Affine Group Schemes, Hopf Algebras, and Rational Representations

By [F1] the morphism $X\to \mathrm{Spec}k$ is quasi-compact and locally of finite type, so its affine open charts $\mathrm{Spec}B$ with $B$ a finitely generated $k$-algebra cover $X$; quasi-compactness extracts finitely many of them, giving $X={\bigcup }_{i=1}^n{U}_i$ with $U_i=\mathrm{Spec}{B}_i$ and $B_i$ finitely generated over $k$.

Each $U_i$ is affine, hence quasi-compact by [F4], and by [F2] every point of $U_i$ lies in some distinguished open $D(f)\subseteq U_i$ with $f\in A$; finitely many such sets cover $U_i$. Collecting the finitely many results over $i=1,\dots ,n$, there are $f_1,\dots ,f_m\in A$ with $X={\bigcup }_{j=1}^{m}D(f_j)$ and $D(f_j)\subseteq U_{i(j)}$ for a suitable index $i(j)$.

Fix $j$ and put $i=i(j)$, so that $D(f_j)\subseteq U_i=\mathrm{Spec}{B}_i$; let $b\in B_i$ be the restriction of the global section $f_j$ to $U_i$. A prime $\mathfrak{q}\in U_i$ has image $\mathfrak{p}=\mathfrak{q}\cap A$ under the open immersion, so $f_j\mathrm{\notin }\mathfrak{p}$ if and only if $b\mathrm{\notin }\mathfrak{q}$, and therefore $D(f_j)=D(b)$ as open subsets of $U_i$. The structure sheaf of $X$ restricted to $U_i$ is that of $U_i$, so by [F3] $A_{f_j}=\mathrm{\Gamma }(D(f_j),{\mathcal{O}}_X)=\mathrm{\Gamma }(D(b),{\mathcal{O}}_{U_i})=(B_i{)}_b$, a principal localization of a finitely generated $k$-algebra and hence again a finitely generated $k$-algebra.

For each $j$ choose finitely many $k$-algebra generators of $A_{f_j}$ and write them as fractions with numerators in $A$. By [F5] choose a unit relation $1={\sum }_j{\lambda }_j{f}_j$ with ${\lambda }_j\in A$. Let $B\subseteq A$ be the $k$-subalgebra generated by all $f_j$, all these numerators, and all ${\lambda }_j$. It is finitely generated, and the natural map $B_{f_j}\to A_{f_j}$ is surjective for each $j$, since its image contains the chosen algebra generators.

Fix $a\in A$. For every $j$, surjectivity in step 4.1 gives $a\mathrm{/}1=b_j\mathrm{/}{f}_j^{r_j}$ in $A_{f_j}$ for some $b_j\in B$; equality of localization fractions gives $f_j^{t_j}(f_j^{r_j}a-b_j)=0$ in $A$ for some $t_j\ge 0$. Thus $f_j^{r_j+t_j}a=f_j^{t_j}{b}_j\in B$. Choose $N\ge 1$ exceeding the finitely many exponents $r_j+t_j$; then $f_j^{N}a\in B$ for all $j$. Expanding $({\sum }_j{\lambda }_j{f}_j{)}^{mN}=1$ shows $1={\sum }_j{c}_j{f}_j^N$ with $c_j\in B$, because each monomial has some $f_j$-exponent at least $N$ and all ${\lambda }_j,f_j$ belong to $B$. Hence $a={\sum }_j{c}_j(f_j^{N}a)\in B$. This argument chooses $N$ separately for each $a$, and proves $A=B$. If $X$ is empty, the same unit-ideal conclusion gives $A=0$, which is finitely generated as well.

For the consequence, if $G$ is a group scheme of finite type over $k$ whose underlying scheme is affine, say $G=\mathrm{Spec}A$, then by definition its structure morphism $G\to \mathrm{Spec}k$ is of finite type, so the argument above makes its coordinate ring finitely generated over $k$. The AC-dependent inputs were affine-chart quasi-compactness [F4] and the unit-ideal relation [F5]; the subsequent selections were finite.

