Lemma 39.7.1. If $(G, m)$ is a group scheme over a field $k$, then the multiplication map $m : G \times _ k G \to G$ is open.

## 39.7 Properties of group schemes over a field

In this section we collect some properties of group schemes over a field. In the case of group schemes which are (locally) algebraic over a field we can say a lot more, see Section 39.8.

**Proof.**
The multiplication map is isomorphic to the projection map $\text{pr}_0 : G \times _ k G \to G$ because the diagram

is commutative with isomorphisms as horizontal arrows. The projection is open by Morphisms, Lemma 29.23.4. $\square$

Lemma 39.7.2. If $(G, m)$ is a group scheme over a field $k$. Let $U \subset G$ open and $T \to G$ a morphism of schemes. Then the image of the composition $T \times _ k U \to G \times _ k G \to G$ is open.

**Proof.**
For any field extension $k \subset K$ the morphism $G_ K \to G$ is open (Morphisms, Lemma 29.23.4). Every point $\xi $ of $T \times _ k U$ is the image of a morphism $(t, u) : \mathop{\mathrm{Spec}}(K) \to T \times _ k U$ for some $K$. Then the image of $T_ K \times _ K U_ K = (T \times _ k U)_ K \to G_ K$ contains the translate $t \cdot U_ K$ which is open. Combining these facts we see that the image of $T \times _ k U \to G$ contains an open neighbourhood of the image of $\xi $. Since $\xi $ was arbitrary we win.
$\square$

Lemma 39.7.3. Let $G$ be a group scheme over a field. Then $G$ is a separated scheme.

**Proof.**
Say $S = \mathop{\mathrm{Spec}}(k)$ with $k$ a field, and let $G$ be a group scheme over $S$. By Lemma 39.6.1 we have to show that $e : S \to G$ is a closed immersion. By Morphisms, Lemma 29.20.2 the image of $e : S \to G$ is a closed point of $G$. It is clear that $\mathcal{O}_ G \to e_*\mathcal{O}_ S$ is surjective, since $e_*\mathcal{O}_ S$ is a skyscraper sheaf supported at the neutral element of $G$ with value $k$. We conclude that $e$ is a closed immersion by Schemes, Lemma 26.24.2.
$\square$

Lemma 39.7.4. Let $G$ be a group scheme over a field $k$. Then

every local ring $\mathcal{O}_{G, g}$ of $G$ has a unique minimal prime ideal,

there is exactly one irreducible component $Z$ of $G$ passing through $e$, and

$Z$ is geometrically irreducible over $k$.

**Proof.**
For any point $g \in G$ there exists a field extension $k \subset K$ and a $K$-valued point $g' \in G(K)$ mapping to $g$. If we think of $g'$ as a $K$-rational point of the group scheme $G_ K$, then we see that $\mathcal{O}_{G, g} \to \mathcal{O}_{G_ K, g'}$ is a faithfully flat local ring map (as $G_ K \to G$ is flat, and a local flat ring map is faithfully flat, see Algebra, Lemma 10.38.17). The result for $\mathcal{O}_{G_ K, g'}$ implies the result for $\mathcal{O}_{G, g}$, see Algebra, Lemma 10.29.5. Hence in order to prove (1) it suffices to prove it for $k$-rational points $g$ of $G$. In this case translation by $g$ defines an automorphism $G \to G$ which maps $e$ to $g$. Hence $\mathcal{O}_{G, g} \cong \mathcal{O}_{G, e}$. In this way we see that (2) implies (1), since irreducible components passing through $e$ correspond one to one with minimal prime ideals of $\mathcal{O}_{G, e}$.

