Groups of Multiplicative Type and Arithmetic Tori

The ring $A{\otimes }_{k}B$ represents pairs of maps from $A,B$ into any commutative $k$-algebra: the unique map is $a\otimes b\mapsto f(a)g(b)$. Therefore $\mathrm{Spec}(A\otimes B)$ is the product of the two affine schemes. Apply F1 to multiplication, identity, and inverse. Their group diagrams reverse to exactly the coassociativity, counit, and inverse identities in F2. Conversely these identities reverse to the group diagrams, so reconstruct a group object. Maps reverse in the same manner, and the two constructions are inverse.

A map $k[t,t^{-1}]\to A$ is determined by an invertible element $a$, the image of $t$. The multiplication and identity diagrams say $\mathrm{\Delta }(a)=a\otimes a$ and $ϵ(a)=1$. The inverse identity gives $S(a)a=1$, so every group-like element is invertible and also respects inverse. Thus characters are exactly group-like elements. Tensoring each structural map with an extension field preserves the identities and all formulas, proving base-change compatibility.

If $a={\sum }_m{a}_m{e}_m$ is group-like, coefficient comparison in $\mathrm{\Delta }(a)=a\otimes a$ gives $a_m{a}_n=0$ when $m\mathrm{\ne }n$ and $a_m=a_m^2$. Over a field at most one coefficient is nonzero; the counit makes exactly one coefficient nonzero, and it is $1$. Thus the group-like elements are precisely $e_m$. A Hopf map $k[N]\to k[M]$ must send $e_n$ to $e_{f(n)}$, with $f(n+n^{\mathrm{\prime }})=f(n)+f(n^{\mathrm{\prime }})$. Every such $f$ gives a Hopf map, proving both natural identifications.

Finite generators of $M$, together with their negatives, give finite algebra generators of $k[M]$. Conversely, take the finite union $T$ of the supports of finite algebra generators. Every product and sum has support in the submonoid generated by $T$, so all $e_m$ can occur only if that monoid is $M$; in particular $T$ generates $M$ as a group. Direct sums become tensor products of group algebras. The algebras for $\mathbb{Z}$ and $\mathbb{Z}\mathrm{/}n\mathbb{Z}$ are respectively $k[t,t^{-1}]$ and $k[t]\mathrm{/}(t^n-1)$, which proves the product formula.

The identity is a closed point of $G$: a $k$-rational point in a finite-type $k$-scheme is closed. To check the latter assertion affine-locally, its residue map has maximal kernel since its image is $k$; if another point were a specialization, every affine neighborhood of that specialization would contain the rational point and contradict maximality. The diagonal of $G$ is the inverse image of the identity under $(g,h)\mapsto g^{-1}h$, so it is closed, and $G$ is separated. More generally, for any quasi-compact separated $k$-scheme $X$, choose a finite affine cover $U_i$. Each intersection $U_i\cap U_j$ is affine, as a closed subscheme of $U_i{\times }_k{U}_j$ pulled back from the diagonal. Its global sections are the kernel of the difference-of-restrictions map from the finite product of the rings of the $U_i$ to the finite product of the rings of their intersections. Tensoring with a $k$-algebra preserves this kernel: exactness of vector-space tensor products is checked on the finitely many independent coefficients of a given tensor. F2 identifies the tensored rings with those of the base-changed cover. Consequently $\mathrm{\Gamma }(X,{\mathcal{O}}_X){\otimes }_{k}R=\mathrm{\Gamma }(X_R,{\mathcal{O}}_{X_R})$ for every $k$-algebra $R$.

Set $A=\mathrm{\Gamma }(G,{\mathcal{O}}_G)$. By step 1.1, $A\otimes K=\mathrm{\Gamma }(G_K,\mathcal{O})$, a finite-type $K$-algebra. Choose its finite algebra generators and write them as finite sums of coefficients times elements of $A$. Let $A^{\mathrm{\prime }}$ be generated over $k$ by those finitely many elements. Then $A^{\mathrm{\prime }}\otimes K\to A\otimes K$ is onto, so $(A\mathrm{/}{A}^{\mathrm{\prime }})\otimes K=0$. A nonzero vector remains nonzero after extension, hence $A^{\mathrm{\prime }}=A$ and $A$ is finite type. F3 gives the canonical map $f:G\to H=\mathrm{Spec}A$. By step 1.1 and F2, $f_K$ is the canonical affine global-sections isomorphism. Its inverse $g_K:H_K\to G_K$ has equal pullbacks to $K{\otimes }_{k}K$, since both are the inverse of the same pulled-back $f$.

