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This is my first question, so my apologies if it is too simple/poorly motivated.

During the course of some recent research I came across a particular variant of the following problem.

Let $G$ contain a normal unipotent subgroup $N$, where $R$ is a unital commutative ring of characteristic zero, $n\geq 4$, and it is assumed that $G/N$ is unipotent. Is it true that $G$ is also unipotent? What if $N$ is isomorphic to a subgroup of $U(n,R)$; the group of upper uni-triangular matrices?

It is somewhat well known, and I believe has already been answered on this site, that for $G$ an algebraic group and $N$ a normal closed subgroup, then the answer is in the affirmative.

Unfortunately, all of the literature I have read over the last few weeks points to results of this kind always being proven when replacing $R$ by an algebraically closed field or by a finite field, or only considering the case $n=3$.

Even if the original question cannot be answered entirely, any non-trivial examples of such $R$ would be enlightening. Otherwise, I would be satisfied to know if there are any similarly general results if $G$ is not taken to be an algebraic group and/or $R$ is a ring of characteristic zero.

Thank you.

Thank you for the helpful comments. Unfortunately, I am no expert with regards to algebraic geometry.

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    $\begingroup$ Is $G$ meant to be a closed $R$-subgroup scheme of ${\rm{GL}}_n$ (whereas ${\rm{GL}}_n(R)$ is an ordinary group of $R$-valued points, not a group scheme), and $N$ a flat closed normal $R$-subgroup scheme of $G$ (so $G/N$ is an algebraic space, possibly not a scheme if $R$ is general)? Perhaps more seriously, the question doesn't make sense unless you have defined what "unipotent" means over $R$. What do you mean by unipotence in your question (or in your hypotheses)? Something with the Lie algebra for smooth $G$ and $N$? Over arbitrary fields unipotence makes sense and the answer is "yes". $\endgroup$ – nfdc23 Apr 25 '16 at 6:00
  • $\begingroup$ I, too, cannot make sense of this question as asked. In some paragraphs the possible "algebraicity" of $G$ is discussed, so it suggests that "unipotencity" is a possible property of a general subgroup of $\text{GL}_n(R)$, as well as their quotient groups. Could you please define this property carefully? I also didn't understand what rings are regarded in "such $R$" in the last paragraph. $\endgroup$ – Uri Bader Apr 25 '16 at 6:45
  • $\begingroup$ @nfdc23 I do realize the ambiguity issue, and that is precisely part of the problem. The accepted notions of unipotent for linear algebraic groups over an algebraically closed field have, as far as I have read, differing but equivalent definitions. If you want, feel free to use any such definition of unipotent, but for argument's sake let us suppose that $G$ is unipotent if it is isomorphic to some (proper) subgroup of upper uni-triangular matrices $\endgroup$ – S.A.K.A. John Apr 25 '16 at 6:54
  • $\begingroup$ "let us suppose that $G$ is unipotent if it is isomorphic..." - isomorphic as what? abstract group? that doesn't fit any definition I know. You also regarded in your question the unipotency of $G/N$ which is not a matrix group in any canonical way afaik. $\endgroup$ – Uri Bader Apr 25 '16 at 7:10
  • $\begingroup$ I should have been clearer: I am not aware of a good theory of unipotence for smooth affine groups over rings. (Over general fields it is crucial that several different-looking definitions are equivalent, and if we didn't have the equivalence among such definitions then we would not be able to prove much about the notion over fields.) It would help to know what the specific motivation is (e.g., how do we know the algebraic space $G/N$ is a scheme, let alone affine?). Note that it isn't known over even the dual numbers $\mathbf{Q}[\epsilon]$ if smooth affine groups are closed in some GL$_N$. $\endgroup$ – nfdc23 Apr 25 '16 at 7:14
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To get some clarity it may help to consider the example where $G=\{\pmatrix{1&t\cr 0&e^u}\mid t, u\in \mathbb R\}$. The adjoint action of its Lie algebra is not right for a unipotent group.

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  • $\begingroup$ Thank you for the example. I am aware that the extension property fails in general, which is why I was wondering if there were instances where it holds over some ring. By the way, it is most likely due to my poor initial wording, but I am unclear as to what is precisely being demonstrated. Is $N=\mathbb{G}_a$ the normal subgroup? $\endgroup$ – S.A.K.A. John Apr 25 '16 at 7:51
  • $\begingroup$ @S.A.K.A. John You have changed the wording again. It still does not make sense. Now $G$ has nothing to do with $R$. Is $G$ an abstract group or not? What does "over some ring" mean? My example shows that such things matter. As a real Lie group my example has only one nontrivial normal subgroup, but it should never be written $\mathbb{G}_a$. $\endgroup$ – Wilberd van der Kallen Apr 25 '16 at 8:18
  • $\begingroup$ In the case where $G$ is abstract, I believe that the last comment from nfdc23 indicates what should be asked if we wish to speak about "over some ring". In your case, where $G$ is given explicitly, I admit that I fail to see why we need to invoke the Lie Algebra to refute unipotence. By the way, if you see a way to edit the question to make it cohesive, then feel free. Even in your example, you use the field $\mathbb{R}$ in order to construct the Lie group. I am beginning to believe that question cannot be properly defined for arbitrary rings. $\endgroup$ – S.A.K.A. John Apr 25 '16 at 13:47

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