In order to prove (2) and (3) it suffices to prove (2) when $k$ is algebraically closed. In this case, let $Z_1$, $Z_2$ be two irreducible components of $G$ passing through $e$. Since $k$ is algebraically closed the closed subscheme $Z_1 \times _ k Z_2 \subset G \times _ k G$ is irreducible too, see Varieties, Lemma 33.8.4. Hence $m(Z_1 \times _ k Z_2)$ is contained in an irreducible component of $G$. On the other hand it contains $Z_1$ and $Z_2$ since $m|_{e \times G} = \text{id}_ G$ and $m|_{G \times e} = \text{id}_ G$. We conclude $Z_1 = Z_2$ as desired. $\square$

Remark 39.7.5. Warning: The result of Lemma 39.7.4 does not mean that every irreducible component of $G/k$ is geometrically irreducible. For example the group scheme $\mu _{3, \mathbf{Q}} = \mathop{\mathrm{Spec}}(\mathbf{Q}[x]/(x^3 - 1))$ over $\mathbf{Q}$ has two irreducible components corresponding to the factorization $x^3 - 1 = (x - 1)(x^2 + x + 1)$. The first factor corresponds to the irreducible component passing through the identity, and the second irreducible component is not geometrically irreducible over $\mathop{\mathrm{Spec}}(\mathbf{Q})$.

Lemma 39.7.6. Let $G$ be a group scheme over a perfect field $k$. Then the reduction $G_{red}$ of $G$ is a closed subgroup scheme of $G$.

**Proof.**
Omitted. Hint: Use that $G_{red} \times _ k G_{red}$ is reduced by Varieties, Lemmas 33.6.3 and 33.6.7.
$\square$

Lemma 39.7.7. Let $k$ be a field. Let $\psi : G' \to G$ be a morphism of group schemes over $k$. If $\psi (G')$ is open in $G$, then $\psi (G')$ is closed in $G$.

**Proof.**
Let $U = \psi (G') \subset G$. Let $Z = G \setminus \psi (G') = G \setminus U$ with the reduced induced closed subscheme structure. By Lemma 39.7.2 the image of

is open (the first arrow is surjective). On the other hand, since $\psi $ is a homomorphism of group schemes, the image of $Z \times _ k G' \to G$ is contained in $Z$ (because translation by $\psi (g')$ preserves $U$ for all points $g'$ of $G'$; small detail omitted). Hence $Z \subset G$ is an open subset (although not necessarily an open subscheme). Thus $U = \psi (G')$ is closed. $\square$

Lemma 39.7.8. Let $i : G' \to G$ be an immersion of group schemes over a field $k$. Then $i$ is a closed immersion, i.e., $i(G')$ is a closed subgroup scheme of $G$.

**Proof.**
To show that $i$ is a closed immersion it suffices to show that $i(G')$ is a closed subset of $G$. Let $k \subset k'$ be a perfect extension of $k$. If $i(G'_{k'}) \subset G_{k'}$ is closed, then $i(G') \subset G$ is closed by Morphisms, Lemma 29.25.12 (as $G_{k'} \to G$ is flat, quasi-compact and surjective). Hence we may and do assume $k$ is perfect. We will use without further mention that products of reduced schemes over $k$ are reduced. We may replace $G'$ and $G$ by their reductions, see Lemma 39.7.6. Let $\overline{G'} \subset G$ be the closure of $i(G')$ viewed as a reduced closed subscheme. By Varieties, Lemma 33.24.1 we conclude that $\overline{G'} \times _ k \overline{G'}$ is the closure of the image of $G' \times _ k G' \to G \times _ k G$. Hence

as $m$ is continuous. It follows that $\overline{G'} \subset G$ is a (reduced) closed subgroup scheme. By Lemma 39.7.7 we see that $i(G') \subset \overline{G'}$ is also closed which implies that $i(G') = \overline{G'}$ as desired. $\square$

Lemma 39.7.9. Let $G$ be a group scheme over a field $k$. If $G$ is irreducible, then $G$ is quasi-compact.

