my question, I will be happy with some precise statement. $\newcommand{\Mod}{\mathrm{Mod}}\newcommand{\Gal}{\mathrm{Gal}}$

Let $\Mod(K/k)$ be the category of $K/k$-vector spaces: Objects are $K$-vector spaces $V$ together with an action of the Galois group $\Gal(K/k)$ on $V$ which is semilinear in the sense that $\sigma(a v) = \sigma(a) \sigma(v)$ for $\sigma \in \Gal(K/k)$, $a \in K$ and $v \in V$. Morphisms are $K$-linear maps which commute with the action.

Now Galois descent is the statement that the functors $$\Mod(k) \to \Mod(K/k), \qquad V \mapsto K \otimes_k V$$ and $$\Mod(K/k) \to \Mod(k), \qquad W \mapsto W^{\Gal(K/k)}$$ form an equivalence of ($k$-linear, symmetric monoidal, …) categories.

This induces equivalences of the respective categories of monoids (i.e. $k$-algebras on the left hand side), which is related to Brauer groups, or of “objects with a blinear form” (i.e. objects $V$ equipped with a map from $V \otimes V$ to the monoidal unit, satisfying some axioms), which is related to the Galois cohomology of the orthogonal group.

Is there some equivalence between categories of non-linear objects, which induces the above equivalence between $k$-vector spaces and $K/k$-vector spaces? For example, is there some ringed topos $(X, \mathcal{O}_X)$ such that $\Mod(K/k)$ is more or less by definition the category of modules in this ringed topos and such that $(X, \mathcal{O}_X)$ is equivalent to $(\mathrm{Set},K)$?

I have to admit that I do not really expect the answer to the above question to be “yes”. For example every proof I can think of, to prove Galois descent is based on the invertibility of the matrix $\{(\sigma_i(a_j)\}_{i,j = 1, \dots, n}$ if $\{\alpha_1, \dots, \alpha_n\}$ is a $k$-basis of $K$ and if $\sigma_1, \dots, \sigma_n$ are the elements of $\Gal(K/k)$. Another reason is that the functor $V \mapsto K \otimes_k V$ does not seem to have a non-linear equivalent. Therefore I would like to know:

Is there any high-level explenation (more convincing than the one given in the above paragraph) why the answer to question 1 has to be “no”?

This question is about Galois theory. So let $K / k$ be a Galois extension of fields. Let us assume that $K / k$ is finite dimensional, though everything can be made to work in the profinite case by adding some continuity conditions here and there. I should perhaps remark that I do not (yet) know Grothendieck's Galois theory. If this theory answers my question, I will be happy with some precise statement. $\newcommand{\Mod}{\mathrm{Mod}}\newcommand{\Gal}{\mathrm{Gal}}$

Let $\Mod(K/k)$ be the category of $K/k$-vector spaces: Objects are $K$-vector spaces $V$ together with an action of the Galois group $\Gal(K/k)$ on $V$ which is semilinear in the sense that $\sigma(a v) = \sigma(a) \sigma(v)$ for $\sigma \in \Gal(K/k)$, $a \in K$ and $v \in V$. Morphisms are $K$-linear maps which commute with the action.

Now Galois descent is the statement that the functors $$\Mod(k) \to \Mod(K/k), \qquad V \mapsto K \otimes_k V$$ and $$\Mod(K/k) \to \Mod(k), \qquad W \mapsto W^{\Gal(K/k)}$$ form an equivalence of ($k$-linear, symmetric monoidal, …) categories.

This induces equivalences of the respective categories of monoids (i.e. $k$-algebras on the left hand side), which is related to Brauer groups, or of “objects with a blinear form” (i.e. objects $V$ equipped with a map from $V \otimes V$ to the monoidal unit, satisfying some axioms), which is related to the Galois cohomology of the orthogonal group.

Is there some equivalence between categories of non-linear objects, which induces the above equivalence between $k$-vector spaces and $K/k$-vector spaces? For example, is there some ringed topos $(X, \mathcal{O}_X)$ such that $\Mod(K/k)$ is more or less by definition the category of modules in this ringed topos and such that $(X, \mathcal{O}_X)$ is equivalent to $(\mathrm{Set},k)$?

I have to admit that I do not really expect the answer to the above question to be “yes”. For example one way to prove Galois descent is to realise that $\Mod(K/k)$ is basically the category of modules over the twisted group algebra $K[\Gal(K/k)]$ and that by linear indpendence of characters, this algebra is isomorphic to $\mathrm{End}_k(K)$, which by Morita theory has the same module category as $k$. So one ringed topos giving rise to $\Mod(K/k)$ would be $(\mathrm{Set},K[\Gal(K/k)])$, which is, of course, not equivalent to $(\mathrm{Set}, K)$. But maybe there is some category of sets with a “semi-linear“ Galois action (whatever that would mean) which could work? Or maybe there is a better suited topos than $(\mathrm{Set},k)$, which gives rise to $\Mod(k)$?

But in fact, every proof I can think of to prove Galois descent is based on the linear independence of characters and hence something very linear-ish. Another thing making me sceptical is that the functor $V \mapsto K \otimes_k V$ does not seem to have a non-linear equivalent. Therefore I would like to know:

Is there any high-level explenation (more convincing than the one given in the above paragraph) why the answer to question 1 has to be “no”?