Replacing $A$ by $B\mathrm{/}I$ along the isomorphism of [F4], the morphism $\mathrm{Spec}\phi$ is the contraction map $\mathrm{Spec}(B\mathrm{/}I)\to \mathrm{Spec}B$ of [F2], which by [F3] is a homeomorphism onto the closed subset $V(I)$.

At a prime $\mathfrak{p}\supseteq I$ of $B$, the sections of the direct image $(\mathrm{Spec}\phi {)}_{\ast }{\mathcal{O}}_{\mathrm{Spec}A}$ over a basic open $D(b)\ni \mathfrak{p}$ are $A_{\phi (b)}=(B\mathrm{/}I{)}_{\phi (b)}$ by [F2], and the basic opens $D(b)\ni \mathfrak{p}$ are cofinal among the neighbourhoods of $\mathfrak{p}$; hence the stalk of the direct image at $\mathfrak{p}$ is $(B\mathrm{/}I{)}_{\mathfrak{p}}$, with stalk map $B_{\mathfrak{p}}\to (B\mathrm{/}I{)}_{\mathfrak{p}}$ induced by localizing $\phi$ at $B\setminus \mathfrak{p}$. Localization of the surjection $\phi$ at each multiplicative set is surjective by [F6], so this stalk map is surjective, and the identification of the source stalk is [F5].

At a prime $\mathfrak{p}⊉I$ of $B$, choose $u\in I\setminus \mathfrak{p}$; then $D(u)\ni \mathfrak{p}$ and the sections of the direct image over $D(u)$ are $(B\mathrm{/}I{)}_{\phi (u)}=(B\mathrm{/}I{)}_0=0$, because $\phi (u)=0$ becomes invertible in the localization. Every smaller basic open containing $\mathfrak{p}$ also lies in $D(u)$ and has zero sections, so the stalk of the direct image at $\mathfrak{p}$ is the zero ring and the stalk map is surjective trivially.

Steps 1.2 and 1.3 compute every stalk of the structure-sheaf map ${\mathcal{O}}_{\mathrm{Spec}B}\to (\mathrm{Spec}\phi {)}_{\ast }{\mathcal{O}}_{\mathrm{Spec}A}$ and show each is surjective, so by the stalk criterion [F7] the sheaf map is surjective. With the homeomorphism onto $V(I)$ from step 1.1, [F1] makes $\mathrm{Spec}\phi$ a closed immersion. Only the first isomorphism theorem, localizations of the given surjection and the stalk criterion were used, all applied to structures already determined by $\phi$; no choice principle is used.

Define a $k$-algebra homomorphism $\bar{\mathrm{\Delta }}\text{\hspace{0.17em}⁣}:B\to A{\otimes }_{k}A$ by $\bar{\mathrm{\Delta }}(x_{ij})={\sum }_l{x}_{il}\otimes x_{lj}$ (The tensor product $M{\otimes }_{R}N$ from the additive group underlying the free $\mathbb{Z}$-module on $M\times N$, elementary tensors, and finite tensor sums). With $X^{(1)}=(x_{ij}\otimes 1)$ and $X^{(2)}=(1\otimes x_{ij})$ one has $\bar{\mathrm{\Delta }}(X)=X^{(1)}{X}^{(2)}$, so [F1] gives $\bar{\mathrm{\Delta }}(d)=\det (X^{(1)})\det (X^{(2)})=(d\otimes 1)(1\otimes d)=d\otimes d$, a unit of $A{\otimes }_{k}A$; the identities $\det (X^{(1)})=d\otimes 1$ and $\det (X^{(2)})=1\otimes d$ are the Leibniz formula (For $n\ge 1$, the determinant over a commutative ring by the Leibniz formula, and $\mathrm{\mid }\det A\mathrm{\mid }$ for a real matrix) applied termwise to the ring homomorphisms $B\to A{\otimes }_{k}A$, $x_{ij}\mapsto x_{ij}\otimes 1$ and $x_{ij}\mapsto 1\otimes x_{ij}$ (Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$). By [F3] there is a unique $k$-algebra homomorphism $\mathrm{\Delta }\text{\hspace{0.17em}⁣}:A\to A{\otimes }_{k}A$ with $\mathrm{\Delta }\lambda =\bar{\mathrm{\Delta }}$.

