Galois Descent for vector spaces

Definition (Internal modules). Let $G$$G$ be a group and let $A$$A$ be a commutative unital ring object in $G\text{-}\mathbf{Set}$$G\text{-}\mathbf{Set}$. The category ${\mathrm{Mod}}_A(G\text{-}\mathbf{Set})$$\operatorname{Mod}_A(G\text{-}\mathbf{Set})$ is defined as follows.

An object is an abelian group object $M$$M$ in $G\text{-}\mathbf{Set}$$G\text{-}\mathbf{Set}$, together with a morphism

$$\mu :A\times M⟶M,$$

called scalar multiplication. Writing $am:=\mu (a,m)$$am:=\mu(a,m)$, we require

$$\begin{array}{rl}a(m+n)& =am+an,\\ (a+b)m& =am+bm,\\ (ab)m& =a(bm),\\ 1_{A}m& =m\end{array}$$

for all $a,b\in A$$a,b\in A$ and $m,n\in M$$m,n\in M$.

Here, an abelian group object consists of $G$$G$-equivariant maps

$${+}_M:M\times M⟶M,0_M:1⟶M,{-}_M:M⟶M$$

satisfying the usual abelian group axioms. Products carry the diagonal $G$$G$-action, and $1$$1$ denotes the terminal $G$$G$-set. In particular, the equivariance of $\mu$$\mu$ means that

$$g(am)=g(a)g(m)$$

for every $g\in G$$g\in G$, $a\in A$$a\in A$, and $m\in M$$m\in M$.

A morphism $f:M\to N$$f:M\to N$ is a morphism of abelian group objects in $G\text{-}\mathbf{Set}$$G\text{-}\mathbf{Set}$ satisfying

$$f\circ {\mu }_M={\mu }_N\circ ({\mathrm{id}}_A\times f).$$

Equivalently, $f$$f$ is an additive map such that

$$f(am)=af(m),f(gm)=gf(m).$$

Identities and composition are inherited from $G\text{-}\mathbf{Set}$$G\text{-}\mathbf{Set}$.

Lemma. The global section functor ${\mathrm{Hom}}_G(1,-)$$\mathrm{Hom}_{G}(1,-)$ induce the left exact functor $\mathrm{\Gamma }:{\mathrm{Mod}}_A(G\text{-}\mathbf{Set})\to A^G\text{-Mod}.$$\Gamma:\operatorname{Mod}_A(G\text{-}\mathbf{Set})\to A^G\text{-Mod}.$

Proof. $\mathrm{\Gamma }$$\Gamma$ preserve equation and limit, hence preserve the model of module over internal ring. $◻$$\square$

Proposition (Galois descent). Let $L/K$$L/K$ be a finite Galois extension with Galois group $G$$G$. Regard $L$$L$ as an internal ring in $G\text{-}\mathbf{Set}$$G\text{-}\mathbf{Set}$ via its natural Galois action. Then the invariants functor

$$\mathrm{\Gamma }:{\mathrm{Mod}}_L(G\text{-}\mathbf{Set})⟶K\text{-}\mathbf{Mod},M⟼M^G,$$

is a $K$$K$-linear strong symmetric monoidal equivalence, where the tensor products on the source and target are ${\otimes }_L$$\otimes_L$ and ${\otimes }_K$$\otimes_K$, respectively.

A quasi-inverse is given by extension of scalars,

$$L{\otimes }_K-:K\text{-}\mathbf{Mod}⟶{\mathrm{Mod}}_L(G\text{-}\mathbf{Set}),$$

with the action

$$g(\ell \otimes v)=g(\ell )\otimes v.$$

In particular, $\mathrm{\Gamma }$$\Gamma$ is exact.

Proof. Define

$$E:K\text{-}\mathbf{Mod}⟶{\mathrm{Mod}}_L(G\text{-}\mathbf{Set}),V⟼L{\otimes }_{K}V,$$

with $g(\ell \otimes v)=g(\ell )\otimes v$$g(\ell\otimes v)=g(\ell)\otimes v$. We have natural maps

$${\eta }_V:V⟶(L{\otimes }_{K}V{)}^G,v⟼1\otimes v,$$

and

$${\epsilon }_M:L{\otimes }_K{M}^G⟶M,\ell \otimes m⟼\ell m.$$

Choosing a $K$$K$-basis of $V$$V$, we see that an element of $L{\otimes }_{K}V$$L\otimes_K V$ is $G$$G$-invariant precisely when all its coefficients belong to $L^G=K$$L^G=K$. Hence ${\eta }_V$$\eta_V$ is an isomorphism.

