Exposé XVII. Unipotent algebraic groups. Extensions between unipotent groups and groups of multiplicative type

by M. Raynaud¹

0. Some notations

In the present chapter, we shall mostly consider algebraic groups defined over a field $k$. The number $p⩾0$ will always denote the characteristic of $k$, $F_p$ the prime field with $p$ elements if $p>0$, $\bar{k}$ an algebraically closed extension of $k$, $q$ a prime number distinct from $p$.

For every $S$-prescheme, $(G_a{)}_S$ (resp. $(G_m{)}_S$) denotes the additive group (resp. the multiplicative group) over $S$ (cf. Exp. I, 4.3). For every integer $n>0$, $({\mu }_n{)}_S$ (resp. $(\mathbb{Z}/n\mathbb{Z}{)}_S$) denotes the group of $n$-th roots of unity (Exp. I, 4.4.4) (resp. the constant group over $S$ associated with the abstract group $\mathbb{Z}/n\mathbb{Z}$ (Exp. I, 4.1)). The group $(\mathbb{Z}/n\mathbb{Z}{)}_S$ is finite and étale over $S$; the group $({\mu }_n{)}_S$ is flat and finite over $S$, and is étale over $S$ if and only if $n$ is invertible on $S$ (Exp. VIII, 2.1).

If $S$ is a prescheme of characteristic $p>0$, for every integer $n>0$ and every $S$-prescheme in groups $G$, we denote by $F_n(G)$ the radicial sub-$S$-prescheme in groups of $G$ equal to the kernel of the $n$-th iterate of the Frobenius morphism relative to $G$ (Exp. VII_A). In particular, if $G=(G_a{)}_S$ we set $F_n(G)=({\alpha }_{p^n}{)}_S$, which is a radicial $S$-group, flat and finite over $S$, representing the following functor: for every $S$-prescheme $S^{\prime }$, $({\alpha }_{p^n}{)}_S(S^{\prime })$ is the set of $x^{\prime }\in \Gamma (S^{\prime },O_{S^{\prime }})$ such that $x^{\prime p^n}=0$.

The group $({\mu }_{p^n}{)}_S$, already defined, is canonically isomorphic to $F_n(G_m{)}_S$.

When there is no ambiguity about the base scheme $S$, we shall simply write $G_a$, $G_m$, ${\alpha }_{p^n}$, etc. instead of $(G_a{)}_S$, $(G_m{)}_S$, $({\alpha }_{p^n}{)}_S$, etc.

If $G$ is an $S$-prescheme in commutative groups, for every integer $n>0$, ${}_{n}G$ is the sub-prescheme in groups of $G$ equal to the kernel of raising to the $n$-th power in $G$.

For the convenience of the reader, we have gathered in an appendix some properties of algebraic groups proved in Exp. VI and VII, as well as elementary properties of Hochschild cohomology, which will be useful in this chapter.

1. Definition of unipotent algebraic groups

Definition 1.1. An algebraic group $G$ defined over an algebraically closed field $k$ is said to be unipotent if $G$ admits a composition series whose successive quotients are isomorphic to algebraic subgroups of $(G_a{)}_k$.

Proposition 1.2. Let $k$ be an algebraically closed field, $K$ an algebraically closed extension of $k$, and $G$ an algebraic group defined over $k$. Then, in order that $G$ be unipotent, it is necessary and sufficient that G_K be unipotent.

The necessity of the condition is clear, since a composition series gives a composition series by extension of the base.

The proof of sufficiency is standard: $K$ is the direct limit of its finitely generated $k$-subalgebras. By Exp. VI_B § 10, one can find a finitely generated $k$-subalgebra $A$ of $K$, a composition series $G_i$ of G_S ($S=\mathrm{Spec}A$), and immersions $u_i:H_i=G_i/G_{i+1}\to (G_a{)}_S$. To prove that $G$ is unipotent, it then suffices to make a $k$-base extension $A\to k$, which is possible because ${\mathrm{Hom}}_{k-alg}(A,k)$ is non-empty, since $A$ is a non-zero $k$-algebra of finite type over the algebraically closed field $k$.

Definition 1.3. Let $G$ be an algebraic group defined over a field $k$. We shall say that $G$ is unipotent if there exists an algebraically closed extension $\bar{k}$ of $k$ such that $G_{\bar{k}}$ is unipotent (Definition 1.1).

By 1.2, the property is independent of the chosen algebraically closed extension $\bar{k}$.

Definition 1.4. Let $G$ be an algebraic group defined over a field $k$ and $H$ an algebraic group defined over an extension $k^{\prime }$ of $k$. We shall say that $H$ is a form of $G$ over $k^{\prime }$ if the algebraic groups $G_{{\bar{k}}^{\prime }}$ and $H_{{\bar{k}}^{\prime }}$ become isomorphic over ${\bar{k}}^{\prime }$ (as above, one sees that the property does not depend on the choice of the algebraically closed extension ${\bar{k}}^{\prime }$ of $k^{\prime }$). We shall also say that $H$ is a twisted form of $G$.

We are now in a position to describe the algebraic subgroups of $G_a$.

Proposition 1.5. Let $k$ be a field of characteristic $p⩾0$. Then an algebraic subgroup $H$ of $(G_a{)}_k$ is of one of the following types:

(i) $H=0$.

(ii) $H=G_a$.

(iii) (If $p>0$) $H$ is an extension of a twisted constant group $(\mathbb{Z}/p\mathbb{Z}{)}^r$ by a radicial group ${\alpha }_{p^n}$ ($r$ and $n$ integers $⩾0$, $r+n>0$). If moreover $k$ is perfect, this extension is necessarily trivial.

Proof. If $H$ is of dimension 1, it is clear that $H=G_a$. Otherwise $H$ is of dimension 0 and consequently is an extension² of an étale group H'' by its identity component $H^{\prime }$, which is a radicial group. To describe $H_{\bar{k}}^{\prime \prime }$, it suffices to know the abstract group $H^{\prime \prime }(\bar{k})$, isomorphic to $H(\bar{k})$.

Now this latter is a finite subgroup of $G_a(\bar{k})$, hence is zero if $p=0$, and of the form $(\mathbb{Z}/p\mathbb{Z}{)}^r$ otherwise, since it is annihilated by $p$. The group $H^{\prime }$ is closed in $G_a$ and is defined by a single equation (the ring k[T] of $G_a$ is principal), which over $\bar{k}$ admits 0 as its only root, so this equation is of the form $T^n=0$. Compatibility with the group law entails that $(T+T^{\prime }{)}^n$ belongs to the ideal generated by $T^n$ and $T^{\prime n}$ in the ring k[T, T'], hence: if $p=0$, one has $n=1$ and $H^{\prime }$ is the trivial group; if $p>0$, one has $n=p^m$ and $H^{\prime }={\alpha }_{p^m}$.

The last assertion of 1.5 follows, more generally, from the lemma:

Lemma 1.6. If $k$ is a perfect field, every extension $H$ of an étale algebraic group H'' by a radicial group $H^{\prime }$ is trivial. Moreover, there is a unique lift of H'' in $H$, namely $H_{red}$.

Indeed, since $k$ is perfect, the reduced $k$-scheme $H_{red}$ is an algebraic subgroup of $H$ (Exp. VI_A, 0.2) which is geometrically reduced, hence smooth over $k$ (Exp. VI_A, 1.3.1), hence étale, since $H$ is of dimension 0. To see that the canonical projection $H_{red}\to H^{\prime \prime }$ is an isomorphism, it suffices to verify it after base extension $k\to \bar{k}$, in which case it suffices to show that one has an isomorphism on the $\bar{k}$-valued points, which is clear. The last assertion follows from the fact that every lift of H'' in $H$, being étale over $k$, is reduced, hence is necessarily contained in $H_{red}$.

Note that ${\alpha }_{p^n}$ is a multiple extension of groups isomorphic to ${\alpha }_p$. From 1.5 we then deduce the corollary:

Corollary 1.7. In order that an algebraic group $G$ defined over an algebraically closed field $k$ be unipotent, it is necessary and sufficient that it possess a composition series whose successive quotients are isomorphic to $G_a$ if $p=0$, and to one of the groups $G_a$, $\mathbb{Z}/p\mathbb{Z}$, ${\alpha }_p$ if $p>0$. (We shall call these the elementary unipotent groups*.)*

2. First properties of unipotent groups

Proposition 2.1. A unipotent algebraic group defined over a field $k$ is affine over $k$.

By fpqc descent of affine morphisms, it suffices to prove 2.1 when $k$ is algebraically closed. In this case, $G$ is by definition a multiple extension of affine algebraic groups, hence is affine, by Exp. VI_B, 9.2 (viii) applied to affine morphisms.³

Proposition 2.2.

i) The property of being unipotent for an algebraic group is invariant under base field extension.

ii) Every algebraic subgroup of a unipotent group is unipotent.

iii) Every algebraic quotient group of a unipotent group is unipotent.

iv) Every extension of a unipotent algebraic group by a unipotent algebraic group is likewise unipotent.