Define $\bar{\epsilon }\text{\hspace{0.17em}⁣}:B\to k$ by $\bar{\epsilon }(x_{ij})={\delta }_{ij}$; then $\bar{\epsilon }(d)=\det (I_n)=1$ is a unit, so by [F3] there is a unique $k$-algebra homomorphism $\epsilon \text{\hspace{0.17em}⁣}:A\to k$ with $\epsilon \lambda =\bar{\epsilon }$.

Let $M=(m_{ij})$ be the matrix over $A$ with entries $m_{ij}=d^{-1}(\mathrm{adj}X{)}_{ij}$, and define the $k$-algebra homomorphism $\bar{S}\text{\hspace{0.17em}⁣}:B\to A$ by $\bar{S}(x_{ij})=m_{ij}$. Since $X\mathrm{adj}(X)=\mathrm{adj}(X)X=d{I}_n$ over $A$ by [F1], multiplying by $d^{-1}$ gives $XM=I_n=MX$; multiplicativity of the determinant gives $\bar{S}(d)\cdot d=\det (M)\det (X)=\det (MX)=\det (I_n)=1$, so $\bar{S}(d)=d^{-1}$ is a unit and [F3] yields a unique $k$-algebra homomorphism $S\text{\hspace{0.17em}⁣}:A\to A$ with $S\lambda =\bar{S}$.

Coassociativity holds on the generators: $(\mathrm{\Delta }\otimes \mathrm{id})\mathrm{\Delta }(x_{ij})={\sum }_{l,k}{x}_{ik}\otimes x_{kl}\otimes x_{lj}=(\mathrm{id}\otimes \mathrm{\Delta })\mathrm{\Delta }(x_{ij})$ by associativity of matrix multiplication in [F1]. Both sides are $k$-algebra homomorphisms $A\to A{\otimes }_{k}A{\otimes }_{k}A$ agreeing on all $x_{ij}$, hence on $d$ and on $d^{-1}$, so they agree on $A$ by [F3].

The counit identities hold on the generators: $(\epsilon \otimes \mathrm{id})\mathrm{\Delta }(x_{ij})={\sum }_l{\delta }_{il}{x}_{lj}=x_{ij}={\sum }_l{x}_{il}{\delta }_{lj}=(\mathrm{id}\otimes \epsilon )\mathrm{\Delta }(x_{ij})$, with the canonical identification $k{\otimes }_{k}A\cong A\cong A{\otimes }_{k}k$; agreement on generators and on $d^{-1}$ as in step 2.1 extends this to $A$.

The antipode identities hold on the generators: $m_A(S\otimes \mathrm{id})\mathrm{\Delta }(x_{ij})={\sum }_{l}S(x_{il})x_{lj}=(MX{)}_{ij}={\delta }_{ij}=\epsilon (x_{ij})$ and $m_A(\mathrm{id}\otimes S)\mathrm{\Delta }(x_{ij})=(XM{)}_{ij}={\delta }_{ij}$, so $m_A(S\otimes \mathrm{id})\mathrm{\Delta }=u_A\epsilon =m_A(\mathrm{id}\otimes S)\mathrm{\Delta }$ on generators, both sides being $k$-algebra homomorphisms $A\to A$; agreement on the generators extends the identity to $A$ as in step 2.1.

