Categorification of Artin's Lemma

Lemma (Isometries from perfect pairings). Let $\mathcal{C}$$\mathcal C$ be a monoidal category, and let $X$$X$ admit a left dual $X^{\vee }$$X^\vee$, with evaluation and coevaluation

$${\mathrm{ev}}_X:X^{\vee }\otimes X⟶\mathbf{1},{\mathrm{coev}}_X:\mathbf{1}⟶X\otimes X^{\vee }.$$

The duality adjunction gives natural bijections

$${\mathrm{Hom}}_{\mathcal{C}}(U\otimes X,\mathbf{1})\cong {\mathrm{Hom}}_{\mathcal{C}}(U,X^{\vee }).$$

A pairing $b_X:X\otimes X\to \mathbf{1}$$b_X:X\otimes X\to\mathbf1$ therefore determines a morphism

$${\beta }_X:X\stackrel{{\mathrm{id}}_X\otimes {\mathrm{coev}}_X}{\to }X\otimes X\otimes X^{\vee }\stackrel{b_X\otimes {\mathrm{id}}_{X^{\vee }}}{\to }{X}^{\vee },$$

uniquely characterized by

$${\mathrm{ev}}_X\circ ({\beta }_X\otimes {\mathrm{id}}_X)=b_X.$$

We call $b_X$$b_X$ perfect if ${\beta }_X$$\beta_X$ is an isomorphism.

Suppose $b_X$$b_X$ is perfect. Let $Y$$Y$ be any object with a pairing $b_Y:Y\otimes Y\to \mathbf{1}$$b_Y:Y\otimes Y\to\mathbf1$. If $f:X\to Y$$f:X\to Y$ satisfies

$$b_Y\circ (f\otimes f)=b_X,$$

then $f$$f$ is a split monomorphism. In particular, if $f$$f$ is an epimorphism in $\mathcal{C}$$\mathcal C$, it is an isomorphism.

Proof. Let $\gamma :Y\to X^{\vee }$$\gamma:Y\to X^\vee$ correspond under the duality adjunction to

$$Y\otimes X\stackrel{{\mathrm{id}}_Y\otimes f}{\to }Y\otimes Y\stackrel{b_Y}{\to }\mathbf{1}.$$

Then

$${\mathrm{ev}}_X\circ (\gamma f\otimes {\mathrm{id}}_X)=b_Y\circ (f\otimes f)=b_X.$$

By uniqueness, $\gamma f={\beta }_X$$\gamma f=\beta_X$. Hence

$$r:={\beta }_X^{-1}\gamma$$

satisfies $rf={\mathrm{id}}_X$$rf=\operatorname{id}_X$. If $f$$f$ is also an epimorphism, the equality $(fr)f=f$$(fr)f=f$ implies $fr={\mathrm{id}}_Y$$fr=\operatorname{id}_Y$.

$◻$$\square$

Example (Vector spaces and modules). Let $X,Y$$X,Y$ be modules over a commutative ring $A$$A$, equipped with $A$$A$-bilinear pairings $b_X,b_Y$$b_X,b_Y$, with $b_X$$b_X$ perfect. Suppose an $A$$A$-linear map $f:X\to Y$$f:X\to Y$ preserves the pairings:

$$b_X(x,x^{\prime })=b_Y(f(x),f(x^{\prime })).$$

If $x\in \ker f$$x\in\ker f$, then for every $x^{\prime }\in X$$x'\in X$,

$$b_X(x,x^{\prime })=b_Y(0,f(x^{\prime }))=0.$$

Since $b_X$$b_X$ is perfect, the induced map $X\to {\mathrm{Hom}}_A(X,A)$$X\to\operatorname{Hom}_A(X,A)$ is injective, so $x=0$$x=0$. Thus $f$$f$ is injective, and it is an isomorphism whenever it is also surjective. In fact, this argument only requires $b_X$$b_X$ to be nondegenerate in the first variable.

Proposition (A Galois-coordinate criterion). Let a finite group $G$$G$ act on a commutative ring $B$$B$ by automorphisms, and put $A=B^G$$A=B^G$. Consider the canonical map

$$\mathrm{\Phi }:B{\otimes }_{A}B⟶\prod _{g\in G}B,u\otimes v⟼(ug(v){)}_g.$$

The source is a $B$$B$-algebra through the first tensor factor, and the target has the diagonal $B$$B$-algebra structure.

If ${\delta }_1\in \mathrm{im}\mathrm{\Phi }$$\delta_1\in\operatorname{im}\Phi$, where ${\delta }_1$$\delta_1$ is the characteristic function of the identity, then $\mathrm{\Phi }$$\Phi$ is an isomorphism. In particular, surjectivity of $\mathrm{\Phi }$$\Phi$ suffices.

Proof. Choose

$$p=\sum _{i=1}^m{x}_i\otimes y_i\text{with}\mathrm{\Phi }(p)={\delta }_1.$$

Thus

$$\begin{array}{}\text{(1)}& \sum _i{x}_{i}g(y_i)={\delta }_{g,1}(g\in G).\end{array}$$

For each $g\in G$$g\in G$,

$$\mathrm{\Phi }(\sum _i{x}_i\otimes g^{-1}(y_i))={\delta }_g.$$

Since the ${\delta }_g$$\delta_g$ form a $B$$B$-basis of the target, $\mathrm{\Phi }$$\Phi$ is surjective.