Proof. i) follows immediately from 1.3 and 1.2. To establish the other properties, we may suppose the field $k$ algebraically closed. Then iv) is evident from Definition 1.3.

Let then $G$ be a unipotent algebraic group, $G^{\prime }$ an algebraic subgroup of $G$, G'' an algebraic quotient group of $G$, and $G_i$ ($i=1,\cdots ,n$) a composition series of $G$ such that $H_i=G_i/G_{i+1}$ is an elementary unipotent group (1.7).

To prove ii), consider the composition series of $G^{\prime }$ induced by that of $G$: $G_i^{\prime }=G_i\cap G^{\prime }$. The group $G_i^{\prime }/G_{i+1}^{\prime }$ is identified with an algebraic subgroup of $H_i$, hence is isomorphic to a subgroup of $G_a$, and consequently $G^{\prime }$ is unipotent.

To prove iii), consider the composition series of G'' image of that of $G$: $G_i^{\prime \prime }=$ image of $G_i$ in G''. The group $G_i^{\prime \prime }/G_{i+1}^{\prime \prime }$ is then a quotient of $H_i$, and it suffices to prove the lemma:

Lemma 2.3. If $H$ is an elementary unipotent group (1.7) defined over a field $k$, every algebraic quotient group H'' of $H$ is zero or is isomorphic to $H$ over $k$.

Proof. If $H=G_a$ in characteristic 0, or if $H={\alpha }_p$ or $\mathbb{Z}/p\mathbb{Z}$ ($p>0$), it suffices to note that it follows from 1.5 that $H$ has no algebraic subgroups other than 0 and $H$ (observe that a non-zero algebraic subgroup of ${\alpha }_p$ is defined by a $k$-algebra of $k$-rank at least $p$ (1.5 iii)), hence is equal to ${\alpha }_p$).

Now let $H=G_a$ ($p>0$), so that one has an exact sequence:

$$0⟶N⟶G_a⟶H^{\prime \prime }⟶0.$$

If $N=G_a$, then $H^{\prime \prime }=0$. Otherwise, proceeding by induction on the length of a composition series of $N$, one may assume that $N={\alpha }_p$, or that $N$ is a form of $(\mathbb{Z}/p\mathbb{Z}{)}^r$ (1.5).

a) If $N\simeq {\alpha }_p$, the proof of 1.5 iii) shows that $N$ is necessarily the kernel of the Frobenius morphism $F$ in $G_a$, and one concludes using the exact sequence:

0 → α_p → G_a ─F→ G_a → 0.

b) If $N\simeq \mathbb{Z}/p\mathbb{Z}$, it is immediate that there exists $a\in k\ast$ such that $N$ is the closed subscheme of $G_a=\mathrm{Spec}k[X]$ defined by the equation $X^p-aX=0$. One then concludes using the exact sequence:

0 → N → G_a ─P→ G_a → 0,

where $P$ is the Artin–Schreier morphism $x\mapsto x^p-ax$.

c) If $N$ is a form of $(\mathbb{Z}/p\mathbb{Z}{)}^r$, there exists a finite Galois extension $k^{\prime }$ of $k$ that trivializes $N$. By b) and an evident induction on $r$, $G_a/N$ is a form of $G_a$ trivialized by $k^{\prime }$. It then suffices to apply the following lemma:

Lemma 2.3 bis. Let $k$ be a field, $G$ a $k$-algebraic group which is a form of $G_a$, trivialized by a finite separable extension $k^{\prime }$ of $k$. Then $G$ is isomorphic to $G_a$.

Indeed, the group of $k^{\prime }$-automorphisms of the algebraic group $(G_a{)}_{k^{\prime }}$ is the group of non-zero homotheties $(k^{\prime }{)}^{\times }$ (Bible, Exp. 9, Lemma 1), and the Galois cohomology group $H^1(k^{\prime }/k,G_m)$ is zero (one may suppose that $k^{\prime }$ is a Galois extension of $k$), by Hilbert's Theorem 90. Lemma 2.3 bis then follows from the classification of $k$-forms of an algebraic group (cf. J.-P. Serre, Cohomologie galoisienne, Chap. III, 1.3).

This completes the proof of 2.2.

Proposition 2.4. Let $k$ be a field, $M$ a $k$-group of multiplicative type (Exp. VIII) and of finite type, $U$ a unipotent $k$-algebraic group. Then:

i) ${\mathrm{Hom}}_{k-gr}(M,U)=e$ (a fortiori ${\mathrm{Hom}}_{k-gr}(M,U)=e$).

ii) ${\mathrm{Hom}}_{k-gr}(U,M)=e$.

To prove i) we must establish that for every prescheme $S$ over $k$, ${\mathrm{Hom}}_{S-gr}(M_S,U_S)=e$. But this follows from the following lemma:

Lemma 2.5. Let $S$ be a prescheme, $M$ an $S$-group of multiplicative type and of finite type over $S$, $U$ an $S$-prescheme in groups of finite presentation over $S$ with unipotent fibers. Then ${\mathrm{Hom}}_{S-gr}(M,U)=e$.

Indeed, under the hypotheses made, in order that an $S$-morphism of groups $u:M\to U$ be the unit homomorphism, it suffices that the restriction of $u$ to the fibers of $M$ over the points of $S$ be the zero morphism (Exp. IX, 5.2). We are thus reduced to the case where $S$ is the spectrum of a field, which we may further assume to be algebraically closed. In view of Definition 1.1, we may restrict to $U=G_a$, in which case the property has already been proved (Exp. XII, 4.4.1).

Let us now prove 2.4 ii). Let $u:U\to M$ be a $k$-morphism of groups. The image $u(U)$ is representable by an algebraic subgroup U'' of $M$ (Exp. VI_B, 5.4).⁴ The group U'' is unipotent as a quotient of a unipotent group (2.2 iii)) and is of multiplicative type as a subgroup of a group of multiplicative type

(cf. Bible, Exp. 4, Th. 2 cor. 1, or Exp. IX, 6.8), so U'' is the trivial group by 2.4 i).

Remark 2.6. Keeping the notations of 2.4, it is no longer true in general that the functor ${\mathrm{Hom}}_{k-gr}(U,M)$ is equal to $e$. Thus, take a prescheme $S$ such that $\Gamma (S,O_S)$ contains a non-zero element $ϵ$ such that ${ϵ}^2=0$ (for example the spectrum of the algebra of dual numbers of a ring $A$). For every $S^{\prime }$ over $S$, the map $u\mapsto 1+{ϵ}_{S^{\prime }}u$ defines a homomorphism, functorial in $S^{\prime }$, from the additive group $\Gamma (S^{\prime },O_{S^{\prime }})$ to the multiplicative group $\Gamma (S^{\prime },O_{S^{\prime }}^{\times })$, hence defines an $S$-morphism of groups $(G_a{)}_S\to (G_m{)}_S$, and since $ϵ\ne 0$, this morphism is not zero.

Let us recall (Exp. VII_A § 3) that when $G$ is an $S$-prescheme in commutative groups, finite and flat over $S$, one has Cartier duality at one's disposal, and $G$ is reflexive in the sense of Exp. VIII § 1. More precisely, the functor $G\mapsto {\mathrm{Hom}}_{S-gr}(G,G_m)$ is representable

by an $S$-prescheme in commutative groups $D(G)$, finite and flat over $S$, and the canonical morphism $G\to D(D(G))$ is an isomorphism. In particular, one obtains:

(i) $D({\mu }_n)\simeq \mathbb{Z}/n\mathbb{Z}$ and consequently $D(\mathbb{Z}/n\mathbb{Z})\simeq {\mu }_n$.

(ii) If $S$ is of characteristic $p>0$, $D({\alpha }_p)\simeq {\alpha }_p$.

3. Unipotent groups acting on a vector space

Let us recall (Exp. I, 4.6.1) that if $S$ is a prescheme and $M$ a sheaf of O_S-modules, one denotes by $W(M)$ the $S$-functor $(Sch/S{)}^{\circ }\to (Ens)$ defined by the condition $W(M)(S^{\prime })=\Gamma (S^{\prime },M\otimes O_{S^{\prime }})$ for every $S$-prescheme $S^{\prime }$.

Moreover, let us recall that if an $S$-group acts on an $S$-functor $V$, one defines the $S$-functor $V^G$ of invariants of $V$ under $G$ as the subfunctor of $V$ whose set of points with values in some $S^{\prime }$ over $S$ is the set of $x\in V(S^{\prime })$ such that $x_{S^{\prime \prime }}$ is fixed under $G(S^{\prime \prime })$ for every S'' over $S^{\prime }$.

This being so, one has the following lemma:

Lemma 3.1. Let $S$ be a prescheme, $G$ an $S$-prescheme in groups, affine over $S$, defined by the quasi-coherent O_S-algebra $A$. Suppose that $G$ acts on a quasi-coherent sheaf of O_S-modules $M$, and let $\mu :M\to A{\otimes }_{O_S}M$ be the comorphism defining the action of $G$ on $M$ (Exp. I, 4.7.2), and $\nu :M\to A{\otimes }_{O_S}M$ the morphism $x\mapsto \mu (x)-1\otimes x$. Then:

i) $M^G(S)=\Gamma Ker(\nu )$.

ii) If $S$ is the spectrum of a field $k$, $M^G$ is of the form $W(N)$, where $N$ is a vector subspace of $M$.

iii) If $S$ is the spectrum of a field $k$, every element $x$ of $M$ is contained in a finite-dimensional $k$-vector subspace of $M$, stable under the action of $G$.