By [F2] the ring maps $\mathrm{\Delta },\epsilon ,S$ are comorphisms of morphisms $m\text{\hspace{0.17em}⁣}:{\mathrm{GL}}_n{\times }_k{\mathrm{GL}}_n\to {\mathrm{GL}}_n$, $e\text{\hspace{0.17em}⁣}:\mathrm{Spec}k\to {\mathrm{GL}}_n$ and $i\text{\hspace{0.17em}⁣}:{\mathrm{GL}}_n\to {\mathrm{GL}}_n$, where ${\mathrm{GL}}_n=\mathrm{Spec}A$: the identifications $\mathrm{Spec}(A{\otimes }_{k}A)\cong {\mathrm{GL}}_n{\times }_k{\mathrm{GL}}_n$ and $\mathrm{Spec}(A{\otimes }_{k}A{\otimes }_{k}A)\cong {\mathrm{GL}}_n{\times }_k{\mathrm{GL}}_n{\times }_k{\mathrm{GL}}_n$ hold. The comorphisms of $m\circ (m\times \mathrm{id})$ and $m\circ (\mathrm{id}\times m)$ are $(\mathrm{\Delta }\otimes \mathrm{id})\mathrm{\Delta }$ and $(\mathrm{id}\otimes \mathrm{\Delta })\mathrm{\Delta }$, equal by step 2.1, so the two composites agree since $\mathrm{Spec}$ is a contravariant equivalence; the identities $m\circ (e\times \mathrm{id})=\mathrm{id}=m\circ (\mathrm{id}\times e)$ and $m\circ (i,\mathrm{id})=e\circ p=m\circ (\mathrm{id},i)$ follow in the same way from steps 3.1 and 3.2. Thus ${\mathrm{GL}}_n$ is a $k$-group scheme, and it is of finite type because $A$ is the finitely generated $k$-algebra $k[x_{ij},d^{-1}]$ and [F4] applies.

For every commutative unital $k$-algebra $R$ there is a natural bijection between $k$-algebra homomorphisms $A\to R$ and $n\times n$ matrices $N$ over $R$ with unit determinant: a map restricts to $B\to R$ giving $N=(N_{ij})$ with $\det (N)=\phi (d)$ a unit, and conversely a matrix with unit determinant gives $B\to R$, $x_{ij}\mapsto N_{ij}$, which sends $d$ to a unit and factors uniquely through $A$ by [F3]; by [F1] the unit-determinant matrices are exactly the invertible ones. Under this bijection the group law induced by $m$ is matrix multiplication, $(\phi \psi )(x_{ij})={\sum }_l\phi (x_{il})\psi (x_{lj})$, the identity is $I_n$, and $i$ induces matrix inversion, so ${\mathrm{GL}}_n(R)$ is the group of invertible matrices over $R$.

If $V$ is a $k$-vector space with basis $e_1,\dots ,e_n$, then [F5] identifies ${\mathrm{Aut}}_R(V{\otimes }_{k}R)$ with the invertible $n\times n$ matrices over $R$ naturally in $R$, and step 5.1 identifies the latter with ${\mathrm{GL}}_n(R)$; hence the functor $R\mapsto {\mathrm{Aut}}_R(V{\otimes }_{k}R)$ is naturally identified with ${\mathrm{GL}}_n$.

For $n=1$ the constructions specialize: $d=x_{11}$, $\mathrm{adj}(X)=1$, so with $t=x_{11}$ one has $A=k[t,t^{-1}]$, $\mathrm{\Delta }(t)=t\otimes t$, $\epsilon (t)=1$ and $S(t)=t^{-1}$, and step 5.1 identifies the points with $R^{\times }$; this is ${\mathbf{G}}_m$. For $V=0$, the singleton functor $R\mapsto {\mathrm{Aut}}_R(0)$ is represented by $\mathrm{Spec}k$, whose identity, multiplication and inverse are the unique possible maps; it is of finite type since its one-point space is quasi-compact. This supplies ${\mathrm{GL}}_0$ without a determinant formula. The coordinate and point constructions are choice-free; [F4] uses AC for the finite-type assertion when $n\ge 1$.

