# Flatness of the closure of a closed subgroup of the generic fiber of an algebraic group inside an integral model of the ambient group

$$\DeclareMathOperator\GL{GL}$$Let $$K$$ be a number field and $$R$$ its ring of integers. Let $$G$$ be a connected reductive closed subgroup of $$\GL_{n,K}$$. On p55 of Brian Conrad's notes Reductive group schemes, he claims that the schematic closure $$\overline{G}$$ of $$G$$ inside $$\GL_{n,R}$$ is a flat closed $$R$$-subgroup of $$\GL_{n,R}$$.

First, I assume by schematic closure, he means the reduced induced structure on the topological closure. I can see that this closure $$\overline{G}$$ is an $$R$$-subgroup of $$\GL_{n,R}$$, and that it satisfies $$\overline{G}(R) = \{g\in \GL_n(R) : g\cap \GL_{n,K}\in G\}$$, where in the brackets we view $$g$$ as a closed subscheme of $$\GL_{n,R}$$.

Why is $$\overline{G}$$ necessarily flat? (Is this obvious?) Is this also true if $$R$$ is an arbitrary domain?

• This follows from $R$ being a Dedekind ring. See Hartshorne AG Prop. 9.7. Dec 27, 2021 at 10:54
• I think Brian does not mean to pass to the reduced structure on the topological closure. Dec 27, 2021 at 13:57

$$\DeclareMathOperator\GL{GL}$$A colleague asked me this question some time ago and here is the answer I sent him.

Let $$R$$ be a domain with fraction field $$K$$. Let $$A$$ be the $$R$$-algebra underlying the group scheme $$\GL_{n,R}$$ over $$R$$.

Suppose given a closed subgroup scheme $$H$$ of $$\GL_{n,K}$$. Let $$B$$ be the underlying $$K$$-algebra of $$H$$. We are then given by assumption a surjection $$\phi:A_K\to B$$ and an injection $$\lambda:A\to A_K$$ (given by $$a \mapsto a\otimes_K 1$$). Let $$\mu = \phi\circ\lambda$$ and consider the $$R$$-module $$\mu(A)$$, which is a sub-$$R$$-module of $$B$$. The following fact can be checked from the definitions: $$\mu(A)$$ has a natural $$R$$-algebra structure, compatible with the $$R$$-algebra structure of B via $$\lambda$$. In fact the R-algebra $$\mu(A)$$ is (by definition) the underlying $$R$$-algebra of the scheme-theoretic image of $$H$$ in $$\GL_{n,R}$$. Furthermore, we have $$\mu(A)_K\simeq B$$ and $$\mu(A)$$ is (clearly) torsion free. Now consider the Hopf algebra structure of B, which is given by a morphism of $$K$$-modules $$c:B\to B\otimes_K B$$. It is easy to see (diagram chasing) that if $$x\in \mu(A)\subseteq B$$, then $$c(x)$$ lies in the image of the natural map $$\mu(A)\otimes_R\mu(A)\to B\otimes_K B$$. So if the natural map $$\mu(A)\otimes_R\mu(A)\to B\otimes_K B$$ is an injection, we obtain a natural map $$\mu(A)\to \mu(A)\otimes_R\mu(A)$$ and this then (checking this is elementary) defines a Hopf algebra structure on $$\mu(A)$$. In particular, if the natural map $$\mu(A)\otimes_R\mu(A)\to B\otimes_K B$$ is an injection, then the scheme-theoretic closure of $$H$$ in $$\GL_{n,R}$$ is indeed a subgroup scheme, which is reduced if $$H$$ is reduced (which is what would happen if $$K$$ is of characteristic 0).

So the basic condition, which must be satisfied, is that the natural map $$\mu(A)\otimes_R\mu(A)\to B\otimes_K B$$ is injective. This will be true iff $$\mu(A)\otimes_R\mu(A)$$ is torsion free. For example, if $$\mu(A)$$ happens to be flat over $$R$$ then $$\mu(A)\otimes_R\mu(A)$$ will also be flat and thus torsion free. This is what will happen if $$R$$ is a Dedekind domain (where torsion free is equivalent to flat) or if $$R$$ is a valuation ring (where again torsion free is equivalent to flat). So if we work over either of these two rings, then the scheme-theoretic closure of $$H$$ in $$\GL_{n,R}$$ is indeed a subgroup scheme and it is even flat over $$R$$.

• Does the schematic closure literally just mean the closure, or does it (as @stupid_question_bot suggests) mean taking the reduced subscheme? Dec 27, 2021 at 13:57
• @LSpice I'm not sure what you mean by "just the closure" (what is the scheme structure?) But in this case because $G\hookrightarrow GL_{n,R}$ is quasi-compact, the topological image is dense inside the scheme-theoretic image (stacks 01R8), and because $G$ is reduced the minimality of the scheme-theoretic image implies that the scheme theoretic image is reduced, so it's the reduced induced structure on the topological closure. Dec 27, 2021 at 18:46
• The scheme-theoretic closure of $H$ in ${\rm GL}_{n,R}$ is by definition the scheme-theoretic image of the morphism $H\to {\rm GL}_{n,R}$ induced by the morphism ${\rm GL}_{n,K}\to {\rm GL}_{n,R}$. Dec 27, 2021 at 20:29
• Great! Just one question - why is $\mu(A)\otimes_R \mu(A)\rightarrow B\otimes_K B = B\otimes_R B$ injective equivalent to the source being torsion-free? I can see that injectivity is equivalent to $Tor_1^R(B/\mu(A),\mu(A)) = 0$. Is this group somehow related to the torsion of $\mu(A)\otimes_R\mu(A)$? Dec 29, 2021 at 1:42
• f $M$ is an $R$-module, then the kernel of the natural map of $R$-modules $M\to M_K$ is the torsion submodule of $M$. This follows from the fact that $M_K$ is naturally isomorphic to the localisation of $M$ at $R\backslash 0$ (see eg Th. 4.4 in Matsumura, Commutative Ring Theory). Hence if $m\in M$ is mapped to $0$ in $M_K$ then $m/1=0/1$ in $M_{R\backslash 0}$ and thus there exists $r\in R\backslash 0$ such that $r\cdot m=0$ by the definition of the localisation of a module at a multiplicative set. Now apply this to $\mu(A)\otimes_R\mu(A)$. Dec 29, 2021 at 10:59