The map ${\epsilon }_M$$\varepsilon_M$ is $L$$L$-linear and $G$$G$-equivariant. We prove its bijectivity using an auxiliary function space.

Consider the $L$$L$-linear map

$$\mathrm{\Phi }:L{\otimes }_{K}M⟶\mathrm{Map}(G,M),\mathrm{\Phi }(a\otimes m)(g)=ag(m),$$

where the target has pointwise addition and scalar multiplication.

First, $\mathrm{\Phi }$$\Phi$ is an isomorphism. Indeed, the standard Galois isomorphism

$$L{\otimes }_{K}L\stackrel{\sim }{\to }\prod _{g\in G}L,a\otimes b⟼(ag(b){)}_{g\in G},$$

is an isomorphism of $(L,L)$$(L,L)$-bimodules if the $g$$g$-th factor on the right is given the usual left action and the twisted right action

$$a\cdot b=ag(b).$$

Denote this bimodule by $L_g$$L_g$. Tensoring over the right $L$$L$-action with $M$$M$ gives

$$L{\otimes }_{K}M\cong (L{\otimes }_{K}L){\otimes }_{L}M\cong \prod _{g\in G}(L_g{\otimes }_{L}M).$$

For each $g$$g$, semilinearity gives an isomorphism

$$L_g{\otimes }_{L}M\stackrel{\sim }{\to }M,a\otimes m⟼ag(m),$$

with inverse $m\mapsto 1\otimes g^{-1}(m)$$m\mapsto1\otimes g^{-1}(m)$. Their product is precisely $\mathrm{\Phi }$$\Phi$.

Now equip these spaces with the auxiliary $G$$G$-actions

$$h\cdot (a\otimes m)=a\otimes h(m),(h\cdot f)(g)=f(gh).$$

In particular, the first $L$$L$-factor in the auxiliary tensor product is fixed. With these actions, $\mathrm{\Phi }$$\Phi$ is $G$$G$-equivariant, since

$$\mathrm{\Phi }(a\otimes h(m))(g)=a(gh)(m)=(h\cdot \mathrm{\Phi }(a\otimes m))(g).$$

Let $B=K[G]$$B=K[G]$, and let $M_0$$M_0$ denote the underlying $K$$K$-vector space of $M$$M$. Restriction to the basis $G\subset B$$G\subset B$ identifies

$${\mathrm{Hom}}_K(B,M_0)\cong \mathrm{Map}(G,M),$$

where the left $B$$B$-action on the Hom is

$$(b\cdot f)(x)=f(xb).$$

This is exactly the right translation action above.

We therefore have the following chain of isomorphisms of $K$$K$-vector spaces:

$$\begin{array}{rl}L{\otimes }_K{M}^G& \cong (L{\otimes }_{K}M{)}^G\\ & \underset{\sim }{\overset{{\mathrm{\Phi }}^G}{\to }}\mathrm{Map}(G,M{)}^G\\ & \cong {\mathrm{Hom}}_B(K,{\mathrm{Hom}}_K(B,M_0))\\ & \stackrel{(\ast )}{\cong }{\mathrm{Hom}}_K(B{\otimes }_{B}K,M_0)\\ & \cong {\mathrm{Hom}}_K(K,M_0)\\ & \cong M_0.\end{array}$$

The first isomorphism uses flatness of $L/K$$L/K$: taking $G$$G$-invariants is the kernel of the $K$$K$-linear map

$$M⟶\prod _{g\in G}M,m⟼(g(m)-m{)}_{g\in G},$$

and $L{\otimes }_K-$$L\otimes_K-$ preserves this kernel and the finite product.