(Coefficient criterion.) Let $X,H$ be $k$-vector spaces, let $h_1,\dots ,h_n\in H$ be linearly independent and let $x_1,\dots ,x_n\in X$ satisfy ${\sum }_i{x}_i\otimes h_i=0$ in $X{\otimes }_{k}H$. Then $x_1=\cdots =x_n=0$. Indeed, by [F3] the element ${\sum }_i{e}_{(x_i,h_i)}$ of the free module on $X\times H$ is a finite $k$-linear combination of finitely many bilinearity generators; let $X_0\subseteq X$ and $H_0\subseteq H$ be the spans of the initial $x_i,h_i$ together with all vectors occurring in that finite witness, so that $X_0,H_0$ are finite-dimensional and the same combination exhibits ${\sum }_i{x}_i\otimes h_i=0$ already in $X_0{\otimes }_k{H}_0$. By [F4] the independent list $h_1,\dots ,h_n$ extends to a finite basis $v_1,\dots ,v_N$ of $H_0$ with $v_i=h_i$ for $i\le n$, and $X_0$ has a finite basis $u_1,\dots ,u_M$; the universal property [F3] gives an isomorphism $X_0{\otimes }_k{H}_0\to k^{M\times N}$ with $u_a\otimes v_b\mapsto E_{ab}$ (both composites with the canonical maps are the identity on spanning sets). The image of ${\sum }_i{x}_i\otimes h_i$ is the matrix whose $i$-th column is the coordinate vector of $x_i$ for $i\le n$ and whose other columns vanish, so this matrix is zero by the assumed relation, and each $x_i$ is zero.

(Part (b).) Let $\mathfrak{a}\subseteq A$ be a Hopf ideal and let $q\text{\hspace{0.17em}⁣}:A\to A\mathrm{/}\mathfrak{a}$ be the quotient map. Since $\mathrm{\Delta }(\mathfrak{a})\subseteq \ker (A{\otimes }_{k}A\to A\mathrm{/}\mathfrak{a}{\otimes }_{k}A\mathrm{/}\mathfrak{a})$, there are unique $k$-algebra homomorphisms $\bar{\mathrm{\Delta }}\text{\hspace{0.17em}⁣}:A\mathrm{/}\mathfrak{a}\to A\mathrm{/}\mathfrak{a}{\otimes }_{k}A\mathrm{/}\mathfrak{a}$, $\bar{\epsilon }\text{\hspace{0.17em}⁣}:A\mathrm{/}\mathfrak{a}\to k$ and $\bar{S}\text{\hspace{0.17em}⁣}:A\mathrm{/}\mathfrak{a}\to A\mathrm{/}\mathfrak{a}$ with $\bar{\mathrm{\Delta }}q=(q\otimes q)\mathrm{\Delta }$, $\bar{\epsilon }q=\epsilon$ and $\bar{S}q=qS$, by the universal property of the quotient ring [F2] and because $q\otimes q$, $\epsilon$ and $qS$ kill $\mathfrak{a}$. The three Hopf identities for $(\bar{\mathrm{\Delta }},\bar{\epsilon },\bar{S})$ hold because they hold for $(\mathrm{\Delta },\epsilon ,S)$ and $q,q\otimes q,q\otimes q\otimes q$ are surjective; this makes $A\mathrm{/}\mathfrak{a}$ a Hopf algebra with $q$ a morphism, and any Hopf structure with that property must satisfy the three displayed identities, so it is unique. If $g\text{\hspace{0.17em}⁣}:A\to C$ is a Hopf morphism with $\mathfrak{a}\subseteq \ker g$, then $g$ induces $\bar{g}\text{\hspace{0.17em}⁣}:A\mathrm{/}\mathfrak{a}\to C$ with $\bar{g}q=g$ by [F2]; since $q$ is surjective and $g$ preserves $\mathrm{\Delta },\epsilon ,S$, so does $\bar{g}$, and $\bar{g}$ is the unique such map.