Proof. iii) is included for the record and has already been proved in Exp. VI_B, 11.2.

i) It is clear that $M^G(S)$ contains $\Gamma Ker(\nu )$. To establish the converse, one may suppose $S$ affine with ring $B$. Let $m\in M^G(S)$. Then for every $B$-algebra $B^{\prime }$ and every $u$ in ${\mathrm{Hom}}_{B-alg}(A,B^{\prime })$ ($u$ corresponds to an element of $G(\mathrm{Spec}{B}^{\prime })$), one has:

(u ⊗ 1_M) ν(m) = 0   in   B' ⊗_B M.

Taking in particular $B^{\prime }=A$ and $u$ the identity of $A$, one finds $\nu (m)=0$.

ii) Let $N$ be the kernel of $\nu$, equal to $M^G(k)$ by i). Every $k$-prescheme $S$ is flat over $k$, so one has:

M^G(S) = Γ Ker(ν ⊗_k S) = Γ(N ⊗_k S).

So $M^G$ is isomorphic to the functor $W(N)$.

Proposition 3.2. Let $G$ be a unipotent algebraic group defined over a field $k$, acting on a $k$-vector space $V$. Then if $V\ne 0$, one has $V^G\ne 0$.

In view of 3.1 ii) and iii), one may suppose $k$ algebraically closed and $V$ of finite dimension over $k$.

Let $0\to G^{\prime }\to G\to G^{\prime \prime }\to 0$ be an exact sequence of algebraic groups and suppose that $G$ acts on a sheaf $V$ (for the fpqc topology). Of course $V^G\subset V^{G^{\prime }}$,

and the quotient presheaf $G/G^{\prime }$ acts naturally on $V^{G^{\prime }}$. But $V^{G^{\prime }}$ is a sheaf (as kernel of the well-known pair of morphisms $V\Rightarrow \mathrm{Hom}(G^{\prime },V)$); consequently, the sheaf associated with $G/G^{\prime }$, that is, G'', acts on $V^{G^{\prime }}$, and it is immediate to verify that $(V^{G^{\prime }}{)}^{G^{\prime \prime }}=V^G=(V^{G^{\prime }}{)}^{G/G^{\prime }}$.

This remark allows us to reduce, in order to prove 3.2, to the case where $G$ is an elementary unipotent group (1.7).

a) $G=G_a$, $p=0$. It follows from (Bible 4 prop. 4) that a morphism from $G_a$ into the linear group $GL(V)$ is given by an exponential map:

T ↦ Σ_{q=0}^∞ T^q n^q / q!

where $n$ is a nilpotent endomorphism of $V$. But then $V\ne 0\Rightarrow Kern\ne 0$, and it is clear that every vector of $V$ annihilated by $n$ is left fixed by $G$.

Suppose now $p>0$.

b) $G={\alpha }_p$. The group ${\alpha }_p$ being a radicial group of height 1 (Exp. VII_A § 7), giving a representation of ${\alpha }_p$ in $V$ amounts to giving a representation of the $p$-Lie algebra ${\alpha }_p$ in $gl(V)$ (App. II 2.2), that is, here, to giving an element $X$ of $\mathrm{End}(V)$ such that $X^p=0$ (App. II 2.1). But then $V\ne 0\Rightarrow W=Ker(X)\ne 0$, and still by (App. II 2.2), one has $W=V^{{\alpha }_p}$.

c) $G=\mathbb{Z}/p\mathbb{Z}$. A representation of $G$ in $V$ is equivalent to giving an element $x$ of $\mathrm{Aut}(V)$ such that $x^p=1$, i.e. $(1-x{)}^p=0$, so $x$ is of the form $1+n$ with $n$ nilpotent, and $W=Kern$ is left fixed by $x$.

d) $G=G_a$. Let $G_i$, $i\in I$, be the increasing filtered family of étale algebraic subgroups of $G_a$, hence isomorphic to $(\mathbb{Z}/p\mathbb{Z}{)}^{r_i}$ (prop. 1.5), and let $V_i=V^{G_i}$. Since $V$ is of finite, non-zero dimension, and $V_i$ is non-zero by c), the decreasing filtered family of the $V_i$ is stationary, and $W={\bigcap }_{i\in I}{V}_i\ne 0$. Now one has the lemma:

Lemma 3.3. The family of étale subgroups of $(G_a{)}_S$ ($S$ an $S$-prescheme of characteristic $p>0$) is schematically dense in $G$ (Exp. IX, 4.1).

By Exp. IX, 4.4, it suffices to prove the lemma when $S$ is the spectrum of the prime field $F_p$. In this case, it suffices to consider the family of étale subgroups $G_n$ ($n⩾1$) with equation $X^{p^n}-X=0$, which is schematically dense in $(G_a{)}_{F_p}$ since it contains every closed point.

This being so, let us return to the proof of 3.2 d). If $w\in W$, its stabilizer in $G$ is an algebraic subgroup of $G_a$ that majorizes $G_i$ for every $i\in I$, hence is equal to $G_a$ (3.3), and consequently $W=V^G$.

One deduces immediately from 3.2 the

Corollary 3.4. Let $k$ be a field, $G$ a unipotent $k$-algebraic group acting on a finite-dimensional $k$-vector space $V$. Then $V$ possesses a sequence of vector subspaces $V_i$, defined over $k$, stable under $G$,

0 = V_0 ⊂ V_1 ⊂ … ⊂ V_n = V,

such that $G$ acts trivially on $V_{i+1}/V_i$. One may further suppose $V_{i+1}/V_i$ of dimension 1.

We shall now summarize and complement the properties of unipotent groups already proved, in the following theorem:

Theorem 3.5. Let $G$ be an algebraic group defined over a field $k$. The following properties are equivalent:

i) $G$ is unipotent.

ii) $G$ possesses a composition series, defined over $k$, whose successive quotients are isomorphic to $G_a$ if $p=0$ (resp. to ${\alpha }_p$, $G_a$, or twisted $(\mathbb{Z}/p\mathbb{Z}{)}^r$ (1.4) if $p>0$).

iii) As in ii), but one further assumes the composition series to be central.

iv) $G$ possesses a characteristic composition series (Exp. VI_B) defined over $k$, whose successive quotients are isomorphic to $(G_a{)}^r$ if $p=0$ (resp. to $({\alpha }_p{)}^r$, twisted $(G_a{)}^s$, or twisted $(\mathbb{Z}/p\mathbb{Z}{)}^t$, taken in this order, if $p>0$).

v) $G$ is isomorphic to an algebraic subgroup of the group $Trigstr(n{)}_k$ of strictly upper-triangular matrices of the linear group $GL(n{)}_k$, for some suitable integer $n⩾0$.

vi) $G$ is affine and, for every linear representation of $G$ in a non-zero finite-dimensional $k$-vector space $V$, one has $V^G\ne 0$.

Proof.

i) ⇒ vi) by 2.1 and 3.2.

vi) ⇒ v). Since the algebraic group $G$ is affine, $G$ is an algebraic subgroup of a suitable linear group $GL(V)$ (Exp. VI_B, 11.3). Applying 3.4 to the representation of $G$ in $V$ defined by this embedding, one finds v).

v) ⇒ iii). One knows that the algebraic group $Trigstr(n)$ possesses a central composition series with successive quotients isomorphic to $G_a$. The composition series induced on $G$ gives property iii), taking 1.5 into account.

iii) ⇒ ii) ⇒ i) and iv) ⇒ i) is clear.

We shall prove i) ⇒ iv) shortly, but first let us note some consequences of what has been proved.

Definition 3.6. We shall say that a $p$-Lie algebra $g$ ($p>0$) (cf. Exp. VII_A § 5) is unipotent if the map $x\mapsto x^{(p)}$ is nilpotent, i.e. if for every $x\in g$, there exists an integer $n>0$ such that $x^{(p^n)}=0$.

Corollary 3.7. A unipotent algebraic group $G$ is nilpotent (Exp. VI_B § 8); its Lie algebra $g$ is nilpotent (Bourbaki, Groupes et algèbres de Lie, Chap. 1 § 4) and is isomorphic to a Lie algebra of nilpotent endomorphisms of a finite-dimensional vector space.

In characteristic $p>0$, $g$ is a unipotent $p$-Lie algebra (3.6).

Since i) ⇒ v), it suffices to prove 3.7 when $G=Trigstr(n)$. We have already used the fact that $Trigstr(n)$ is a nilpotent algebraic group. Moreover, the Lie algebra $h$ of $Trigstr(n)$ consists of the upper-triangular endomorphisms of $V$ having zeros on the principal diagonal. They are therefore nilpotent and consequently $h$ is nilpotent (Bourbaki, loc. cit. Chap. 1 § 4 cor. 3). If $p>0$, since the $p$-th power in the $p$-Lie algebra $gl(V)=\mathrm{End}(V)$ coincides with the $p$-th power of the endomorphisms of $V$ (Exp. VII_A, 6.4.4), one sees that $h$ is unipotent.