**Proof.**
Suppose that $k \subset K$ is a field extension. If $G_ K$ is quasi-compact, then $G$ is too as $G_ K \to G$ is surjective. By Lemma 39.7.4 we see that $G_ K$ is irreducible. Hence it suffices to prove the lemma after replacing $k$ by some extension. Choose $K$ to be an algebraically closed field extension of very large cardinality. Then by Varieties, Lemma 33.14.2, we see that $G_ K$ is a Jacobson scheme all of whose closed points have residue field equal to $K$. In other words we may assume $G$ is a Jacobson scheme all of whose closed points have residue field $k$.

Let $U \subset G$ be a nonempty affine open. Let $g \in G(k)$. Then $gU \cap U \not= \emptyset $. Hence we see that $g$ is in the image of the morphism

Since the image of this morphism is open (Lemma 39.7.1) we see that the image is all of $G$ (because $G$ is Jacobson and closed points are $k$-rational). Since $U$ is affine, so is $U \times _{\mathop{\mathrm{Spec}}(k)} U$. Hence $G$ is the image of a quasi-compact scheme, hence quasi-compact. $\square$

Lemma 39.7.10. Let $G$ be a group scheme over a field $k$. If $G$ is connected, then $G$ is irreducible.

**Proof.**
By Varieties, Lemma 33.7.13 we see that $G$ is geometrically connected. If we show that $G_ K$ is irreducible for some field extension $k \subset K$, then the lemma follows. Hence we may apply Varieties, Lemma 33.14.2 to reduce to the case where $k$ is algebraically closed, $G$ is a Jacobson scheme, and all the closed points are $k$-rational.

Let $Z \subset G$ be the unique irreducible component of $G$ passing through the neutral element, see Lemma 39.7.4. Endowing $Z$ with the reduced induced closed subscheme structure, we see that $Z \times _ k Z$ is reduced and irreducible (Varieties, Lemmas 33.6.7 and 33.8.4). We conclude that $m|_{Z \times _ k Z} : Z \times _ k Z \to G$ factors through $Z$. Hence $Z$ becomes a closed subgroup scheme of $G$.

To get a contradiction, assume there exists another irreducible component $Z' \subset G$. Then $Z \cap Z' = \emptyset $ by Lemma 39.7.4. By Lemma 39.7.9 we see that $Z$ is quasi-compact. Thus we may choose a quasi-compact open $U \subset G$ with $Z \subset U$ and $U \cap Z' = \emptyset $. The image $W$ of $Z \times _ k U \to G$ is open in $G$ by Lemma 39.7.2. On the other hand, $W$ is quasi-compact as the image of a quasi-compact space. We claim that $W$ is closed. If the claim is true, then $W \subset G \setminus Z'$ is a proper open and closed subset of $G$, which contradicts the assumption that $G$ is connected.

Proof of the claim. Since $W$ is quasi-compact, we see that points in the closure of $W$ are specializations of points of $W$ (Morphisms, Lemma 29.6.5). Thus we have to show that any irreducible component $Z'' \subset G$ of $G$ which meets $W$ is contained in $W$. As $G$ is Jacobson and closed points are rational, $Z'' \cap W$ has a rational point $g \in Z''(k) \cap W(k)$ and hence $Z'' = Zg$. But $W = m(Z \times _ k W)$ by construction, so $Z'' \cap W \not= \emptyset $ implies $Z'' \subset W$. $\square$

Proposition 39.7.11. Let $G$ be a group scheme over a field $k$. There exists a canonical closed subgroup scheme $G^0 \subset G$ with the following properties

$G^0 \to G$ is a flat closed immersion,

$G^0 \subset G$ is the connected component of the identity,

$G^0$ is geometrically irreducible, and

$G^0$ is quasi-compact.

