The Stacks project

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. $\square$

Proof. For any field extension $K/k$ 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. $\square$

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. $\square$

Proof. For any point $g \in G$ there exists a field extension $K/k$ 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 10.39.17). The result for $\mathcal{O}_{G_ K, g'}$ implies the result for $\mathcal{O}_{G, g}$, see Algebra, Lemma 10.30.5. 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. $\square$

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

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. $\square$

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. $\square$

Proof. Suppose that $K/k$ 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. $\square$

Proof. By Varieties, Lemma 33.7.14 we see that $G$ is geometrically connected. If we show that $G_ K$ is irreducible for some field extension $K/k$, 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.

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 $Z'' \subset W$.

By the claim $W \subset G$ is an open and closed subset of $G$. Now $W \cap Z' = \emptyset$ since otherwise by the argument given in the preceding paragraph we would get $Z' = Zg$ for some $g \in W(k)$. Then as $Z$ is a subgroup we could even pick $g \in U(k)$ which would contradict $Z' \cap U = \emptyset$. Hence $W \subset G$ is a proper open and closed subset which contradicts the assumption that $G$ is connected. $\square$

Proof. Let $G^0$ be the connected component of the identity with its canonical scheme structure (Morphisms, Definition 29.26.3). To show that $G^0$ is a closed subsgroup scheme we will use the criterion of Lemma 39.4.4. The morphism $e : \mathop{\mathrm{Spec}}(k) \to G$ factors through $G^0$ as we chose $G^0$ to be the connected component of $G$ containing $e$. Since $i : G \to G$ is an automorphism fixing $e$, we see that $i$ sends $G^0$ into itself. By Varieties, Lemma 33.7.13 the scheme $G^0$ is geometrically connected over $k$. 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. Thus $m|_{G^0 \times _ k G^0} : G^0 \times _ k G^0 \to G$ factors through $G^0$ by Morphisms, Lemma 29.26.1. Hence $G^0$ is 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. $\square$

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. $\square$

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 universally closed 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. $\square$

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. $\square$