Corollary 3.8. Let $k$ be an algebraically closed field and $G$ a smooth and affine algebraic group over $k$. Then the following properties are equivalent:

i) $G$ is unipotent.

ii) $G(k)$ consists of unipotent elements (Bible 4 prop. 4 cor. 1), that is, $G$ is unipotent in the sense of the Bible.

i) ⇒ ii) because $G$ is isomorphic to an algebraic subgroup of a group $Trigstr(n)$ by 3.2 i) ⇒ v).

ii) ⇒ i). Let $G$ be an algebraic group, unipotent in the sense of the Bible, and let $G^0$ denote its identity component. The maximal tori of $G^0$, being made up of unipotent elements, are trivial, so $G^0$ is equal to its Cartan subgroups. Consequently, $G^0$ is solvable (Bible 6 Th. 6), hence triangularizable (Bible 6, Th. 1). In brief, $G^0$ is an algebraic subgroup of a group $Trigstr(n)$, and is therefore unipotent in the sense of the present Exposé.

The group $(G/G^0)(k)$ is a finite group made up of unipotent elements; it is therefore zero if $p=0$ and equal to a finite $p$-group if $p>0$ (Bible 4 prop. 4). But then $G/G^0$ is unipotent in the sense of the present Exposé as a multiple extension of groups isomorphic to $\mathbb{Z}/p\mathbb{Z}$. This proves that $G$ is unipotent.

End of the proof of 3.5. Let us prove that i) ⇒ iv).

a) $p>0$. Consider the increasing sequence of algebraic subgroups of $G$:

{e} ⊂ F(G) ⊂ F_2(G) ⊂ ⋯ ⊂ F_n(G) ⊂ G⁰ ⊂ G.

One thus obtains a characteristic composition series of $G$ (App. II 1), and for $n$ large enough, $G/F_n(G)$ is smooth (App. II 3.1), so that the successive quotients are, in order:

(1) radicial groups of height 1,

(2) a smooth and connected group,

(3) an étale group.

To prove i) ⇒ iv) it therefore suffices to prove the:

Lemma 3.9. Let $G$ be a unipotent algebraic group defined over a field $k$ of characteristic $p>0$. Then $G$ possesses a characteristic composition series, defined over $k$, whose successive quotients are isomorphic to:

i) $({\alpha }_p{)}^r$ if $G$ is radicial.

ii) Twisted $(G_a{)}^r$ if $G$ is smooth and connected.

iii) Twisted $(\mathbb{Z}/p\mathbb{Z}{)}^r$ if $G$ is étale.

Proof. i) The group $G$ is radicial. Filtering $G$ by the $F_n(G)$, one reduces to the case where $G$ is radicial of height 1. Since $G$ is nilpotent (3.5 i) ⇒ iii)), and since the center of an algebraic group is representable (Exp. VIII, 6.5 e)), one may consider the ascending central series of $G$, evidently characteristic in $G$, which reduces us to the case where $G$ is moreover commutative.

Let $g=Lie(G)$. The $p$-th power morphism $\pi$ is therefore additive on $g$ (Exp. VII_A); we shall reduce to the case where it is zero. For every prescheme $S$ over $k$, set $g_S=g{\otimes }_k{O}_S$, and let $h_S$ be the subsheaf of abelian groups of $g_S$ image of $g_S$ under ${\pi }_S$. Finally, let ${\bar{h}}_S$ be the subsheaf of O_S-modules of $g_S$ generated by $h_S$. It is clear that ${\bar{h}}_S={\bar{h}}_k{\otimes }_k{O}_S$ and that ${\bar{h}}_S$⁵ is a characteristic sub-$p$-Lie algebra of $g_S$ (i.e. is stable under the $S$-functor ${\mathrm{Aut}}_{p-Lie}(g)$). It then follows from App. II 2.2 that ${\bar{h}}_k$ is the Lie algebra of an algebraic subgroup $H$ of $G$ that is characteristic in $G$.

Moreover, taking 3.7 and Lemma 3.9 bis below into account, if $G\ne e$, $H$ is distinct from $G$, because ${\bar{h}}_k$ is distinct from $g$. On the other hand, if $G/H=G^{\prime \prime }$, one has $Lie{G}^{\prime \prime }=g/{\bar{h}}_k$ (App. II 2.2), and consequently, the $p$-th power is zero in Lie G''. Proceeding by induction on $\dim LieG$, we are thus reduced to the case where Lie G is a $p$-Lie algebra in which the $p$-th power is zero. But then Lie G'' is isomorphic to $Lie({\alpha }_p{)}^r$ for some suitable integer $r⩾0$ (App. II 2.1), and consequently (App. II 2.2), G'' is isomorphic to $({\alpha }_p{)}^r$. It remains to prove the:

Lemma 3.9 bis. Let $k$ be a field of characteristic $p>0$, $g$ a commutative, unipotent (3.6), finite-dimensional $p$-Lie algebra over $k$, and $h$ the sub-$p$-Lie algebra of $g$ generated by the image of the $p$-th power in $g$. Then if $g\ne 0$, one has $h\ne g$.

Indeed, since $g$ is commutative, $h$ is simply the $k$-vector subspace of $g$ generated by the $X^{(p)}$ ($X\in g$). If $g\ne 0$ and is unipotent, there exists $X\in g$, $X\ne 0$, such that $X^{(p)}=0$.

Let $X_1,\cdots ,X_n$ be a basis of a supplement in $g$ of the line kX. The Lie algebra $h$ is then the $k$-vector subspace of $g$ generated by $X_1^{(p)},\cdots ,X_n^{(p)}$, hence has dimension at most $n=\dim g-1$.

Proof of 3.9 ii). $G$ is smooth and connected. In this case, the descending central series of $G$ is representable by characteristic smooth connected algebraic subgroups $G_i$ (Exp. VI_B, 8.3 and 7.4), and $G_i=0$ for $i$ large enough, since $G$ is nilpotent (3.5 i) ⇒ iii)). It suffices to prove 3.9 for the groups $G_i/G_{i+1}$, which reduces us to the case where moreover $G$ is commutative. For every integer $n>0$, let $G_n$ be the algebraic subgroup of $G$ which is the image of $G$ under the $p^n$-th power morphism. The group $G_n$ is therefore smooth, connected and characteristic, and it follows from Definition 1.1 of unipotent groups that $G_n=0$ for $n$ large enough. Replacing $G$ by $G_n/G_{n+1}$, one may further suppose that $G$ is annihilated by raising to the $p$-th power. But then, by (J.-P. Serre, Groupes algébriques et corps de classes, Chap. VII, prop. 11), $G$ is a form of $(G_a{)}^r$ for some suitable integer $r$.

Proof of 3.9 iii). $G$ is étale. Proceeding as in ii), one reduces to the case where $G$ is commutative, then to the case where $G$ is annihilated by $p$, but then $G_{\bar{k}}$ is isomorphic to $(\mathbb{Z}/p\mathbb{Z}{)}^r$.

Proof of 3.5 i) ⇒ iv) in the case b) $p=0$. The group $G$ is then smooth and connected, and proceeding as in 3.9 ii), one reduces to the case where $G$ is moreover commutative. One then has the following more precise result:

Lemma 3.9 ter. Let $k$ be a field of characteristic 0, $G$ a commutative unipotent $k$-algebraic group, $g=LieG$. Then there exists a canonical isomorphism:

$$\exp :W(g)\stackrel{\sim }{\to }G.$$

The morphism exp is the unique homomorphism $W(g)\to G$ that induces the identity on Lie algebras.

Since $G$ is unipotent, $G$ is realized as an algebraic subgroup of $Trigstr(n)$ for some suitable integer $n$ (3.5 i) ⇒ v)). The choice of such an embedding allows one to identify $g$ with a Lie subalgebra of $gl(n)$ consisting of nilpotent endomorphisms. Whence a $k$-morphism:

exp : W(g) ⟶ GL(n),    T ↦ Σ_{i ⩾ 0} T^i / i!.

Since $G$ is commutative, the morphism exp is a homomorphism. Let $G^{\prime }$ be the algebraic group image of $W(g)$ under the morphism exp. If one canonically identifies Lie W(g) with $g$, the linear tangent map of exp is simply the injection $g\to gl(n)$. Consequently $Lie(G\cap G^{\prime })=g\cap g=g$. Since $G\cap G^{\prime }$ is smooth (Exp. VI_B, 1.6.1) and $G$ is connected (being a multiple extension of groups $G_a$ (3.5 i) ⇒ ii)), one necessarily has $G=G^{\prime }$. The kernel $Ker(\exp )$ is a unipotent étale group, hence is the trivial group, and consequently exp is an isomorphism of $W(g)$ onto $G$.