**Proof.**
Let $G^0$ be the connected component of the identity with its canonical scheme structure (Morphisms, Definition 29.26.3). By Varieties, Lemma 33.7.13 we see that $G^0$ is geometrically connected. Thus $G^0 \times _ k G^0$ is connected (Varieties, Lemma 33.7.4). Thus $m(G^0 \times _ k G^0) \subset G^0$ set theoretically. To see that this holds scheme theoretically, note that $G^0 \times _ k G^0 \to G \times _ k G$ is a flat closed immersion. By Morphisms, Lemma 29.26.1 it follows that $G^0 \times _ k G^0$ is a closed subscheme of $(G \times _ k G) \times _{m, G} G^0$. Thus we see that $m|_{G^0 \times _ k G^0} : G^0 \times _ k G^0 \to G$ factors through $G^0$. Hence $G^0$ becomes a closed subgroup scheme of $G$. By Lemma 39.7.10 we see that $G^0$ is irreducible. By Lemma 39.7.4 we see that $G^0$ is geometrically irreducible. By Lemma 39.7.9 we see that $G^0$ is quasi-compact.
$\square$

Lemma 39.7.12. Let $k$ be a field. Let $T = \mathop{\mathrm{Spec}}(A)$ where $A$ is a directed colimit of algebras which are finite products of copies of $k$. For any scheme $X$ over $k$ we have $|T \times _ k X| = |T| \times |X|$ as topological spaces.

**Proof.**
By taking an affine open covering we reduce to the case of an affine $X$. Say $X = \mathop{\mathrm{Spec}}(B)$. Write $A = \mathop{\mathrm{colim}}\nolimits A_ i$ with $A_ i = \prod _{t \in T_ i} k$ and $T_ i$ finite. Then $T_ i = |\mathop{\mathrm{Spec}}(A_ i)|$ with the discrete topology and the transition morphisms $A_ i \to A_{i'}$ are given by set maps $T_{i'} \to T_ i$. Thus $|T| = \mathop{\mathrm{lim}}\nolimits T_ i$ as a topological space, see Limits, Lemma 32.4.6. Similarly we have

by the lemma above and the fact that limits commute with limits. $\square$

The following lemma says that in fact we can put a “algebraic profinite family of points” in an affine open. We urge the reader to read Lemma 39.8.6 first.

Lemma 39.7.13. Let $k$ be an algebraically closed field. Let $G$ be a group scheme over $k$. Assume that $G$ is Jacobson and that all closed points are $k$-rational. Let $T = \mathop{\mathrm{Spec}}(A)$ where $A$ is a directed colimit of algebras which are finite products of copies of $k$. For any morphism $f : T \to G$ there exists an affine open $U \subset G$ containing $f(T)$.

**Proof.**
Let $G^0 \subset G$ be the closed subgroup scheme found in Proposition 39.7.11. The first two paragraphs serve to reduce to the case $G = G^0$.

Observe that $T$ is a directed inverse limit of finite topological spaces (Limits, Lemma 32.4.6), hence profinite as a topological space (Topology, Definition 5.22.1). Let $W \subset G$ be a quasi-compact open containing the image of $T \to G$. After replacing $W$ by the image of $G^0 \times W \to G \times G \to G$ we may assume that $W$ is invariant under the action of left translation by $G^0$, see Lemma 39.7.2. Consider the composition

The space $\pi _0(W)$ is profinite (Topology, Lemma 5.23.9 and Properties, Lemma 28.2.4). Let $F_\xi \subset T$ be the fibre of $T \to \pi _0(W)$ over $\xi \in \pi _0(W)$. Assume that for all $\xi $ we can find an affine open $U_\xi \subset W$ with $F \subset U$. Since $\psi : T \to \pi _0(W)$ is proper as a map of topological spaces (Topology, Lemma 5.17.7), we can find a quasi-compact open $V_\xi \subset \pi _0(W)$ such that $\psi ^{-1}(V_\xi ) \subset f^{-1}(U_\xi )$ (easy topological argument omitted). After replacing $U_\xi $ by $U_\xi \cap \pi ^{-1}(V_\xi )$, which is open and closed in $U_\xi $ hence affine, we see that $U_\xi \subset \pi ^{-1}(V_\xi )$ and $U_\xi \cap T = \psi ^{-1}(V_\xi )$. By Topology, Lemma 5.22.4 we can find a finite disjoint union decomposition $\pi _0(W) = \bigcup _{i = 1, \ldots , n} V_ i$ by quasi-compact opens such that $V_ i \subset V_{\xi _ i}$ for some $i$. Then we see that

the right hand side of which is a finite disjoint union of affines, therefore affine.