Take a finite affine open cover $V_i$ of $G$. The opens $W_i=g_K^{-1}((V_i{)}_K)$ in $H_K$ have equal pullbacks under the two projections of $H_{K\otimes K}$. They are saturated for $H_K\to H$: any two points above the same point of $H$ can be compared using a point in the fibre product, since the tensor product of their residue fields over the residue field of that point is a nonzero ring and has a prime ideal by A1. Thus $W_i$ is the inverse image of a subset $U_i\subset H$, and F1 makes $U_i$ open. These opens cover $H$, and they are quasi-compact and separated because $H$ is a Noetherian affine scheme. For any vector space $R$, the equalizer of $R\otimes K\rightrightarrows R\otimes K\otimes K$ is $R$, where the arrows insert $1$ in the first or second field factor. Indeed a $k$-linear retraction $\lambda :K\to k$ with $\lambda (1)=1$, applied to one field factor of an equality, shows a fixed tensor equals $r\otimes 1$. The ring map of $g_K{\mathrm{\mid }}_{(U_i{)}_K}$ from $\mathrm{\Gamma }(V_i,\mathcal{O})$ into $\mathrm{\Gamma }(U_i,\mathcal{O})\otimes K$ has equal pullbacks by step 2.1 and the base-change formula in step 1.1. It therefore lands in $\mathrm{\Gamma }(U_i,\mathcal{O})$ by this equalizer computation, and F3 descends it to $g_i:U_i\to V_i$. On overlaps these maps agree: their pullbacks agree, the underlying point maps are equal by surjectivity of the field projection, and on preimages of affine target opens the ring maps are equal by the injectivity of extension of scalars. They glue to $g:H\to G$. The same uniqueness argument applied to $fg$ and $gf$ proves they are identities since they become identities over $K$. Hence $f$ is an isomorphism and $G$ is affine.

If $G$ is fpqc locally diagonalizable over $\mathrm{Spec}k$, there is a nonempty covering scheme $S^{\mathrm{\prime }}$ over which it is diagonalizable. Choose a point of $S^{\mathrm{\prime }}$ and take its residue field $K$; the diagonalizable isomorphism pulls back to $G_K\cong D_K(M)$. The preceding steps prove affineness. Conversely a field extension giving a diagonalizable group is an fpqc cover of $\mathrm{Spec}k$. Therefore the two multiplicative-type formulations give exactly the same full class. The same argument with $D(M)={\mathbf{G}}_m^r$ applies to fpqc forms of tori.

For $a\in A$, write $\mathrm{\Delta }(a)={\sum }_{i=1}^s{v}_i\otimes w_i$ with the $w_i$ linearly independent. Coassociativity, followed by coefficient functionals on the third tensor factor, shows $\mathrm{\Delta }(v_i)\in V\otimes A$, where $V$ is the span of the $v_i$. The counit gives $a\in V$. Adding the finitely many resulting spaces gives a finite-dimensional right coideal $V$ containing the prescribed subset. In a basis $v_1,\dots ,v_d$ write $\mathrm{\Delta }(v_j)={\sum }_i{v}_i\otimes c_{ij}$. Coassociativity and the counit give $\mathrm{\Delta }(c_{ij})={\sum }_{\mathrm{\ell }}{c}_{i\mathrm{\ell }}\otimes c_{\mathrm{\ell }j}$ and $ϵ(c_{ij})={\delta }_{ij}$. Their finite span $C$ is a subcoalgebra; applying $ϵ$ on the first factor gives $v_j={\sum }_iϵ(v_i)c_{ij}$, so $V\subset C$.

Inside $K[M]$, any finite-dimensional subcoalgebra $C_K$ is spanned by monomials. Indeed, if $\sum a_m{e}_m\in C_K$, apply the $e_m$ coefficient functional to the second tensor factor of its comultiplication; this produces $a_m{e}_m\in C_K$. All monomials appearing in a finite basis therefore belong to $C_K$ and span it. The dual basis consists of orthogonal idempotents, with sum the unit, because $\mathrm{\Delta }(e_m)=e_m\otimes e_m$ and $ϵ(e_m)=1$. Hence $(C_K{)}^{\ast }\cong K^d$ where $d=\dim C$, and finite dimension identifies this dual with $C^{\ast }{\otimes }_{k}K$.