If $h:W(g)\to G$ is another homomorphism such that $Lie(h)$ is the identity map of $g$, the morphism $h-\exp$ is a homomorphism ($G$ being commutative) whose linear tangent map is therefore zero. Since $k$ is of characteristic 0 and $W(g)$ is connected, it again follows from Cartier's theorem that one necessarily has $h=\exp$.

Proposition 3.10. Let $G$ be an algebraic group defined over an algebraically closed field $k$. Then the following properties are equivalent:

i) Every morphism of a group of multiplicative type $M$ into $G$ is the zero morphism.

i bis) $G$ has no non-zero subgroups of multiplicative type.

ii) a) if $p=0$: $G(k)$ contains no points of finite order other than $e$;

b) if $p\ne 0$: for every prime number $q\ne p$, $x\in G(k)$ and $x^q=e$ imply $x=e$,

   *for every `X ∈ g = Lie G` such that `X^{(p)} = X`, one has `X = 0`.*

ii bis) a) if $p=0$: as in ii) a);

   *b) if `p ≠ 0`: every finite subgroup of `G(k)` is a `p`-group,*

          *for every `X ∈ g = Lie G` such that `X^{(p)} = X`, one has `X = 0`.*

Proof.

i) ⇔ i bis), since the image of a group of multiplicative type is of multiplicative type (Exp. IX, 2.7).

ii) ⇔ ii bis). Because if an ordinary finite group $H$ has order not a power of $p$, there exists a prime number $q\ne p$, and an element $x$ of $H$ distinct from $e$, such that $x^q=e$ (Sylow's theorem, cf. J.-P. Serre, Corps locaux, Chap. IX § 2).

i) ⇒ ii) follows from the following lemma:

Lemma 3.11. Let $G$ be an algebraic group defined over a field $k$.

a) If $k$ contains the $n$-th roots of unity, with $n$ an integer prime to $p$, one has:

Hom_{k-gr}(μ_n, G) ≃ Hom_{k-gr}(ℤ/nℤ, G) ≃ _n G(k)

(points of order $n$ of $G(k)$).

b) If $p>0$, Hom_{k-gr}(μ_p, G) ≃ {X ∈ g = Lie G, such that X^{(p)} = X}.

Indeed, to prove a) we note that ${\mu }_n$ is then isomorphic over $k$ to $\mathbb{Z}/n\mathbb{Z}$, and b) is a consequence of App. II 2.1.

ii) ⇒ i bis). By Lemma 3.11, ii) is equivalent to the fact that for every prime number $r$, $G$ does not contain any subgroups ${\mu }_r$, which entails i bis) by reason of the:

Lemma 3.12. Let $G$ be a diagonalizable $S$-group, of finite type over $S$ and distinct from the trivial group. Then there exists a prime number $r$ and a subgroup of $G$ isomorphic to $({\mu }_r{)}_S$.

Let $G=D_S(M)$, where $M$ is a finitely generated abelian group, hence extension of a free group M'' by a finite group $M^{\prime }$. If $M^{\prime \prime }\ne 0$, it is clear that $M$ admits quotients isomorphic to $\mathbb{Z}/r\mathbb{Z}$ for every integer $r$. If $M^{\prime \prime }=0$, $M^{\prime }$ admits a quotient isomorphic to $\mathbb{Z}/r\mathbb{Z}$ for every prime number $r$ dividing the order of $M$.

One then deduces the lemma by duality.

We have seen (Prop. 2.4) that a unipotent group satisfies the equivalent conditions of 3.10. The aim of the next section is to prove the converse.

4. A characterization of unipotent groups

As announced, we shall show that an algebraic group $G$ defined over an algebraically closed field $k$, which does not contain any non-zero subgroup of multiplicative type, is unipotent. In fact, it suffices that it does not contain certain particular "elementary" subgroups of multiplicative type, which depend on the hypotheses made on $G$. Before stating the general theorem, let us study in detail some special cases.

4.1. Smooth, connected, affine algebraic groups

Proposition 4.1.1. Let $k$ be a field, $G$ a smooth, connected, affine $k$-algebraic group, $g=LieG$. Then the following properties are equivalent:

i) $G$ is unipotent.

ii) $G$ possesses a central composition series whose successive quotients are forms of $G_a$.

iii) $G$ possesses a central, characteristic composition series whose successive quotients are forms of $(G_a{)}^r$.

iv) There exists an integer $n>1$ such that $G_{\bar{k}}$ does not contain any subgroup isomorphic to ${\mu }_n$.

v) Every maximal torus of $G$ is the trivial group.

Suppose moreover that $G$ is an algebraic subgroup of a linear group $GL(n)$. Then the preceding conditions are also equivalent to:

vi) $g\subset gl(n)$ consists of nilpotent endomorphisms.

vii) $g$ is nilpotent and its center contains no non-zero semisimple endomorphism.

Proof. ii) ⇒ i) is clear, and i) ⇒ iii) was seen in 3.9. The implication iii) ⇒ ii) will follow from the following lemma:

Lemma 4.1.2. Let $k$ be a field, $G$ a $k$-algebraic group which is a form of $(G_a{)}^r$. Then:

a) $G$ is realized as an algebraic subgroup of the group $(G_a{)}^n$ for some suitable integer $n$.

b) $G$ possesses a composition series whose successive quotients are forms of $G_a$.

Indeed, by hypothesis, there exists an extension $k^{\prime }$ of $k$ such that $G_{k^{\prime }}$ is isomorphic to $(G_{a,k^{\prime }}{)}^r$. By the principle of finite extension (EGA IV 9.1.1), one may suppose that $k^{\prime }$ is a finite extension of $k$. But then, for a), it suffices to consider the canonical closed immersion (EGA V⁶):

G ⟶ ∏_{k'/k} (G_{k'})/k' ⥲ (G_{a, k})^n   (with n = r deg(k'/k)).

To prove b), in view of a), one may suppose that $G$ is a closed subgroup of $G^{\prime }=(G_a{)}^n$. If $G\ne 0$, there exists a hyperplane $h$ of $g^{\prime }=Lie{G}^{\prime }$ not containing $g$. Let $H$ be the vector subgroup $W(h)$ of $W(g^{\prime })=G^{\prime }$. Since $H$ is defined by an equation in $G^{\prime }$, $H\cap G$ is defined by an equation in $G$, and one has the inequalities:

dim G − 1 ⩽ dim(G ∩ H)
         ⩽ dim Lie(G ∩ H) = dim_k(g ∩ h) = dim_k(g) − 1 = dim G − 1.

Hence dim(G ∩ H) = dim Lie(G ∩ H), and consequently $G\cap H$ is smooth. The group $G_1=(G\cap H{)}^0$

is an algebraic subgroup of $G$, smooth and connected, such that $G/G_1$ is smooth, connected, of dimension 1, hence a form of $G_a$ (4.1 i) ⇒ iii)). One finishes by induction on the dimension of $G$.

Before continuing the proof of 4.1, let us note that the equivalence i) ⇔ ii) and 2.3 bis entail the following corollary:

Corollary 4.1.3. If $k$ is a perfect field, a smooth and connected $k$-algebraic group is unipotent if and only if it possesses a composition series whose successive quotients are isomorphic to $G_a$.

Continuation of the proof of 4.1.

i) ⇒ iv) by 2.4 i).

iv) ⇒ v). By Exp. XIV, 4.1, $G$ possesses a maximal torus $T$ defined over $k$. Now if $r=\dim T$, $({}_{n}T{)}_{\bar{k}}$ is isomorphic to $({\mu }_n{)}^r$. Hence $r=0$.

v) ⇒ i) as remarked in the proof of 3.8.

i) ⇒ vi). By 3.4, $G$ is in fact contained in an algebraic subgroup of $GL(n)$ isomorphic to $Trigstr(n)$, hence $g$ consists of nilpotent endomorphisms.

vi) ⇒ vii). Indeed, $g$ is nilpotent by Engel's theorem (Bourbaki, Groupes et algèbres de Lie, Chap. I § 4 cor. 3).

vii) ⇒ v). Let $T$ be a maximal sub-torus of $G$ (Exp. XIV, 1.1), $t$ its Lie algebra. The embedding of $G$ in $GL(n)$ defines a representation of $T$ which is necessarily semisimple (this is seen on an algebraic closure $\bar{k}$ of $k$, and one applies Exp. I, 4.7.3).

Hence if $X\in t$, $X$ is a semisimple endomorphism in $gl(n)$. One sees immediately that this entails that the map

ad X : Y ↦ [X, Y]

is a semisimple endomorphism of $gl(n)$, hence of $g$. Since, moreover, this endomorphism is nilpotent ($g$ being nilpotent), ad X is zero, so $X$ is central. But then $t$ is central and consists of semisimple endomorphisms, hence is zero by hypothesis; a fortiori, $T$ is the trivial group.

Remark 4.1.4.

a) We shall give below (4.3.1) an infinitesimal characterization of unipotent groups in characteristic $p>0$, which is independent of an embedding in $GL(n)$.

b) When $k$ is perfect, conditions ii) and iii) of 4.1.1 simplify by reason of the following lemma:

Lemma 4.1.5. If $k$ is a perfect field, every $k$-algebraic group $G$ which is a form of $(G_a{)}^r$ is isomorphic to $(G_a{)}^r$.

