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Let $G$ be an absolutely simple simply connected and connected algebraic group defined over a global field $k$ with ring of integers $\mathcal{O}$. Fix an embedding of $G$ into $GL_n$. Given $v$ a non-archimedean valuation of $k$, let $\mathfrak{p}_v\subset \mathcal{O}$ the associated maximal ideal, $\mathcal{O}_v$ its completion, $\mathfrak{m}_v$ the corresponding maximal ideal , $\mathbb{F}_v=\mathcal{O}_v/\mathfrak{m}_v$ the residue field and $G_v$ the reduction of $G$ modulo $\mathfrak{p_v}$. Let $G(\mathbb{F}_v)$ be the image of $G(\mathcal{O}_v)$ in $GL_n(\mathcal{O}_v/\mathfrak{m}_v)$. I am trying to understand why the following statement holds:

  • For all but a finite number of valuations $v$, $G(\mathbb{F}_v)$ is perfect and is a central extension of a finite simple group of Lie type $H(\mathbb{F}_v)$, where $H$ is of the Lie type of $G$ (either twisted or untwisted).

In trying to understand why this holds I have thought there are several steps:

  1. The reduction of $G$ modulo $\mathfrak{p_v}$ is absolutely simple simply connected and of the same Lie type as $G$ for almost every $v$.

  2. $G(\mathbb{F}_v)=G_v(\mathbb{F}_v)$ for almost every $v$.

  3. If $H$ is an absolutely simple simply connected algebraic group defined over a finite field $\mathbb{F}_q$, then for almost every $q$, $H(\mathbb{F_q})$ is perfect and the quotient by its center is a finite simple group.

If this is a right way to prove the statement, I would appreciate help concerning (1.). Any other proof or idea is also very welcome.

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  • $\begingroup$ Lemma 4.9 on page 18 of the following paper will partially help you with (1): Springer, T. A. Reductive groups. Automorphic forms, representations and L-functions (Proc. Sympos. Pure Math., Oregon State Univ., Corvallis, Ore., 1977), Part 1, pp. 3–27, Proc. Sympos. Pure Math., XXXIII, Amer. Math. Soc., Providence, R.I., 1979. $\endgroup$ – Mikhail Borovoi Feb 18 '16 at 17:01
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    $\begingroup$ The assertion is false for most classical split absolutely simple $G$ of adjoint type (step 3 fails), say realized inside ${\rm{GL}}(\mathfrak{g})$ (as a closed subgroup over $\mathbf{Z}$) via the adjoint representation. Indeed, your definition of the notation $G(\mathbf{F}_v)$ then coincides with the group of $\mathbf{F}_v$-points, and if $f:\widetilde{G} \rightarrow G$ is the simply connected central cover then $G(\mathbf{F}_v)$ has commutator subgroup the image of $\widetilde{G}(\mathbf{F}_v)$ away from a few cases. If $G$ is simply connected what you want to prove is true. $\endgroup$ – nfdc23 Feb 19 '16 at 3:26
  • $\begingroup$ @nfdc23: You are right, I forgot to write the assumption G simply connected. Thanks $\endgroup$ – Javier Garcia Feb 19 '16 at 7:57
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This is basically an instance of "spreading out" principles in algebraic geometry, combined with arguments specific to simply connected semisimple groups over general fields away from a few low-rank examples over fields of size 2 and 3. The formulation of the question is a bit imprecise, since the notation $G_v$ isn't really defined for all $v$, and making precise what it means for all but finitely many $v$ implicitly entails many of the steps below. We'll work directly with group schemes over rings since that is the natural framework to systematically discuss and analyze all questions of this type.

First, the spreading out. Let $R$ be a domain with fraction field $K$ (for example, $K$ a global field and $R$ a ring of $S$-integers for some non-empty finite set $S$ of places of $K$ containing the archimedean places) and $G$ a connected reductive $K$-group. By general spreading-out results in EGA IV, there exists a nonzero $r \in R$ such that $G$ is the generic fiber of a smooth affine $R[1/r]$-group scheme $\mathscr{G}$ whose geometric fibers are connected; we may and do rename $R[1/r]$ as $R$.

By Corollary 3.1.11 of the article "Reductive group schemes" in the Proceedings of the 2011 Luminy summer school on SGA3 (applied to $K$ viewed as the direct limit of $R[1/r]$'s) we can replace $R$ with $R[1/r]$ for some nonzero $r \in R$ to arrange that all geometric fibers of $\mathscr{G}$ over ${\rm{Spec}}(R)$ are reductive, and even semisimple when $G_{\overline{K}}$ is semisimple.

There is a finite separable extension $K'/K$ such that $G_{K'}$ is split (as $G_{K'_s}$ is split). Fix an $R$-finite domain $R' \subset K'$ with fraction field $K'$, so $K'$ is the direct limit of subrings $R'[1/r]$ for nonzero $r \in R$ (as $R' \otimes_R K = K'$). Hence, by inverting some such $r$ (again using the EGA IV spreading-out principles, for properties such as being an open immersion) we can arrange that $\mathscr{G}_{R'}$ admits an "open cell" with fiberwise maximal split $R'$-torus and root groups over $R'$ recovering the usual ones on the generic fiber. That already tells us the root datum on every fiber, and the root datum detects being simply connected.

