Classical Complex Algebraic Actions and Affine Embeddings

Equations for $X$ in the first $m$ variables and for $Y$ in the last $n$ variables cut out exactly $X\times Y$. Multiplying polynomial functions in separate variables gives the displayed algebra map. Every polynomial in the $m+n$ coordinates is a sum of products of such polynomials, so this map is surjective.

For injectivity write a kernel element as ${\sum }_{i=1}^r{f}_i\otimes h_i$ with the $h_i$ linearly independent, by eliminating redundant terms in a finite expression. At each $x$, the polynomial function ${\sum }_i{f}_i(x)h_i$ on $Y$ is zero. Linear independence in the function space therefore forces every $f_i(x)=0$. By F1 all $f_i$ are zero. If a factor is empty its coordinate ring and that of the product are zero, so the same conclusion holds. Iteration proves the three-factor formula. Only finite expressions and finite-dimensional elimination were used.

Pull back an action along $a$ and apply F1. The two action identities evaluated on $f$ give respectively $f(ghx)=f(g(hx))$ and $f(ex)=f(x)$, precisely the displayed identities for $\delta$. Conversely F2 reconstructs a unique morphism from an algebra map $\delta$; F1 and equality of pullbacks turn the displayed identities back into the action identities. Thus this is a bijection, also for the empty $X$.

For this action define $b(x,g)=g^{-1}x$. Inversion is a morphism and $b(b(x,g),h)=h^{-1}{g}^{-1}x=(gh{)}^{-1}x=b(x,gh)$, while $b(x,e)=x$. Pullback gives exactly $c=\tau (S\otimes \mathrm{id})\delta$, with the right-comodule identities asserted. Conversely a right-comodule algebra map reconstructs $b$ by F2 and its identities by F1; $a(g,x)=b(x,g^{-1})$ reconstructs the original left action. This also proves that conversion of either coaction to the other is inverse, since inversion squared is the identity.

Evaluation gives $r(g)f(x)=f(g^{-1}x)$ and $r(g)r(h)f(x)=f(h^{-1}{g}^{-1}x)=r(gh)f(x)$, with $r(e)=\mathrm{id}$ and inverse $r(g^{-1})$. Finally $u(gx)=gu(x)$ is equivalent, by F2, to ${\delta }_X{u}^{\ast }=({\mathrm{id}}_H\otimes u^{\ast }){\delta }_Y$; conversion in step 2.1 gives the corresponding $c$ identity. AC is used only through F2, not through any selection of a vector-space basis.

Write $c(v)={\sum }_{i=1}^n{v}_i\otimes h_i$ with the $h_i$ linearly independent, eliminating redundancies from a finite tensor expression, and put $W_v=\mathrm{span}(v_1,\dots ,v_n)$. The counit gives $v={\sum }_i\epsilon (h_i)v_i\in W_v$. With $q:V\to V\mathrm{/}{W}_v$, coassociativity gives ${\sum }_{i}c(v_i)\otimes h_i={\sum }_i{v}_i\otimes \mathrm{\Delta }(h_i)$. Apply $q$ to the first factor: the right side is zero and independence of the $h_i$ gives $(q\otimes \mathrm{id})c(v_i)=0$. Thus $c(v_i)\in W_v\otimes H$, since the kernel of $q\otimes \mathrm{id}$ is $W_v\otimes H$ over a field. Indeed write any finite tensor expression with independent second-factor coefficients; its image under $q\otimes \mathrm{id}$ is zero exactly when each first-factor coefficient lies in $W_v$. This proves the kernel assertion without an infinite basis.

Sums of finitely many $W_v$ are finite-dimensional subcomodules, so they contain any prescribed finite subset; sums of two such subcomodules also show directedness. Evaluating coassociativity at $g,h$ gives $r(g)r(h)=r(gh)$, and evaluating the counit gives $r(e)=\mathrm{id}$, so $r(g^{-1})$ is the inverse of $r(g)$. Choose a finite basis of $W$ and write $c(w_j)={\sum }_i{w}_i\otimes a_{ij}$. The matrix entries $a_{ij}$ are regular functions on $G$, and its determinant is invertible at each point. Its inverse determinant is regular: the inverse matrix is the regular matrix $(S(a_{ij}))$, so determinants multiply to 1 as functions in $H$. Therefore $g\mapsto (a_{ij}(g))$ is a morphism to $GL(W)$, proving rationality. Only finite bases and finite elimination were used.