The lemma follows from 3.9 ter if the characteristic $p$ of $k$ is zero, and from 2.3 bis if $r=1$. In the general case ($p>0$), realize $G$ as an algebraic subgroup of $(G_a{)}^n$ for some suitable integer $n$ (4.1.2), and argue by induction on the integer $n-r$. If $r=n$, one indeed has $G=(G_a{)}^r$. Otherwise the quotient group $G_a^n/G$ is a non-zero smooth connected unipotent group which, taking 4.1.1 i) ⇒ ii) and 2.3 bis into account,

possesses a composition series with quotients isomorphic to $G_a$. One deduces that there exists an algebraic subgroup G_1 of $(G_a{)}^n$, smooth and connected, containing $G$, such that $G_a^n/G_1=G_a$. By induction it suffices to show that G_1 is isomorphic to $(G_a{)}^{n-1}$. Now it is immediate to verify that a homomorphism from $\mathrm{Spec}k[X_1,\cdots ,X_n]=G_a^n$ to $G_a=\mathrm{Spec}k[T]$ is defined by an additive polynomial of the form:

$${\Sigma }_{i,j}{a}_{i,j}{X}_i^{p^j}.$$

Since G_1 is smooth, the linear part of this polynomial is non-zero. Possibly making a linear change of the coordinates $X_i$, we may assume that G_1 is an algebraic subgroup of $(G_a{)}^n$ defined by the equation:

(*)   P(X) = −X_1 + Σ_{j=1}^q a_j X_1^{p^j} + Q(X_2, …, X_n) = 0,
                                        with   Q(X_2, …, X_n) = Σ_{i>1, j>0} b_{i,j} X_i^{p^j}.

Proceed then by induction on the degree of $P$. If $degP=1$, it is clear that $G_1\stackrel{\sim }{\to }{G}_a^{n-1}$. Otherwise, since $k$ is perfect, one has $Q(X)=Q_1(X{)}^p$ and we may define an endomorphism $v$ of $G_a^n$ by the formulas:

X_i ↦ X_i   for i > 1,        X_1 ↦ Σ_{j=1}^q a_j^{1/p} X_1^{p^{j−1}} + Q_1(X).

It is clear that $v$ induces an isomorphism on G_1 and that $v(G_1)$ has equation in $G_a^n$:

(*)_1   P_1(X) = −X_1 + Σ_{j=1}^q a_j^{1/p} X_1^{p^j} + Q_1(X_2, …, X_n) = 0.

Let us then distinguish two cases:

1st case. $deg(Q)=degP>p^q$. Then $deg{P}_1<degP$ and one wins by the induction hypothesis.

2nd case. $degP=p^q$ ($a_q\ne 0$). One then has $deg{P}_1=degP=p^q$ and $deg{Q}_1<p^q$. (One cannot have $Q=0$, otherwise G_1 would not be connected.) If the polynomial Q_1 has no linear part, one can iterate the preceding transformation. Continuing the process, one finally obtains an equation of the form:

(*)_s   P_s(X) = −X_1 + Σ_{j=1}^q a_j^{1/p^s} X_1^{p^j} + Q_s(X),

where $Q_s(X)=Q(X_2,\cdots ,X_n{)}^{1/p^s}$ is an additive polynomial having a non-zero linear part, and moreover $deg{Q}_s<p^q$. Suppose, for example, that the coefficient of X_2 in $Q_s$ is non-zero, and let $-L$ be the linear part of $P_s(X)$. Possibly making a linear change of coordinates, the equation of G_1 becomes:

P'(X) = −L + Σ_{j=1}^q a_j^{1/p^s} X_1^{p^j} + Q'(L, X_3, …, X_n),

where $Q^{\prime }$ is an additive polynomial without linear part, and $deg(Q^{\prime })<p^q$. But then we are reduced to the first case.⁷

4.2. Radicial groups

Proposition 4.2.1. Let $G$ be a radicial algebraic group defined over a field $k$ of characteristic $p>0$. Then the following conditions are equivalent:

i) $G$ is unipotent.

ii) $G$ possesses a central composition series with successive quotients isomorphic to ${\alpha }_p$.

iii) $G$ possesses a central and characteristic composition series with successive quotients isomorphic to $({\alpha }_p{)}^r$.

iv) $G_{\bar{k}}$ contains no subgroup isomorphic to ${\mu }_p$.

v) $g=LieG$ is a unipotent $p$-Lie algebra (3.6).

Proof. iii) ⇒ ii) ⇒ i) is clear, i) ⇒ iii) is 3.9 i), and i) ⇒ iv) by 2.4 i).

We shall need the following lemma on abelian $p$-Lie algebras:

Lemma 4.2.2. Let $g$ be an abelian, finite-dimensional $p$-Lie algebra over a perfect field $k$. Then $g$ can be written uniquely as a direct sum of a sub-$p$-Lie algebra $r$ on which the $p$-th power is bijective, and a sub-$p$-Lie algebra $u$, unipotent (3.6). (The algebra $r$ will be called the reductive part of $g$ and $u$ the unipotent part*.) The formation of $r$ and $u$ is compatible with extension of the field $k$. If moreover $k$ is algebraically closed, $r$ admits a basis $e_i$ such that $e_i^{(p)}=e_i$.*

The proof of this lemma is easy and left to the reader (cf. Bourbaki, Groupes et algèbres de Lie, Chap. I § 1 exercise 23). Let us simply say that $u$ is the kernel of a suitable iterate of the map $X\mapsto X^{(p)}$, and that $r$ is the image of the same iterate.

This being so, let us prove iv) ⇒ v). Possibly making an extension of the base field, one may suppose $k$ algebraically closed.

Let then $X$ be an element of $g$ and $h$ the sub-$p$-Lie algebra generated by $X$ in $g$. The algebra $h$ is evidently commutative, and its reductive part (4.2.2) is zero, otherwise by loc. cit., $h$ would contain an element $Y\ne 0$ such that $Y^{(p)}=Y$, and consequently (App. II 2.1 and 2.2) $G$ would contain a subgroup isomorphic to ${\mu }_p$ contrary to hypothesis. Hence $h$, and consequently $g$, is a unipotent $p$-Lie algebra.

v) ⇒ i). This is the least trivial implication of 4.2.1.

a) Case where $G$ is of height 1 (Exp. VII_A, 4.1.3). Since $G$ is radicial, it is affine, hence isomorphic to an algebraic subgroup of a linear group $GL(V)$ (Exp. VI_B, § 11). This embedding identifies $g$ with a sub-$p$-Lie algebra of $\mathrm{End}(V)$, the $p$-th power of $X$ in $g$ coinciding with the $p$-th power of the endomorphism $X$ (Exp. VII_A, 6.4.4). Since $g$ is unipotent by hypothesis, $g$ is therefore an algebra of nilpotent endomorphisms of $V$, and by Engel's theorem (Bourbaki, Groupes et algèbres de Lie, Chap. I § 4 th. 1) is a Lie subalgebra of the Lie algebra $h$ of the group of strictly upper-triangular matrices $Trigstr(n)$ with respect to a suitable basis of $V$. Since $G$ is of height 1, one then deduces from App. II 2.2 that $G$ itself is an algebraic subgroup of $Trigstr(n)$, hence is unipotent (3.5 v) ⇒ i)).

b) General case. Proceed by induction on the height $h$ of $G$⁸. The case $h=1$ has just been treated. Suppose $h>1$, and set $G^{\prime }=FG$ and $G^{\prime \prime }=G/G^{\prime }$. The group $G^{\prime }$ is of height 1 and $Lie{G}^{\prime }=LieG$ is unipotent, hence $G^{\prime }$ is unipotent by a).

To show that $G$ is unipotent, it therefore suffices to prove that G'' is unipotent (2.2). But G'' is of height $h-1$, hence, by induction hypothesis, it suffices to show that Lie G'' is unipotent. Since iv) ⇒ v), it suffices to show that $G_{\bar{k}}^{\prime \prime }$ does not contain any group isomorphic to ${\mu }_p$. Let then a subgroup of $G_{\bar{k}}^{\prime \prime }$ isomorphic to ${\mu }_p$, $H$ its inverse image in $G_{\bar{k}}$. The group $G^{\prime }$ being unipotent, we shall prove in § 5 that the extension:

e ⟶ G'_{k̄} ⟶ H ⟶ μ_p ⟶ e

is necessarily trivial (the proof given of this fact is independent of the results of the present section). Briefly, the group ${\mu }_p$ lifts in $H$, but being of height 1 it is necessarily contained in $G_{\bar{k}}^{\prime }=F{G}_{\bar{k}}$, whence a contradiction, $G^{\prime }$ being unipotent.

