Lecture 5: Group schemes 1

Functor of points

Let $X$ be an object of $\cC$. For an object $Y$ of $\cC$, let $h_X(Y)=\Hom_{\cC}(Y, X)$. Then $h_X$ defines a contravariant functor from $\cC$ to the category of sets. Yoneda's lemma says that $X$ is determined from $h_X$, in a precise sense.

Suppose $G$ is a group object of $\cC$. One then verifies that $h_G(Y)$ inherits the structure of a group; for example, the multiplication map $h_G(Y) \times h_G(Y) \to h_G(Y)$ is induced from $m$. Furthermore, if $f \colon Y \to Y'$ is a morphism in $\cC$, then the induced map $f^* \colon h_G(Y') \to h_G(Y)$ is a group homomorphism. Conversely, if $G$ is an object of $\cC$ such that each $h_G(Y)$ is endowed with the structure of a group and each $f^*$ is a group homomorphism then $G$ naturally has the structure of a group object of $\cC$. In other words, giving $G$ a group structure is the same as lifting $h_G$ to a functor from $\cC$ to the category of groups. Said yet again, the Yoneda embedding is an equivalence between group objects in $\cC$ and group objects in the functor category $\operatorname{Fun}(\cC, \mathrm{Set})$ which are representable. This point of view allows one to define group objects even if $\cC$ doesn't have finite products.

The group object $G$ is commutative if and only if $h_G(Y)$ is commutative for all $Y$.

Kernels and cokernels

The category of group objects in $\cC$ has a zero object 1, namely the object $\pt$ endowed with its unique group structure. One therefore has a definition for kernel and cokernel of a map of group objects, namely fiber (co)product with the zero object (if it exists).

Let $f \colon G \to H$ be a homomorphism of group objects. Since $\ker(f)$ is defined by what maps to it look like, one has a good description of its functor of points: $(\ker{f})(T)$ is equal to the kernel of the map $f \colon G(T) \to  H(T)$. In contrast, $\coker(f)$ is defined by what maps out of it look like, and so its functor of points does not admit an easy description in general. In particular, it is not true that $(\coker{f})(T)$ is the cokernel of $f \colon G(T) \to H(T)$.

Group schemes

The connection with Hopf algebras

Fix a field $k$. The category of affine schemes over $k$ is anti-equivalent to the category of $k$-algebras. One therefore finds that an affine group scheme $G$ over $k$ correspond to a $k$-algebra $A$ equipped with all the structure of a group object, but with the arrows going in the opposite direction:

• The multiplication map $G \times G \to G$ turns into a comultiplication map $A \to A \otimes A$.

• The inversion map $G \to G$ turns into the antipode map $A \to A$.

• The identity section $\pt \to G$ turns into the counit $A \to k$.

One can furthermore translate the group axioms: for example, associativity of $G$ means the comultiplication on $A$ is coassociative.

A (commutative) $k$-algebra $A$ equipped with a comultiplication, counit, and antipode satisfying the necessary axioms is called a (commutative) Hopf algebra. It is revealing to think of a Hopf algebra not as an algebra with comultiplication, counit, and antipode, but as a vector space with multiplication, unit, comultiplication, counit, and antipode. In this way, the data becomes completely symmetric with respecting to flipping all the arrows.

Examples

In what follows, we write $T=\Spec(R)$ for a test scheme.

The additive group. Let $\bG_a=\Spec(k[t])$. We have $\Hom(T, \bG_a)=R$. Regarding $R$ as an additive group, this shows that $\bG_a$ naturally has the structure of a commutative group scheme. It is called the <em>additive group</em>. The comultiplication on $k[t]$ is given by $t \mapsto t \otimes 1 + 1 \otimes t$.

The multiplicative group. Let $\bG_m=\Spec(k[t,t^{-1}])$. We have $\Hom(T, \bG_m)=R^{\times}$, the group of units in $R$. Again, this shows that $\bG_m$ naturally has the structure of a commutative group scheme. The comultiplication is given by $t \mapsto t \otimes t$.

The constant group. Let $\Gamma_0$ be an ordinary group. Let $\Gamma$ be the disjoint union of $\Spec(k)$'s indexed by $\Gamma_0$. We have $\Hom(T, \Gamma)=\Hom(\pi_0(T), \Gamma_0)$, which is a group; therefore $\Gamma$ is a group scheme, which we call the <em>constant group scheme</em> on $\Gamma_0$. In fact, $\Gamma=\Spec(A)$, where $A$ is the ring of functions $\Gamma_0 \to k$. We can identify $A \otimes A$ with the ring of functions $\Gamma_0 \times \Gamma_0 \to k$, and then comultiplication takes a function $f$ to the function $(x, y) \mapsto f(xy)$. In the future, we do not notationally distinguish between $\Gamma$ and $\Gamma_0$.

Roots of unity. Let $\mu_n=\Spec(k[t]/(t^n-1))$. We have that $\Hom(T, \mu_n)$ equal to the set of elements $x \in R$ such that $x^n=1$. This is obviously a commutative group under multiplication, and so $\mu_n$ is a commutative group scheme. It is the kernel of the multiplication-by-$n$ map on $\bG_m$.