The isomorphism $(\ast )$$(*)$ is precisely the bimodule tensor–Hom adjunction, applied to the $(K,B)$$(K,B)$-bimodule

$${}_K{B}_B,$$

the trivial left $B$$B$-module $K$$K$, and the left $K$$K$-module $M_0$$M_0$:

$${\mathrm{Hom}}_B(K,{\mathrm{Hom}}_K(B,M_0))\cong {\mathrm{Hom}}_K(B{\otimes }_{B}K,M_0).$$

Explicitly, it sends $F:K\to {\mathrm{Hom}}_K(B,M_0)$$F:K\to\operatorname{Hom}_K(B,M_0)$ to

$$b\otimes c⟼F(c)(b).$$

Together with $B{\otimes }_{B}K\cong K$$B\otimes_B K\cong K$ and evaluation at $1\in K$$1\in K$, this identifies an invariant function $f:G\to M$$f:G\to M$ with $f(e)$$f(e)$.

Finally, if $m\in M^G$$m\in M^G$, then

$$\mathrm{\Phi }(a\otimes m)(g)=ag(m)=am.$$

Thus the composite above sends $a\otimes m$$a\otimes m$ to $am$$am$: it is exactly the underlying $K$$K$-linear map of ${\epsilon }_M$$\varepsilon_M$.

Consequently, ${\epsilon }_M$$\varepsilon_M$ is bijective. Since it is already $L$$L$-linear and $G$$G$-equivariant for the stated semilinear structures, it is an isomorphism in ${\mathrm{Mod}}_L(G\text{-}\mathbf{Set})$$\operatorname{Mod}_L(G\text{-}\mathbf{Set})$.

The functor $\mathrm{\Gamma }$$\Gamma$ is $K$$K$-linear, since restriction to invariant subspaces preserves $K$$K$-linear combinations of morphisms.

We next establish its strong symmetric monoidal structure. For $M,N\in {\mathrm{Mod}}_L(G\text{-}\mathbf{Set})$$M,N\in\operatorname{Mod}_L(G\text{-}\mathbf{Set})$, there is a natural $K$$K$-linear map

$${\alpha }_{M,N}:M^G{\otimes }_K{N}^G⟶(M{\otimes }_{L}N{)}^G,m\otimes n⟼m\otimes n,$$

where $M{\otimes }_{L}N$$M\otimes_L N$ carries the diagonal semilinear action. The unit comparison is the canonical isomorphism

$$K\stackrel{\sim }{\to }{L}^G.$$

To prove that ${\alpha }_{M,N}$$\alpha_{M,N}$ is an isomorphism, extend scalars to $L$$L$. The composite

$$L{\otimes }_K(M^G{\otimes }_K{N}^G)\stackrel{L{\otimes }_K{\alpha }_{M,N}}{\to }L{\otimes }_K(M{\otimes }_{L}N{)}^G\stackrel{{\epsilon }_{M{\otimes }_{L}N}}{\to }M{\otimes }_{L}N$$

agrees with the composite

$$L{\otimes }_K(M^G{\otimes }_K{N}^G)\stackrel{\sim }{\to }(L{\otimes }_K{M}^G){\otimes }_L(L{\otimes }_K{N}^G)\stackrel{{\epsilon }_M{\otimes }_L{\epsilon }_N}{\to }M{\otimes }_{L}N$$

where the first isomorphism sends

$$a\otimes (m\otimes n)⟼(a\otimes m)\otimes (1\otimes n).$$

Indeed, both composites send $a\otimes (m\otimes n)$$a\otimes(m\otimes n)$ to $am\otimes n$$am\otimes n$.

Since ${\epsilon }_M$$\varepsilon_M$, ${\epsilon }_N$$\varepsilon_N$, and ${\epsilon }_{M{\otimes }_{L}N}$$\varepsilon_{M\otimes_L N}$ are isomorphisms, so is $L{\otimes }_K{\alpha }_{M,N}$$L\otimes_K\alpha_{M,N}$. Faithful flatness of $L/K$$L/K$ then implies that ${\alpha }_{M,N}$$\alpha_{M,N}$ is an isomorphism.

The associativity, unit, and symmetry compatibility diagrams commute by the defining formulas on pure tensors. Thus $\mathrm{\Gamma }$$\Gamma$ is strong symmetric monoidal.

Finally, both categories are abelian. In ${\mathrm{Mod}}_L(G\text{-}\mathbf{Set})$$\operatorname{Mod}_L(G\text{-}\mathbf{Set})$, kernels and cokernels are computed on the underlying $L$$L$-modules and carry the induced semilinear $G$$G$-actions. Since $\mathrm{\Gamma }$$\Gamma$ is an additive equivalence of abelian categories, it preserves kernels and cokernels, and hence is exact. $◻$$\square$