F1 makes $A=\mathbb{C}[X]$ a right comodule whose evaluated representation is exactly $f\mapsto f\circ g^{-1}$. Apply F2 to any finite subset of $A$ to obtain the required finite-dimensional stable subspace with algebraic $G$-action. By the definition in Classical complex affine algebraic actions and rational modules, this is rationality and local finiteness.

Since $c$ is an algebra map, evaluating its second factor at any $g$ gives $(g(fh))(x)=f(g^{-1}x)h(g^{-1}x)$ and $g1=1$. Thus each $g$ acts by an algebra automorphism, with inverse $g^{-1}$. The hypotheses of F1 and F2 cover reducible and empty $X$ and disconnected $G$, so none of the excluded extra hypotheses is needed. AC is inherited solely from F1's affine reconstruction route.

For a right comodule expand uniquely $c(v)={\sum }_m{p}_m(v)\otimes t^m$, with finitely many nonzero terms for each $v$. Coassociativity and independence of Laurent monomials give $c(p_m(v))=p_m(v)\otimes t^m$, hence $p_n{p}_m=0$ for $n\mathrm{\ne }m$ and $p_m^2=p_m$. The counit gives $v={\sum }_m{p}_m(v)$. Therefore $V={⨁}_m{V}_m$, where $V_m=p_m(V)$; uniqueness follows by applying $p_n$ to any finite relation between homogeneous vectors. Evaluation identifies $V_m$ with the character eigenspace. Conversely such a direct sum defines $c$ by this finite formula, which satisfies both comodule identities. Each vector spans, together with its finitely many components, a finite-dimensional stable algebraic subspace, so the resulting action is rational.

For completeness an arbitrary rational $T$-module has such a coaction. On a finite-dimensional stable algebraic subspace, expand the regular matrix coefficients of the action in Laurent monomials; the group and identity laws give the comodule identities. On two overlapping subspaces these coactions agree on the intersection because all evaluations agree, and Laurent polynomials are determined by their values on $T$. Thus they glue to $c$ on the union $V$. A linear map intertwines comodules precisely when it preserves each $V_m$, by coefficient comparison. Equivariant rational-module maps also intertwine comodules because their evaluations agree.

For a comodule algebra, $c(ab)=c(a)c(b)$ and $c(1)=1\otimes 1$ imply $A_m{A}_n\subseteq A_{m+n}$ and $1\in A_0$. Conversely these grading laws make the coaction in step 1.1 a unital algebra map. F1 reconstructs the action on the affine set with coordinate ring $A$. If the algebra is given abstractly, first use F2 to realize it; reducedness makes its presentation ideal radical. This is the only additional realization required, and is where AC is used. Finally evaluating the coordinate action gives $f(t^{-1}x)=t^{m}f(x)$, hence replacing $t$ by $t^{-1}$ gives $f(tx)=t^{-m}f(x)$.

Choose finitely many algebra generators of $A$ by F1 and put them in a finite-dimensional rational submodule $W$ by F2. Then $W$ generates $A$. Choose a finite basis $w_1,\dots ,w_N$ of $W$. Its action has regular matrix entries; the dual action has transpose-inverse matrix, whose entries are regular because inversion $g\mapsto g^{-1}$ is a morphism. Thus $W^{\ast }$ is a finite-dimensional rational module.

The coordinate map $\pi :\mathbb{C}[z_1,\dots ,z_N]\to A$, $z_i\mapsto w_i$, is surjective. Its kernel $I$ is radical since $A$ is reduced. F4 identifies $\mathbb{C}[V(I)]$ with $\mathbb{C}[z_1,\dots ,z_N]\mathrm{/}I$, and $\pi$ identifies that quotient with $A$. F3 gives mutually inverse morphisms $X\leftrightarrow V(I)$ from this algebra isomorphism. The morphism to $W^{\ast }\cong {\mathbb{A}}^N$ is precisely evaluation because its coordinates are $w_i(x)$; hence evaluation is an isomorphism onto the closed set $V(I)$. If $X$ is empty, $A=0$, and $I$ is the unit ideal so the conclusion still holds.

For $g\in G$, $x\in X$ and $w\in W$, $(g\mathrm{e}\mathrm{v}(x))(w)=\mathrm{e}\mathrm{v}(x)(g^{-1}w)=(g^{-1}w)(x)=w(gx)=\mathrm{e}\mathrm{v}(gx)(w)$ by the inverse-pullback action on functions. This proves equivariance and invariance of the image. The only choice beyond finite-dimensional selection is the AC inherited by F3 and F4 in step 2.1; local finiteness itself needs no infinite basis.