4.3. Connected affine groups in characteristic $p>0$

Proposition 4.3.1. Let $G$ be a connected affine algebraic group over a field $k$ of characteristic $p>0$. Then the following conditions are equivalent:

i) $G$ is unipotent.

ii) $G$ possesses a composition series with successive quotients isomorphic to ${\alpha }_p$ and $G_a$ (taken in this order).

iii) $G$ admits a characteristic composition series with successive quotients isomorphic to $({\alpha }_p{)}^r$ and $(G_a{)}^s$ (taken in this order).

iv) $G_{\bar{k}}$ contains no subgroup isomorphic to ${\mu }_p$.

v) $g=LieG$ is unipotent (3.6).

vi) $g$ is nilpotent, and the reductive part of the center of $g$ (4.2.2) is trivial.

vi bis) $g$ is nilpotent, and every subgroup of multiplicative type of the identity component of the center of $G$ is zero.

vi ter) $G$ is nilpotent, and every subgroup of multiplicative type of the identity component of the center of $G$ is zero.

Proof. It is clear that iii) ⇒ ii) ⇒ i). To establish i) ⇒ iii), we shall need the following lemma:

Lemma 4.3.2. Let $k$ be a field of characteristic $p>0$, $n\in \mathbb{N}$, $k^{\prime }$ a radicial extension of $k$ such that $(k^{\prime }{)}^{p^n}$ is contained in $k$; for every $k$-prescheme $X$ (resp. every $k^{\prime }$-prescheme $X^{\prime }$), denote by $X^{(p^n)}$ (resp. $X^{\prime \varphi }$) the $k$-prescheme deduced from $X$ (resp. $X^{\prime }$) by the base change:

F^n : k ⟶ k,   x ↦ x^{p^n},   (resp.   φ : k' ⟶ k,   x' ↦ x'^{p^n}).

Then, for every $k$-prescheme $X$, there exists a functorial isomorphism:

$$(X_{k^{\prime }}{)}^{\varphi }\stackrel{\sim }{\to }{X}^{(p^n)}.$$

Consequently, if $X$ and $Y$ are two $k$-preschemes such that there exists a $k^{\prime }$-isomorphism $u^{\prime }:X_{k^{\prime }}\stackrel{\sim }{\to }{Y}_{k^{\prime }}$, then there exists a $k$-isomorphism $v:X^{(p^n)}\stackrel{\sim }{\to }{Y}^{(p^n)}$. If moreover $X$ and $Y$ are equipped with structures of $k$-preschemes in groups and if $u^{\prime }$ is a $k^{\prime }$-homomorphism, then $v$ is a $k$-homomorphism.

The lemma follows simply from the transitivity of base changes and from the fact that the composite morphism $k\to k^{\prime }\stackrel{\varphi }{\to }k$ is equal to $F^n$.

Continuation of the proof of 4.3.1.

i) ⇒ iii). Proceed by induction on $\dim G$. If $\dim G=0$, since $G$ is connected, it is radicial, and one applies 3.9 i). If $\dim G>0$, there exists an integer $m⩾0$ such that the quotient $G/F_{m}G$ is a smooth group (App. II 3.1), evidently connected and non-zero. Applying 4.1.1 i) ⇒ iii) to the latter, one sees that there exists an algebraic subgroup $G^{\prime }$ of $G$ which is characteristic and connected, and such that the quotient $G^{\prime \prime }=G/G^{\prime }$ is a form of $G_a^r$ ($r>0$). By 4.1.5, if $K$ is a perfect closure of $k$, one has $G_K^{\prime \prime }\stackrel{\sim }{\to }(G_{a,K}{)}^r$. Since G'' is of finite type over $k$, there exists a finite radicial extension $k^{\prime }$ of $k$ such that $G_{k^{\prime }}^{\prime \prime }\stackrel{\sim }{\to }(G_{a,k^{\prime }}{)}^r$ (Exp. VI_B § 10). Let $n>0$ be such that $(k^{\prime }{)}^{p^n}\subset k$. Keeping the notations of 4.3.2, one deduces that there exists a $k$-isomorphism of algebraic groups:

(G_{a, k})^r = (G_{a, k'})^{rφ} ⥲ (G'')^{(p^n)}.

Consider then the Frobenius homomorphism relative to G'' (Exp. VII_A § 4)

$$F^n:G^{\prime \prime }⟶(G^{\prime \prime }{)}^{(p^n)}.$$

Since G'' (and hence also $(G^{\prime \prime }{)}^{(p^n)}$) is smooth over $k$, and $F^n$ is radicial, $F^n$ is an epimorphism for the fpqc topology, so that $(G^{\prime \prime }{)}^{(p^n)}$ is identified with $G^{\prime \prime }/F_n(G^{\prime \prime })$. Finally we have shown that $G^{\prime \prime }/F_n(G^{\prime \prime })$ is isomorphic, as an algebraic group, to $(G_a{)}^r$.

The inverse image $G_n^{\prime }$ of $F_n(G^{\prime \prime })$ in $G$ is a subgroup of $G$, connected, characteristic, of dimension strictly less than that of $G$, to which we may apply the induction hypothesis.

i) ⇒ iv) by 2.4 i).

iv) ⇒ i). Consider $G$ as an extension of a smooth connected group G'' by a radicial group $G^{\prime }$ (App. II 3.1). The group $G^{\prime }$ is unipotent (4.2.1 iv) ⇔ i)). It suffices to see that G'' is unipotent, and for that it suffices to show that $G_{\bar{k}}^{\prime \prime }$ does not contain a subgroup isomorphic to ${\mu }_p$ (4.1.1 i) ⇔ iv)). Now if $G_{\bar{k}}^{\prime \prime }$ contained a subgroup ${\mu }_p$, the latter would lift in $G_{\bar{k}}$, by the result (5.1) — already used — proved in § 5, whence a contradiction with iv).

i) ⇒ v) by 3.7.

v) ⇒ vi). Indeed, since $(adX{)}^{p^r}=ad(X^{(p^r)})$ (VII_A, 5.2), ad X is nilpotent if $g$ is unipotent, hence $g$ is nilpotent by Engel's theorem (Bourbaki, Groupes et algèbres de Lie, Chap. I § 4). On the other hand, if $g$ is unipotent, evidently so is its center, whose reductive part is then trivial (4.2.2).

vi) ⇒ iv). Indeed, if $G_{\bar{k}}$ contains a subgroup isomorphic to ${\mu }_p$, there exists a non-zero element $X$ of its Lie algebra such that $X^{(p)}=X$ (App. II 2.1), hence $X^{(p^r)}=X$ for every $r>0$. Since ad X is nilpotent (because $g$ is nilpotent and $(adX{)}^{p^r}=ad{X}^{(p^r)}$), necessarily $adX=0$, so $X$ belongs to the reductive part of the center of $g_{\bar{k}}$, whence a contradiction with vi).

i) ⇒ vi ter) follows from 2.4 i) and from 3.5 i) ⇒ iii).

vi ter) ⇒ vi bis). Indeed, if $G$ is nilpotent, so is its subgroup F G. It also follows from App. II 2.2 that F G is nilpotent if and only if $LieFG=LieG$ (Exp. VII_A) is nilpotent.

vi bis) ⇒ vi). Let $Z$ be the identity component of the center of $G$, and let $r$ be the reductive component of the center of $g$. We must show that $r=0$. Now it is immediate that $r$ is a characteristic sub-$p$-Lie algebra of $g=LieFG$ (i.e. stable under the functor ${\mathrm{Aut}}_{p-Lie}(g)$); hence $r$ is the Lie algebra of a characteristic radicial subgroup $R$ of F G (App. II 2.2). On the other hand, it follows from the last assertion of 4.2.2 and from App. II 2.1 that $R$ is a form of $({\mu }_p{)}^r$. The group $R$ being characteristic in F G, which is itself a characteristic subgroup of $G$ (App. II 1), $R$ is a fortiori invariant in $G$, hence is central, $G$ being connected (Exp. IX, 5.5). Hence by hypothesis vi bis), $R$ is zero, and so therefore is $r$.

4.4. Étale groups

The following proposition is an easy consequence of Sylow's theorems and of the structure of finite $q$-groups (cf. J.-P. Serre, Corps locaux, Chap. IX § 1).

Proposition 4.4.1. Let $G$ be a finite étale algebraic group defined over an algebraically closed field $k$. Then in order that $G$ be unipotent, it is necessary and sufficient that for every prime number $q$ distinct from the characteristic $p$ of $k$, $G$ does not contain a subgroup isomorphic to ${\mu }_q$.

4.5. Abelian varieties

Let $G$ be an abelian variety defined over an algebraically closed field $k$. Then the following conditions are equivalent:

i) $G$ is unipotent.

ii) $G=0$.

iii) There exists an integer $n$, prime to the characteristic $p$ of $k$, such that $G$ does not contain a subgroup isomorphic to ${\mu }_n$.

Indeed, if $G$ is an abelian variety of dimension $d$, one knows (cf. S. Lang, Abelian varieties, Chap. IV § 3 th. 6) that the group ${}_{n}G(k)$ ($n$ an integer prime to $p$) is isomorphic to $(\mathbb{Z}/n\mathbb{Z}{)}^{2d}$, hence is isomorphic to $({\mu }_n{)}^{2d}$. Whence iii) ⇒ ii), and ii) ⇒ i) ⇒ iii) are obvious.

