6
$\begingroup$

$\DeclareMathOperator\GL{GL}$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?

$\endgroup$
2
  • 3
    $\begingroup$ This follows from $R$ being a Dedekind ring. See Hartshorne AG Prop. 9.7. $\endgroup$ 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
    Dec 27, 2021 at 13:57

1 Answer 1

9
$\begingroup$

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

$\endgroup$
8
  • $\begingroup$ Does the schematic closure literally just mean the closure, or does it (as @stupid_question_bot suggests) mean taking the reduced subscheme? $\endgroup$
    – LSpice
    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. $\endgroup$ 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 ${\rm GL}_{n,K}\to {\rm GL}_{n,R}$. $\endgroup$ 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 $\mu(A)\otimes_R\mu(A)$? $\endgroup$ Dec 29, 2021 at 1:42
  • 1
    $\begingroup$ 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 $\mu(A)\otimes_R\mu(A)$. $\endgroup$ Dec 29, 2021 at 10:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.