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$\DeclareMathOperator\GL{GL}$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?

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

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$\DeclareMathOperator\GL{GL}$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 $R$.

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  • $\begingroup$ Does the schematic closure literally just mean the closure, or does it (as @stupid_question_bot suggests) mean taking the reduced subscheme? $\endgroup$
    – LSpice
    Commented Dec 27, 2021 at 13:57
  • $\begingroup$ @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. $\endgroup$
    – stupid_question_bot
    Commented Dec 27, 2021 at 18:46
  • $\begingroup$ 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 ${\rm GL}_{n,K}\to {\rm GL}_{n,R}$. $\endgroup$
    – Damian Rössler
    Commented Dec 27, 2021 at 20:29
  • $\begingroup$ 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 $\mu(A)\otimes_R\mu(A)$? $\endgroup$
    – stupid_question_bot
    Commented Dec 29, 2021 at 1:42
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    $\begingroup$ 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 $\mu(A)\otimes_R\mu(A)$. $\endgroup$
    – Damian Rössler
    Commented Dec 29, 2021 at 10:59

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