4.6. General case

If $G$ and $H$ are two algebraic groups defined over an algebraically closed field $k$, we shall denote by $P(G,H)$ the property: "there is no algebraic subgroup of $G$ isomorphic to $H$". One then obtains the following characterizations of unipotent groups:

Theorem 4.6.1. Let $G$ be an algebraic group defined over an algebraically closed field $k$ of characteristic $p$. Then:

i) If $G$ is smooth, affine, and connected:

G is unipotent ⇐⇒ ∃ n > 1 such that P(G, μ_n) is true ⇐⇒ P(G, G_m) is true.

ii) If $G$ is smooth and connected:

G is unipotent ⇐⇒ ∃ n prime to p such that P(G, μ_n) is true.

iii) If $G$ is smooth:

G is unipotent ⇐⇒ for every prime number n ≠ p, P(G, μ_n) is true.

iv) $G$ affine connected and $p>0$: $G$ unipotent ⇐⇒ $P(G,{\mu }_p)$ is true.

v) $G$ connected and $p>0$:

G unipotent ⇐⇒ ∃ n prime to p such that P(G, μ_n) is true and P(G, μ_p) is true.

vi) $G$ an arbitrary algebraic group:

G is unipotent ⇐⇒ for every prime number n, P(G, μ_n) is true.

Proof. i) follows from 4.1.1, and iv) from 4.3.1. We shall prove vi); ii), iii), v) are proved analogously and are left to the reader.

Let then $G$ be an algebraic group. If $G$ is unipotent, $P(G,{\mu }_n)$ is true for every $n>1$ (2.4 i)). Conversely, suppose $P(G,{\mu }_n)$ is true for every prime number $n$, and let us show that $G$ is unipotent. Let $G^0$ be the identity component of $G$. If $G^0$ is smooth, it follows from a classical theorem of Chevalley⁹ that $G^0$ is an extension of an abelian variety $A$ by an affine, connected, smooth group $L$. If $G$ is not smooth, which presupposes $p>0$ (Exp. VI_B, 1.6.1), there exists an integer $n>0$ such that $G^{\prime \prime }=G^0/F_n(G)$ is smooth (App. II 3.1). Then G'' is an extension of an abelian variety $A$ by a smooth connected linear group L''. Denote by $L$ the inverse image of L'' in $G^0$, which is still affine and connected, since $F_n(G)$ is radicial. In all cases, $G$ therefore possesses a composition series:

$$0\subset L\subset G^0\subset G$$

such that $L$ is affine and connected, $G^0/L=A$ is an abelian variety, and $G/G^0$ an étale group.

If $P(G,{\mu }_n)$ is true, a fortiori $P(L,{\mu }_n)$ is true, hence $L$ is unipotent (4.1.1 and 4.3.1). If $A$ is non-zero, there exists a prime number $n$ and a subgroup of $A$ isomorphic to ${\mu }_n$ (4.5); by 5.1 below, this subgroup lifts in $G$, which contradicts the hypothesis $P(G,{\mu }_n)$; hence $A=0$. Finally if $G/G^0$ is not unipotent, there exists an integer $q$ and a subgroup of $G/G^0$ isomorphic to ${\mu }_q$ (4.4.1). One deduces as above that $P(G,{\mu }_q)$ is not true; hence $G/G^0$ is unipotent, and consequently so is $G$.

5. Extension of a group of multiplicative type by a unipotent group

5.1. Statement of the theorem

Definition 5.1.0. Let $k$ be a field, $G$ a $k$-algebraic group. Following the terminology introduced by Rosenlicht (Questions of rationality for solvable algebraic groups over non perfect fields, Annali di Math. 61 (1963)), we shall say that $G$ is "$k$-solvable" if it satisfies the following equivalent conditions:

i) $G$ possesses a composition series with successive quotients isomorphic to $G_a$.

ii) $G$ possesses a characteristic composition series $(G_i)$ such that the successive quotients $G_i/G_{i+1}$ are commutative and possess a composition series with successive quotients isomorphic to $G_a$.

The fact that i) ⇒ ii) is proved in loc. cit. In fact, one may take as composition series $(G_i)$ the composition series introduced in the proof of 3.9 ii).

Theorem 5.1.1. Let $k$ be a field, $U$ a unipotent $k$-algebraic group, $H$ a $k$-group of multiplicative type, $E$ a $k$-algebraic group extension of $H$ by $U$, so that one has the exact sequence:

$$1⟶U⟶E⟶H⟶1.$$

Then:

i) The extension $E$ is trivial in each of the following cases:

a) $k$ is algebraically closed.

b) $k$ is perfect and one of the groups $U$ or $H$ is connected.

c) $U$ is $k$-solvable.

d) $U$ is smooth and $H$ is connected.

ii) If $H^{\prime }$ and H'' are two lifts of $H$ in $E$, then $H^{\prime }$ and H'' are conjugate by an element of $U(k)$ in each of the following cases:

a) $k$ is algebraically closed and $H$ is smooth.

b) $k$ is algebraically closed and $U$ is smooth.

(We shall indicate in the course of the proof other cases where the conclusion of ii) is true without assuming $k$ algebraically closed.)

If $U$ is an algebraic group (resp. a commutative algebraic group) defined over a field $k$, we denote by $H^1(k,U)$ (resp. $H^i(k,U)$, $i⩾0$) the pointed set (resp. the $i$-th group) of Galois cohomology of $k$ with values in $U$ (cf. J.-P. Serre, Cohomologie Galoisienne, Lecture Notes Maths. n° 5, Springer).

If $S$ is a prescheme, $H$ an $S$-prescheme in commutative groups, $G$ an $S$-prescheme in groups acting on $H$, we denote by $H^i(G,H)$¹⁰ the $i$-th Hochschild cohomology group of $G$ with values in $H$ (App. I 1).

To prove 5.1.1, we shall proceed in several steps.

5.2. Proof of 5.1.1 i) and ii) in the case $U$ smooth and $H$ étale

Lemma 5.2.1. With the notations of 5.1.1, if $H$ is étale, the canonical morphism $E\to H$ possesses a section $s:H\to E$, defined over $k$, in the following cases:

a) $k$ is algebraically closed.

b) $k$ is perfect and $U$ is smooth and connected.

c) $U$ is "$k$-solvable".

a) follows simply from the fact that $E(k)\to H(k)$ is surjective. One has b) ⇒ c) by 4.1.2 b) and 2.3 bis. It therefore suffices to treat case c). Let $x$ be a point of $H$; $x$ is both an open and closed part of $H$, and the induced subscheme is isomorphic to $\mathrm{Spec}K$, where $K$ is a finite separable extension of $k$. Let $X$ be the inverse image in $E$ of the scheme $x$. The $K$-scheme $X$ is a principal homogeneous $K$-space under the group U_K. But U_K possesses a composition series with successive quotients isomorphic to $(G_a{)}_K$, hence $H^1(K,U_K)=0$ (J.-P. Serre, Cohomologie Galoisienne, Chap. III, prop. 6), and consequently $X$ is trivial, hence has a $K$-rational point. One thus obtains a $k$-section of $E\to H$ over $x$, for every point $x$ of $H$, whence the existence of a section $H\to E$.

Lemma 5.2.3.¹¹ With the notations of 5.1.1, suppose $H$ étale. Then:

i) The extension $E$ is trivial in each of the following cases:

a) $U$ is commutative and $E\to H$ possesses a section.

b) $k$ is algebraically closed.

c) $k$ is perfect and $U$ is connected.

d) $U$ is "$k$-solvable".

Moreover, in each of the above cases, two lifts $H^{\prime }$ and H'' of $H$ in $E$ are conjugate by an element of $U(k)$.

ii) Let $R$ be the $k$-functor $(Sch/k{)}^{\circ }\to (Ens)$ such that, for every $k$-prescheme $S$, $R(S)$ is the set of lifts of H_S in E_S. Then if $U$ is commutative, $R$ is representable by a non-empty affine $k$-scheme. The group $U$ acts by inner automorphisms on $R$, and this action makes $R$ into a homogeneous space under $U$ (for the fpqc topology (cf. Exp. IV)).

Proof of i). We shall reduce cases b), c), and d) to case a).

Case b) Since $U$ possesses a characteristic composition series with successive commutative quotients (3.5 i) ⇔ iv)), one immediately reduces to the case where $U$ is commutative. Moreover $E\to H$ possesses a section by 5.2.1 a), and we are reduced to case a).

Case d) One proceeds similarly using 5.1.0 ii) and 5.2.1 c).

Case c) Let $E_{red}$ be the reduced subgroup associated with $E$. Since $H$ is smooth, $E_{red}$ is an extension of $H$ by $U_0=U\cap E_{red}$, and one may replace $E$ by $E_{red}$. But $U$, and therefore also U_0, is connected, and it is clear that U_0 is the identity component of $E_{red}$, hence is smooth. But then U_0 is smooth and connected, $k$ is perfect, hence U_0 is $k$-solvable (4.1.4 b)) and we are reduced to case d).