The affineness and splitting data supplied by F1 give $A$ and the extension $K$, so the calculation below takes place in the coordinate algebra of $G$ over the splitting field. For a finite subcoalgebra $C\subset A$, the finite algebra $E=C^{\ast }$ is commutative (commutators vanish after the faithful extension to $K$) and satisfies $E\otimes K\cong K^d$ by F2. For each $a\in E$, let $f_a$ be its minimal polynomial over $k$. The independent powers preceding its degree stay independent under extension, so $f_a$ is also the minimal polynomial over $K$. In $K^d$, that polynomial is the product of the distinct linear factors associated to the coordinate values of $a$. Thus $f_a$ is separable over $k$. Choose a finite algebra generating set of $E$, for example a vector-space basis. Over $k_s$, each generator has a split squarefree minimal polynomial, whose Lagrange interpolation idempotents decompose the algebra into factors on which that generator is a scalar. Repeating with the finitely many generators decomposes $E\otimes k_s$ into factors generated only by scalars, hence copies of $k_s$. Dualizing shows that $C\otimes k_s$ is spanned by group-like elements. Every element of $A$ belongs to a finite subcoalgebra by F2, so $A\otimes k_s$ is spanned by its group-like elements.

Distinct group-like elements in any coalgebra are linearly independent. Otherwise take a shortest relation, write one as $g={\sum }_{i=1}^r{a}_i{g}_i$ with the $g_i$ independent and all $a_i\mathrm{\ne }0$, and compare $\mathrm{\Delta }(g)$ with $g\otimes g$. In the independent tensor family $g_i\otimes g_j$, the off-diagonal coefficients give $a_i{a}_j=0$ for $i\mathrm{\ne }j$, so $r=1$; the diagonal and counit then give $a_1=1$, contrary to distinctness. In a Hopf algebra the group-like elements form an abelian group under multiplication with inverse $S$. Therefore their spanning and independence identify $A\otimes k_s$ with $k_s[M]$ as a Hopf algebra. F3 implies that $M$ is finitely generated.

Choose finite generators $m_1,\dots ,m_r$ of $M$. Their corresponding group-like elements are finite sums of tensors, so all their coefficients lie in a finite separable extension of $k$. Enlarge it inside $k_s$ by adjoining all roots of the finitely many separable minimal polynomials of its generators. This gives a finite normal separable extension $L$, which is Galois by F4. The corresponding group-like elements and their inverses define a Hopf map $L[M]\to A\otimes L$ which becomes the isomorphism of step 2.1 after extension to $k_s$. A map of vector spaces is injective and surjective if it becomes so after field extension: kernels and cokernels tensor exactly, and a nonzero vector stays nonzero. Thus the map is already an isomorphism over $L$.

Every vector of $B$ lies in a finite-dimensional stable $L$-subspace: take the span of its finite $\mathrm{\Gamma }$-orbit. The same holds for any finite set. Apply F1 to these spaces to see $L\otimes B^{\mathrm{\Gamma }}\to B$ is surjective. Any tensor in its kernel involves finitely many invariant vectors, which lie in one finite-dimensional stable space; injectivity in F1 kills that tensor. Thus this map is an isomorphism even when $B$ is infinite-dimensional. The fixed algebra $A$ is closed under multiplication and contains $1$, so the isomorphism respects algebras. For canonical scalar extensions, each tensor involves a finite-dimensional $k$-space; F1 on that space gives $(L\otimes A_0{)}^{\mathrm{\Gamma }}=A_0$.

Tensoring the isomorphism gives $B{\otimes }_{L}B\cong L{\otimes }_k(A{\otimes }_{k}A)$ with canonical semilinear action. Step 1.1 gives $(B{\otimes }_{L}B{)}^{\mathrm{\Gamma }}=A\otimes A$. Equivariance of $\mathrm{\Delta },ϵ,S$ thus restricts them to $A$, with counit valued in $k$ by F3. Their identities hold because they hold after extension and tensoring with a field is faithful. These maps make $A$ a Hopf algebra. An equivariant map sends fixed elements to fixed elements and is determined by them after scalar extension, proving existence and uniqueness of descended maps. Conversely scalar extension of every Hopf map is equivariant.

Let $b_1,\dots ,b_t$ be algebra generators of $B$. Write each as a finite $L$-linear combination of elements of $A$ using step 1.1, and let $A^{\mathrm{\prime }}\subset A$ be the $k$-algebra generated by those finitely many elements. Then $L\otimes A^{\mathrm{\prime }}\to B=L\otimes A$ is onto; consequently $(A\mathrm{/}{A}^{\mathrm{\prime }})\otimes L=0$ as a vector-space quotient. Faithfulness gives $A^{\mathrm{\prime }}=A$. Thus $A$ is finitely generated, and F2 gives the descended finite-type affine group.

If $M$ is finitely generated and the action is continuous, intersect the open stabilizers of a finite generating list. That intersection fixes every element of $M$, so is the kernel of the action and is open and normal. A Krull-open subgroup contains $\mathrm{Gal}(k_s\mathrm{/}L)$ for a finite Galois $L\mathrm{/}k$: take the normal closure of the finite extension defining a basic open neighborhood. Thus the action factors through the finite group $\mathrm{Gal}(L\mathrm{/}k)$. Conversely an action factoring through this group is continuous. For $G$, F3 and F2 show $X^{\ast }(G)$ is finitely generated, and its basis characters under a splitting isomorphism over $L$ are fixed by $\mathrm{Gal}(k_s\mathrm{/}L)$, so its action is continuous.