Define

$$T:B⟶A,T(b)=\sum _{g\in G}g(b).$$

This is $A$$A$-linear, and its values lie in $A$$A$ because $G$$G$ permutes the summands. For every $b\in B$$b\in B$, equation (1) gives

$$\begin{array}{}\text{(2)}& \sum _i{x}_{i}T(y_{i}b)=\sum _{g\in G}(\sum _i{x}_{i}g(y_i))g(b)=b.\end{array}$$

Applying $g$$g$ to equation (1) indexed by $g^{-1}$$g^{-1}$ gives

$$\sum _{i}g(x_i)y_i={\delta }_{g,1},$$

and hence

$$\begin{array}{}\text{(3)}& \sum _{i}T(b{x}_i)y_i=b.\end{array}$$

Consider the maps

$$\mathrm{ev}:B{\otimes }_{A}B⟶A,b\otimes c⟼T(bc),$$

and

$$\mathrm{coev}:A⟶B{\otimes }_{A}B,1⟼p.$$

Equations (2) and (3) are, respectively, the triangle identities

$$({\mathrm{id}}_B\otimes \mathrm{ev})\circ (\mathrm{coev}\otimes {\mathrm{id}}_B)={\mathrm{id}}_B$$

and

$$(\mathrm{ev}\otimes {\mathrm{id}}_B)\circ ({\mathrm{id}}_B\otimes \mathrm{coev})={\mathrm{id}}_B.$$

Thus these maps exhibit $B$$B$ as its own dual in ${\mathrm{Mod}}_A$$\operatorname{Mod}_A$.

Extension of scalars

$$B{\otimes }_A-:{\mathrm{Mod}}_A⟶{\mathrm{Mod}}_B$$

is strong monoidal and therefore preserves this duality. Consequently,

$$M:=B{\otimes }_{A}B$$

has the perfect pairing

$$⟨u\otimes v,u^{\prime }\otimes v^{\prime }{⟩}_M=u{u}^{\prime }T(v{v}^{\prime }).$$

Equip $N:=\prod _{g\in G}B$$N:=\prod_{g\in G}B$ with the pairing

$$⟨(z_g{)}_g,(w_g{)}_g{⟩}_N=\sum _{g\in G}{z}_g{w}_g.$$

Then

$$\begin{array}{rl}⟨\mathrm{\Phi }(u\otimes v),\mathrm{\Phi }(u^{\prime }\otimes v^{\prime }){⟩}_N& =\sum _{g\in G}ug(v)u^{\prime }g(v^{\prime })\\ & =u{u}^{\prime }T(v{v}^{\prime })\\ & =⟨u\otimes v,u^{\prime }\otimes v^{\prime }{⟩}_M.\end{array}$$

By the lemma, $\mathrm{\Phi }$$\Phi$ is a split monomorphism of $B$$B$-modules. Since it is also surjective, it is an isomorphism of $B$$B$-algebras.

$◻$$\square$

Remark. The condition (1) is the classical Galois-coordinate condition of Chase–Harrison–Rosenberg. Unwinding the splitting above gives the explicit inverse

$${\mathrm{\Phi }}^{-1}((z_g{)}_g)=\sum _{g\in G}\sum _i{z}_{g}g(x_i)\otimes y_i.$$

Indeed,

$$\sum _{i}g(x_i)h(y_i)=g(\sum _i{x}_i(g^{-1}h)(y_i))={\delta }_{g,h},$$

which verifies $\mathrm{\Phi }{\mathrm{\Phi }}^{-1}=\mathrm{id}$$\Phi\Phi^{-1}=\operatorname{id}$; equation (3) verifies the other composite.

Theorem (Artin's lemma, in structural form). Let $K$$K$ be a field, let $G\le \mathrm{Aut}(K)$$G\leq\operatorname{Aut}(K)$ be a finite subgroup, and put $F=K^G$$F=K^G$. Then the canonical map

$$\mathrm{\Phi }:K{\otimes }_{F}K\stackrel{\sim }{\to }\prod _{\sigma \in G}K,u\otimes v⟼(u\sigma (v){)}_{\sigma },$$

is an isomorphism of $K$$K$-algebras. In particular,

$$[K:F]=|G|.$$

Proof. Put $R=K{\otimes }_{F}K$$R=K\otimes_F K$. For each $\sigma \in G$$\sigma\in G$, the $K$$K$-algebra homomorphism

$${\phi }_{\sigma }:R⟶K,u\otimes v⟼u\sigma (v),$$

is surjective, since ${\phi }_{\sigma }(u\otimes 1)=u$$\varphi_\sigma(u\otimes1)=u$. Therefore

$${\mathfrak{m}}_{\sigma }:=\ker {\phi }_{\sigma }$$

is a maximal ideal.

If $\sigma \ne \tau$$\sigma\ne\tau$, choose $b\in K$$b\in K$ with $\sigma (b)\ne \tau (b)$$\sigma(b)\ne\tau(b)$. Then

$$1\otimes b-\sigma (b)\otimes 1\in {\mathfrak{m}}_{\sigma }\setminus {\mathfrak{m}}_{\tau }.$$

Thus the ideals ${\mathfrak{m}}_{\sigma }$$\mathfrak m_\sigma$ are pairwise distinct, hence pairwise comaximal. The Chinese remainder theorem gives a surjection

$$R⟶\prod _{\sigma \in G}R/{\mathfrak{m}}_{\sigma }\cong \prod _{\sigma \in G}K.$$

This map is precisely $\mathrm{\Phi }$$\Phi$. The proposition therefore shows that $\mathrm{\Phi }$$\Phi$ is an isomorphism.

No finiteness assumption on $K/F$$K/F$ was needed: the reconstruction identity (2), applied to $B=K$$B=K$, already shows that the finitely many $x_i$$x_i$ span $K$$K$ over $F$$F$. Taking $K$$K$-dimensions now gives

$$[K:F]={\dim }_K(K{\otimes }_{F}K)={\dim }_K\left(\prod _{\sigma \in G}K\right)=|G|.$$

$◻$$\square$