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Saturday, July 26, 2025

Gal(-/F) as a Group valued Sheaf over a Site

What is Site?

Definition of a Site

A site is a pair (C,Cov(C)) where:

  • C is a category;

  • Cov(C) assigns to each object U∈C a collection Cov(U) of families of morphisms

    {Ui→U}i∈I

    called coverings, such that:

Identity. For every U, the singleton family

{idU:U→U}

lies in Cov(U).

Base‐change stability. If

{Ui→U}∈Cov(U)andV→U

is any arrow, then the pullback family

{V×UUi→V}

lies in Cov(V).

Transitivity. If

{Ui→U}∈Cov(U)and for each i,{Vij→Ui}∈Cov(Ui)

then the composed family

{Vij→U}i,j

lies in Cov(U).

Equivalently, Cov(C) is a Grothendieck pretopology on C.

Example of a Site

Consider the coslice category C:=F/Field.

Define

Cov(L)={{Ki↪L}i∈I|⋃i∈IKi=L}.

Here ↪ means ⊆ in set theory.

One checks easily that this defines a Grothendieck topology.

Using covering in this site to define algebraic extension.

Definition and Proposition. An equivalent definition of algebraic extension via covering.

Let L/F be a field extension, then L/F is an algebraic extension iff L could be covered by a family of finite extension. i.e.

We have:

{Ki↪L}i∈I∈Cov(L) such that [Ki:F]<∞

Proof.

If L/F is an algebraic extension, then {F(x)↪L}x∈L is such a covering in Cov(L).

Conversely, assume {Ki↪L}i∈I∈Cov(L),[Ki:F]<∞, for each x∈L, we could find a Ki such that x∈Ki. Then x is algebraic since it lies in an algebraic extension. ◻

The Site we will use

Consider the coslice category C:=F/Field.

We would like to associate a Grothendieck topology to make (C,Cov(C)) become a site and Gal(−/F) become a sheaf

Let us define Cov(C) now.

Definition and Proposition.

Let L/F be a field extension

Cov(L)={{Ki↪L}i∈I|⋃i∈IKi=L and ∀ two elemnts subset X⊆L,∃i∈I:X⊆Ki}.

Here ↪ means ⊆ in set theory.

Proof.

Let us check the axiom of covering.

Identity.

If L↪L then clearly we have L↪L∈Cov(L).

Base Change.

If {Ki↪L}i∈I∈Cov(L) and assume E↪L. Then we have Ki×LE≅Ki∩E.

Hence we have {Ki×LE↪E}i∈I∈Cov(E) since we have (⋃i∈IKi)∩E=L∩E=E and for any finite Y such that Y↪E↪L, there exits i∈I:Y↪Ki, hence Y↪Ki×LE.

Transitivity.

If {Ki↪L}i∈I∈Cov(L) and {Ei,j↪Ki}j∈Ji∈Cov(Ki) then {Ei,j↪L}i,j∈Cov(L)

Easy to see that we have

L=⋃i∈IKi=⋃i∈I⋃j∈JiEi,j

For any two elements set X↪L, ∃i∈I,X↪Ki. Now let {Ei,j↪Ki}j∈Ji∈Cov(Ki), then ∃j∈Ji,X↪Ei,j.

Hence Cov(C) is a covering, (C,Cov(C)) form a site. ◻

Sheaf over Site

Definition.

Let (C,Cov(C)) be any site. A Set,Group,Ab,Ring,R-Mod... valued presheaf F is called a sheaf if for every covering family

{Ui→U}i∈I∈Cov(U)

the diagram

F(U)→p∏iF(Ui)⇉pjpi∏i,jF(Ui∩Uj)

is an equalizer. Here Ui∩Uj=Ui×UUj.

Remark. I would like to add more explanation for this definition.

Firstly,

p=∏iF(Ui→U),pi=∏iF(Ui∩Uj→Ui),pj=∏jF(Ui∩Uj→Uj)

The locally property of sheaf comes from the equalizer arrow p is injective. Hence we have:

p(s)=(s|Ui)i=(t|Ui)i=p(t)⟺s=t

The gluing property comes from, let (si|Ui)i∈∏iF(Ui), if

pi((si)i)=(si|Ui∩Uj)i,j=(sj|Ui∩UJ)i,j=pj((sj)j)

then by the universal property of equalizer, there exists unique s∈F(U) such that p(s)=(si|Ui)i.

Main Result: Gal(−/F):(C,Cov(C))op→Grp is a Sheaf

Gal(−/F) as presheaf

Before we check the sheaf condition, we need to explain why Gal(−/F) is a group valued presheaf.

Firstly, we know that for each object K in C, Gal(K/F)=AutF(K) is a group. For a field extension K↪L, we have the natural restriction map

resKL:Gal(L/F)→Gal(K/F),σ⟼σ|K

Hence it is a presheaf.

The Sheaf Condition

Now let us check the sheaf condition. Let {Ki↪L}i∈I∈Cov(L), then we need to check the equalizer diagram as follows:

Gal(L/F)→p∏iGal(Ki/F)⇉pjpi∏i,jGal(Ki∩Kj/F)

Firstly, Ki∩Kj↪Ki↪L=Ki∩Kj↪Kj↪L, hence we have pi∘p=pj∘p.

We check locally and gluing condition.

Locally.

Let τ,σ be two elements in Gal(L/F), then if τ|Ki=σ|Ki for all Ki↪L∈{Ki↪L}i∈I, we have ∀x∈L,τ(x)=σ(x), since ⋃i∈IKi=L. Hence τ=σ.

Gluing.

Assume that we have a family (σi)i∈∏iGal(Ki/F) such that σi|Ki∩Kj=σj|Ki∩Kj for all i,j. Then we define σ(x)=σ|i(x) whenever x∈Ki. Compatibility on overlaps ensures this is uniquely well–defined. Also, σ is bijective on L since σ|i:Ki→Ki is bijective and ⋃i∈IKi=L.

Now we need to check that σ∈Gal(L/F). Firstly, ∀t∈F,σ(t)=σ|i(t)=t.

For {x,y}⊂L. Hence ∃i∈I,{x,y}↪Ki, we have σ(x+y)=σ|i(x+y)=σ|i(x)+σ|i(y)=σ(x)+σ(y). Similarly we have σ(xy)=σ(x)σ(y). Therefore, we have σ∈Gal(L/F) ◻

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