The group scheme $\alpha_p$. Assume $k$ has characteristic $p$. Let $\alpha_p=\Spec(k[t]/(t^p))$. The set $\Hom(T, \alpha_p)$ is identified with the set of elements $x \in R$ which satisfy $x^p=0$. Since $k$ has characteristic $p$, this is a group under addition. It follows that $\alpha_p$ is a commutative group scheme.

Quotients

We are chiefly interested in finite commutative group schemes over $k$. Note that finite schemes are always affine, so such group schemes are described by finite dimensional commutative and cocommutative Hopf algebras. Examples include the constant group scheme on a finite group, $\mu_n$, and $\alpha_p$. We define the order of such a group scheme $G$, denoted $\# G$, to be the dimension of its Hopf algebra.

We state without proof the following theorem, first proved by Grothendieck.

Proving part 1 is not difficult: it simply amounts to showing that kernels exist in the category of Hopf algebras, which can be checked explicitly. Parts 2 and 3 are more difficult.

The theorem shows that the category of finite commutative group schemes over $k$ is an abelian category.

The étale case

We now study the case where $G$ is étale over $k$. Recall that a finite dimensional $k$-algebra is étale if and only if it is a product of separable extensions of $k$; when $k$ has characteristic 0, this is equivalent to being reduced.

Let $A$ be an étale $k$-algebra and let $k^s$ be the separable closure of $k$. Then $A \otimes k^s$ is a finite product of copies of $k^s$ indexed by some set $I$. The Galois group $G_k$ naturally permutes the set $I$. We have thus defined a functor

We note that $\Phi(A)=X(k^s)$, where $X=\Spec(A)$.

Now let $I$ be a finite $G_k$-set. Let $\ol{A}=\prod_{i \in I} k^s$. Then $G_k$ naturally acts on $\ol{A}$, through its action on both $I$ and $k^s$. Let $A$ be the invariant subalgebra. One easily sees that $A$ is a finite dimensional algebra and étale over $k$. We thus have a functor

We have the following basic result:

We thus see that the study of étale group schemes is equivalent to the study of Galois representations.

The connected-étale sequence

Let $G=\Spec(A)$ be a finite commutative group scheme over $k$. Write $A=\prod A_i$ with each $A_i$ local. There is a unique index, denoted 0, such that the counit of $A$ factors through $A_0$. Let $G^{\circ}=\Spec(A_0)$, a connected closed subscheme of $G$. Since $G^{\circ}$ has a $k$-point, it is geometrically connected, and so $G^{\circ} \times G^{\circ}$ is still connected; it follows that multiplication maps $G^{\circ} \times G^{\circ}$ into $G^{\circ}$, from which one easily sees that $G^{\circ}$ is a subgroup of $G$. We call $G^{\circ}$ the identity component of $G$.

Let $A_{\et}$ be the maximal étale subalgebra of $A$ fix!. Concretely, $A_{\et}=\prod (A_i)_{\et}$, where $(A_i)_{\et}$ is the separable closure of $k$ in $A_i$. Put $G^{\et}=\Spec(A^{\et})$. Formation of $A^{\et}$ respects tensor products, and so if $G$ is a group scheme then so is $G^{\et}$, and the natural map $G \to G^{\et}$ is a homomorphism. The universal property of $A^{\et}$ implies the following: a map from $G$ to an étale group scheme factors uniquely through $G^{\et}$. Note that the natural map $G(\ol{k}) \to G^{\et}(\ol{k})$ is an isomorphism.

The tensor product $A \otimes_{A_{\et}} k$ (where the map $A_{\et} \to k$ is the counit) is the universal quotient of $A$ in which the idempotent defining $A_0$ is identified with 1, and is therefore equal to $A_0$. In other words, $G^{\circ}$ is the fiber product of $G$ with the trivial group over $G^{\et}$. We have thus proved the sequence

is exact. This sequence is called the connected–étale sequence.

Suppose now that $k$ is perfect, i.e., every finite extension of $k$ is separable. Then the separable closure of $k$ in $A_i$ coincides with the algebraic closure, and maps isomorphically onto the residue field of $A_i$. It follows that the map $G^{\red} \to G^{\et}$ is an isomorphism of schemes. Since $k$ is perfect, a product of reduced schemes is still reduced, and so $G^{\red}$ is a closed subgroup of $G$. We have thus shown that, in this case, the connected–étale sequence splits. Furthermore, since there are no non-zero maps from an étale group scheme to a connected group scheme, the splitting is canonical. In other words, we have a canonical decomposition $G=G^{\circ} \times G^{\et}$.

Order invertible implies étale

Let $G=\Spec(A)$ be a finite connected commutative group scheme, so $A$ is a local ring. Let $I \subset A$ be the kernel of the counit map. Then $A=k \oplus I$, where $k$ is the span of the unit. We let $\pi \colon A \to I/I^2$ be the projection map, which is easily seen to be a derivation. Let $x_1, \ldots, x_n$ be elements of $I$ mapping to a basis for $I/I^2$. Define $D_i \colon A \to A$ to be the composition

where the first map is comultiplication, the second is $\id \otimes \pi$, and the third is induced from the map $I/I^2 \to k$ taking $x_i$ to 1 and $x_j$ to 0 for $i \ne j$. This is easily seen to be a derivation.