Thus, now assuming $G_{\overline{K}}$ is semisimple, absolutely simple, and simply connected (all detected by the root datum: irreducible root system and coroots spanning the character lattice of a maximal torus), it follows that the same holds for all fibers of $\mathscr{G}$ over ${\rm{Spec}}(R)$.

In the special case that $K$ is a global field and we did spreading-out relative to some ring of $S$-integers, by enlarging $S$ we see that for each $v \not\in S$ the pullback $\mathscr{G}_{O_v}$ over the $v$-adic completion of $O_{K,S}$ is a smooth affine $O_v$-group. Hence, $\mathscr{G}(O_v) \rightarrow \mathscr{G}(\mathbf{F}_v)$ is surjective. The role of GL$_n$ in the setup is subsumed in the theory of group schemes over rings (namely, again by enlarging $S$ depending on some initial inclusion of $G$ into some ${\rm{GL}}_n$ we ensure that the schematic closure of $G$ inside ${\rm{GL}}_n$ over the $S$-integers coincides with $\mathscr{G}$ as an $O_{K,S}$-group).

In other words, the crutch of GL$_n$-embeddings is just a device to discuss flat affine integral model group schemes (of finite type) without speaking in terms of schemes. But it is precisely the relative smoothness of such $\mathscr{G}$ (or more concretely of its Zariski closure in such a GL$_n$ over $R$ at the cost of passing to some $R[1/r]$) which guarantees that the "ad hoc" definition of $G(\mathbf{F}_v)$ in the question actually agrees with the more conceptual notion $\mathscr{G}(\mathbf{F}_v)$ (of $\mathbf{F}_v$-points of the $O_{K,S}$-group scheme $\mathscr{G}$).

Consequently, we can now forget about the entire setup with group schemes over rings of integers (that has already served its purpose, bringing us to a suitable $R = O_{K,S}$ over which we've performed enough nice spreading-out) and are reduced to problem entirely about connected semisimple groups over fields: if $H$ is a connected semisimple group over a field $F$ such that $H$ is absolutely simple and simply connected, then away from a few low-rank cases over $F$ of size 2 or 3, prove $H(F)$ is perfect and that $H(F)/Z_H(F)$ is a simple finite group (with $Z_H$ denoting the center of the $F$-group $H$). This is an application of the role of Tits systems in the Borel-Tits structure theory for connected reductive groups over fields, and is discussed by a variety of methods in a variety of references (depending on your taste).

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  • $\begingroup$ Thanks for your detailed answer. After a considerable amount of reading I have been able to follow the main ideas and concepts involved in it. However I am still confused on the part related to smoothness. $\endgroup$ – Javier Garcia Jun 17 '16 at 7:15
  • $\begingroup$ Why is it true that i can choose r such that G is the generic fiber of a n R[1/r]-group scheme $\mathcal{G}$? The pullback over the v-adic completion just means considering it as an $\mathcal{O}_v$-scheme? Why is then true that $\mathcal{G}(\mathcal{O}_v)\to\mathcal{G}(\mathbb{F}_v)$ is surjective? I do not know to answer this questions when $k$ has positive characteristic. $\endgroup$ – Javier Garcia Jun 17 '16 at 7:21
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    $\begingroup$ Express $K[G]$ in terms of generators and relations over $K$, and likewise for the group law. This uses finitely many elements of $K$; by writing those as fractions we find a multiple $r \ne 0$ of all denominators occurring. Over $R[1/r]$ we thereby get an affine group scheme $\mathscr{G}$ of finite type with generic fiber $G$. More serious EGA input gives a multiple $r' \ne 0$ of $r$ so that $\mathscr{G}_{R[1/r']}$ is smooth with geometrically connected fibers over $R[1/r']$. "Yes" for the 2nd question. The 3rd is a property of smooth schemes over henselian local rings. $\endgroup$ – nfdc23 Jun 17 '16 at 14:19
  • $\begingroup$ Yes, the only part I still did not get is why can i choose $r'$ such that $\mathcal{G}_{R[1/r']}$ is smooth, could you point me to a reference as well as for the "surjectivity property" for smooth schemes over heneselian local rings? I know they are in EGA, but it is difficult to me to find these results in such a long book, plus most of the times the hypothesis for the theorems depend on general concepts that, although very likely they are trivially satisfied in my setting, I cannot recognize them. $\endgroup$ – Javier Garcia Jun 18 '16 at 8:56
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    $\begingroup$ By renaming $R[1/r]$ as $R$, you want that if $X$ is an $R$-scheme of finite type and $X_K$ is smooth then so is $X_{R[1/r']}$ for some nonzero $r' \in R$ (though fibers being geometrically connected for suitable $r'$ when $X_K$ is geometrically connected is also quite non-obvious, so I wonder why you don't ask about that too). And you want that $X(R) \rightarrow X(k)$ is surjective for smooth $X$ over a henselian local $R$ with residue field $k$. See IV$_4$ 17.7.8(ii) (and 17.7.11(iii)), IV$_3$ 9.7.7, IV$_4$, 18.5.17 (uses that smooth schemes are Zariski-locally etale over an affine space). $\endgroup$ – nfdc23 Jun 21 '16 at 1:44

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