All Questions
42
questions
8
votes
0
answers
278
views
Has the notion of a unipotent group scheme been studied?
The concept of a unipotent algebraic group over a field has been extensively studied and is fundamental in algebraic geometry. However, has the notion of a unipotent group scheme over a general base ...
0
votes
0
answers
113
views
Induced action on infinitesimal thickenings by an algebraic group
Let $X$ be an irreducible locally noetherian $k$-scheme (for $k$ any field), $G$ an algebraic group acting on $X$ via $a:G \times X \to X$ and $x \in X$ a closed point, which is by Zariski's lemma ...
3
votes
0
answers
150
views
Centers and conjugacy classes of groups relative to a pair of group homomorphisms
$\newcommand{\defeq}{\mathbin{\overset{\mathrm{def}}{=}}}$Given a group $G$, its center $\mathrm{Z}(G)$ and set of conjugacy classes $\mathrm{Cl}(G)$ are defined by
\begin{align*}
\mathrm{Z}(G) &\...
1
vote
1
answer
203
views
Lie algebras and pulled back group schemes
Suppose I have an extension of fields $L/K$, a group scheme $G_K$ over $\operatorname {Spec} K$. Let $G_L$ denote the pullback of $G_K$ to $\operatorname{Spec} L$. Then, under what conditions on the ...
5
votes
1
answer
258
views
Group scheme with an isotrivial maximal torus
Let $G$ be a reductive group scheme over a normal ring $A$. Then, we know that Zariski locally it admits a maximal torus.
Let us assume that it admits a maximal torus after a finite surjective (resp. ...
3
votes
0
answers
148
views
Finite commutative group schemes whose exponent coincides with its rank
In group theory, a finite commutative group $G$ contains an element whose order is the exponent of $G$. Thus, If the exponent of $G$ is the same as the order of $G$, it must be that $G$ is cyclic. ...
1
vote
0
answers
319
views
Any finite flat commutative group scheme of $p$-power order is etale if $p$ is invertible on the base
This question is immediately related to Discriminant ideal in a member of Barsotti-Tate Group
dealing with Barsotti–Tate groups and here I
would like to clarify a proof presented by
Anonymous in the ...
1
vote
0
answers
162
views
Discriminant ideal in a member of Barsotti-Tate Group
Let $S = \operatorname{Spec} R$ an affine scheme (in our case latter a complete dvr) and $p$ a prime. Then Barsotti-Tate group or $p$-divisible group $G$ of height $h$ over $S$
is an inductive system
...
2
votes
1
answer
1k
views
Viewing a finite group as a group scheme
I've read books which have this statement, without explanation : 'every finite group is an algebraic group'. I'm trying to understand what exactly they mean. The definition I have in my mind of a ...
2
votes
0
answers
270
views
Étale group scheme exact sequence
Consider the exact sequence of finite flat group schemes over the $2$-adic integers ring $\mathbb{Z}_2$:
$$0\longrightarrow\mathbb{Z}/2\mathbb{Z}\longrightarrow A\longrightarrow\mathbb{Z}/2\mathbb{Z}\...
2
votes
1
answer
363
views
fppf-extension of algebraic groups is an algebraic group
The problem is the following:
Let $k$ be a field and let $N,G,Q: (\operatorname{Sch}/k)_{\operatorname{fppf}} \longrightarrow \operatorname{Grps}$
be fppf-sheaves from the big fppf-site of $\...
3
votes
0
answers
215
views
Derived subgroup of the connected component of an algebraic group
Let $G$ be an affine group variety (smooth) over a field $k$, let $G^0$ be the connected component of $G$.
Is it true that the derived subgroup $[G^0,G^0]$ (Or $[G,G^0]$) is the connected component ...
1
vote
1
answer
167
views
Maps to additive group scheme
Let $\underline{\mathbb{Q}_p/\mathbb{Z}_p}$ be constant p-divisible group over $\mathbb{F}_p$. And let $\mathbb{G}_a$ be the additive group over $\mathbb{F}_p$. Let me prove
$$
Hom(\underline{\mathbb{...
7
votes
1
answer
520
views
Lie Algebra of Automorphism Group of $\mathbb{P}_k^1$
Let $X$ be a scheme over an algebraically closed field $k$ and let $\operatorname{Aut}(X)$ denote the functor sending a $k$-scheme $T$ to the group $\operatorname{Aut}_T(X \times_k T)$ of ...
5
votes
1
answer
410
views
What is $\mathrm{O}_q/\mathrm{SO}_q$ if $q$ is a quadratic $\mathbb{Z}$-form which is degenerate?
Any binary quadratic $\mathbb{Z}$-form $q$ induces a symmetric bilinear form
$$ B_q(u,v) = q(u+v) - q(u) -q(v) \ \ \forall u,v \ \in\mathbb{Z}^2 $$
and it is considered non-degenerate (over $\mathbb{...