Given $M$, choose $L$ as in step 1.1. The action on $L[M]$ respects multiplication and all Hopf maps. By F4, $A=L[M{]}^{\mathrm{Gal}(L\mathrm{/}k)}$ is a finite-type Hopf algebra with $L\otimes A\cong L[M]$. Hence $D(M)=\mathrm{Spec}A$ is of multiplicative type. Its character module identifies with $M$ by F2, and the identification is equivariant because $e_m$ transforms to $e_{\sigma m}$. The fixed algebra equals $(k_s[M]{)}^{{\mathrm{\Gamma }}_k}$: invariance under $\mathrm{Gal}(k_s\mathrm{/}L)$ means every coefficient lies in $L$, since this subgroup fixes every monomial; taking the remaining finite-group invariants gives $A$. Thus the construction is independent of the chosen $L$.

For $G$, the evaluation map $k_s[X^{\ast }(G)]\to \mathcal{O}(G)\otimes k_s$ sends $e_{\chi }$ to its character function and is an equivariant Hopf isomorphism by F2 and F3. The finitely many images of a Hopf generating set of $k_s[X^{\ast }(G)]$ and of their antipodes involve finitely many coefficients in $k_s$, hence lie over a common finite Galois extension. Restricting the evaluation isomorphism to ${\mathrm{\Gamma }}_k$-fixed elements identifies its source with $(k_s[X^{\ast }(G)]{)}^{{\mathrm{\Gamma }}_k}=\mathcal{O}(D(X^{\ast }(G)))$ and its target with $(\mathcal{O}(G)\otimes k_s{)}^{{\mathrm{\Gamma }}_k}=\mathcal{O}(G)$, the latter because a fixed element is defined over one finite Galois subextension, where F4's canonical fixed-algebra clause computes the invariants. Hence $\mathcal{O}(D(X^{\ast }(G)))\cong \mathcal{O}(G)$, and so $G\cong D(X^{\ast }(G))$. Given $G,H$, take a common finite Galois splitting extension. F2 identifies every group map there with a reversed character-module map; this identification respects the actions. F4 says precisely the equivariant maps descend uniquely, giving the displayed bijection. The evaluation identifications commute with maps because both send a monomial to the corresponding pulled-back character. They are therefore natural and prove the anti-equivalence.

A finitely generated abelian group admits a finite presentation: for a surjection ${\mathbb{Z}}^d\to M$, its kernel is finitely generated by induction on $d$, projecting to the last coordinate, whose image is cyclic; lift a generator of that image and use the induction hypothesis for the kernel of the projection. For its integer relation matrix, integer row and column operations give diagonal form. Specifically move a nonzero entry of smallest positive absolute value to the first position and divide entries in its row and column by it with remainder. A nonzero remainder lowers that pivot, so this process terminates. If the pivot does not divide an entry of the remaining block, add the row containing that entry to the pivot row and repeat the division; again the pivot decreases. Thus it eventually divides the whole block, its row and column can be cleared, and induction diagonalizes the smaller block. These invertible operations yield $M\cong {\mathbb{Z}}^r\oplus {⨁}_i\mathbb{Z}\mathrm{/}{n}_i\mathbb{Z}$, with $n_i>1$ after removing unit entries. Consequently $M$ is torsion-free exactly when it is free of finite rank.

If $M$ is free of rank $r$, F1 and F2 give $G_{k_s}\cong D_{k_s}(M)\cong {\mathbf{G}}_m^r$, so $G$ is a torus. Conversely, if $G_K\cong {\mathbf{G}}_m^r$ for some extension $K\mathrm{/}k$, choose a finite Galois splitting field $L\mathrm{/}k$ from F1. The finite-dimensional nonzero $K$-algebra $L{\otimes }_{k}K$ has a maximal ideal: select a proper ideal of largest vector-space dimension. Its residue field $E$ contains both $L$ and $K$, since the maps from these fields are unital. Base change of the splitting isomorphism $G_L\cong D_L(M)$ to $E$ identifies $G_E$ with $D_E(M)$, so F2 computes $X^{\ast }(G_E)=M$, while base change of $G_K\cong {\mathbf{G}}_m^r$ computes that same module as ${\mathbb{Z}}^r$. Hence $M\cong {\mathbb{Z}}^r$ and is torsion-free. The natural bijections of F1 restrict to these objects, proving the asserted equivalence and the exclusion of every nonzero torsion module.