Let $Z$ be a connected component of $G$ which meets $f(T)$. Then $Z$ has a $k$-rational point $z$ (because all residue fields of the scheme $T$ are isomorphic to $k$). Hence $Z = G^0 z$. By our choice of $W$, we see that $Z \subset W$. The argument in the preceding paragraph reduces us to the problem of finding an affine open neighbourhood of $f(T) \cap Z$ in $W$. After translation by a rational point we may assume that $Z = G^0$ (details omitted). Observe that the scheme theoretic inverse image $T' = f^{-1}(G^0) \subset T$ is a closed subscheme, which has the same type. After replacing $T$ by $T'$ we may assume that $f(T) \subset G^0$. Choose an affine open neighbourhood $U \subset G$ of $e \in G$, so that in particular $U \cap G^0$ is nonempty. We will show there exists a $g \in G^0(k)$ such that $f(T) \subset g^{-1}U$. This will finish the proof as $g^{-1}U \subset W$ by the left $G^0$-invariance of $W$.

The arguments in the preceding two paragraphs allow us to pass to $G^0$ and reduce the problem to the following: Assume $G$ is irreducible and $U \subset G$ an affine open neighbourhood of $e$. Show that $f(T) \subset g^{-1}U$ for some $g \in G(k)$. Consider the morphism

which is an open immersion (because the extension of this morphism to $G \times _ k T \to G \times _ k T$ is an isomorphism). By our assumption on $T$ we see that we have $|U \times _ k T| = |U| \times |T|$ and similarly for $G \times _ k T$, see Lemma 39.7.12. Hence the image of the displayed open immersion is a finite union of boxes $\bigcup _{i = 1, \ldots , n} U_ i \times V_ i$ with $V_ i \subset T$ and $U_ i \subset G$ quasi-compact open. This means that the possible opens $Uf(t)^{-1}$, $t \in T$ are finite in number, say $Uf(t_1)^{-1}, \ldots , Uf(t_ r)^{-1}$. Since $G$ is irreducible the intersection

is nonempty and since $G$ is Jacobson with closed points $k$-rational, we can choose a $k$-valued point $g \in G(k)$ of this intersection. Then we see that $g \in Uf(t)^{-1}$ for all $t \in T$ which means that $f(t) \in g^{-1}U$ as desired. $\square$

Remark 39.7.14. If $G$ is a group scheme over a field, is there always a quasi-compact open and closed subgroup scheme? By Proposition 39.7.11 this question is only interesting if $G$ has infinitely many connected components (geometrically).

Lemma 39.7.15. Let $G$ be a group scheme over a field. There exists an open and closed subscheme $G' \subset G$ which is a countable union of affines.

**Proof.**
Let $e \in U(k)$ be a quasi-compact open neighbourhood of the identity element. By replacing $U$ by $U \cap i(U)$ we may assume that $U$ is invariant under the inverse map. As $G$ is separated this is still a quasi-compact set. Set

where $m_ n : G \times _ k \ldots \times _ k G \to G$ is the $n$-slot multiplication map $(g_1, \ldots , g_ n) \mapsto m(m(\ldots (m(g_1, g_2), g_3), \ldots ), g_ n)$. Each of these maps are open (see Lemma 39.7.1) hence $G'$ is an open subgroup scheme. By Lemma 39.7.7 it is also a closed subgroup scheme. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)