(Kernel of $f\otimes f$.) One has $\ker (f\otimes f)=A\otimes K+K\otimes A$. The inclusion $\supseteq$ is [F5]. For $\subseteq$, write an element of $A{\otimes }_{k}A$ as $x={\sum }_{i=1}^n{a}_i\otimes b_i$ and suppose $(f\otimes f)(x)=0$. Choose a maximal linearly independent subfamily $(f(a_j){)}_{j\in J}$ of the finite list $(f(a_1),\dots ,f(a_n))$; by [F4] every remaining $f(a_i)$ is a finite linear combination ${\sum }_{j\in J}{c}_{ij}f(a_j)$ with $c_{ij}\in k$. Then ${\sum }_{j\in J}f(a_j)\otimes (f(b_j)+{\sum }_{i\mathrm{\notin }J}{c}_{ij}f(b_i))=0$ in $B{\otimes }_{k}B$, so step 1.1 gives $w_j:=b_j+{\sum }_{i\mathrm{\notin }J}{c}_{ij}{b}_i\in K$ for every $j\in J$. Moreover $a_i^{\mathrm{\prime }}:=a_i-{\sum }_{j\in J}{c}_{ij}{a}_j\in K$ for $i\mathrm{\notin }J$, and the identity $x={\sum }_{j\in J}{a}_j\otimes w_j+{\sum }_{i\mathrm{\notin }J}{a}_i^{\mathrm{\prime }}\otimes b_i$ shows $x\in A\otimes K+K\otimes A$.

(Part (a).) If $x\in K$, then $(f\otimes f){\mathrm{\Delta }}_A(x)={\mathrm{\Delta }}_{B}f(x)=0$, so ${\mathrm{\Delta }}_A(x)\in \ker (f\otimes f)=A\otimes K+K\otimes A$ by step 2.1; further ${\epsilon }_A(x)={\epsilon }_B(f(x))=0$ and $f(S_A(x))=S_B(f(x))=0$, so $S_A(x)\in K$. Hence $K$ is a Hopf ideal of $A$. The same finite-relation argument identifies $U{\otimes }_{k}H$ with its image in $X{\otimes }_{k}H$ for every inclusion $U\subseteq X$: a zero relation has a finite witness; in the resulting finite-dimensional spaces extend a basis of the span of the first factors in $U$ to a basis of the ambient first-factor space, and use the coordinate tensor matrices of step 1.1. Applying this in both factors makes $f(A)\otimes f(A)\to B\otimes B$ injective. For $y=f(x)\in f(A)$ one has ${\mathrm{\Delta }}_B(y)=(f\otimes f){\mathrm{\Delta }}_A(x)\in f(A)\otimes f(A)$, ${\epsilon }_B(y)={\epsilon }_A(x)$ and $S_B(y)=f(S_A(x))\in f(A)$, so $f(A)$ is a Hopf subalgebra of $B$ with the induced structure maps.

(Part (c).) By step 3.1 the image $f(A)$ is a Hopf subalgebra of $B$ and $K$ is a Hopf ideal, so by step 1.2 the quotient $A\mathrm{/}K$ is a Hopf algebra; the map $\bar{\iota }\text{\hspace{0.17em}⁣}:A\mathrm{/}K\to f(A)$, $a+K\mapsto f(a)$, given by the first isomorphism theorem [F2], is a $k$-algebra isomorphism preserving the three structure maps, since $f$ does and $q$ is surjective. Composing this isomorphism with the inclusion $f(A)\hookrightarrow B$ factors $f$ as a surjection followed by an injection of Hopf algebras; any such factorization is unique because the quotient map is an epimorphism and the inclusion is a monomorphism, which also forces the middle isomorphism to be unique.