# Jesko Hüttenhain

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bio website nullteilerfrei.de location Germany age 28 member for 2 years, 9 months seen 2 hours ago profile views 645

I'm a PhD student with research interests in Algebraic Geometry, Algebraic Complexity Theory and Geometric Invariant Theory.

# 589 Actions

 Dec29 accepted Any affine algebraic group is linear. Dec29 comment Any affine algebraic group is linear. This looks really good! Thanks for tending to this very old question of mine and giving a satisfying answer. Dec6 awarded Taxonomist Dec6 awarded Good Question Dec2 accepted Infinite group with no maximal normal solvable subgroup Dec1 comment Infinite group with no maximal normal solvable subgroup I don't understand your last comment. Note that I do not require the subgroup to be proper; if $G$ is solvable, it satisfies all conditions because $G$ is a subgroup of itself. It is also clearly maximal with these properties. Dec1 comment Infinite group with no maximal normal solvable subgroup If the group were abelian, it would be its own, unique, maximal, normal, solvable subgroup - so I am not expecting a counterexample there. More precisely, the group must certainly not be solvable. Dec1 asked Infinite group with no maximal normal solvable subgroup Nov11 awarded Benefactor Nov11 accepted Again: Ample and very ample line bundles Nov10 comment Again: Ample and very ample line bundles @mercio: Post that as an answer. Would be a shame if the 400 rep would just go to waste. Nov9 comment Again: Ample and very ample line bundles I am using Hartshorne's definition, but you are right, the global sections of $I(1)^\sim$ already contain $x$. It was a silly question after all. Still, I am yet to find the example I am looking for. It's fine to know that there are abstract examples, but I would like to see this in coordinates at least once. Sigh. Nov6 asked Again: Ample and very ample line bundles Nov5 comment Is a finite normal subgroup of a reductive algebraic group central? @MarcvanLeeuwen: I think you misunderstood. I ment: Can you give an example of an infinite algebraic group $G$ which is not connected and such that a finite, normal subgroup $N\subseteq G$ exists which is not central? Because yes, I understand that a finite group $G$ is a counterexample, but I would like to know if there is an infinite counterexample. Nov5 comment Is a finite normal subgroup of a reductive algebraic group central? @MarcvanLeeuwen: Can you give an example when we assume $G$ to be infinite, but not connected? Nov4 comment Is a finite normal subgroup of a reductive algebraic group central? I will call $G$ reductive if its unipotent radical is trivial. Equivalently, you can use the definition that every finite-dimensional $G$-module is semisimple, i.e. a direct sum of irreducible $G$-modules. Nov4 asked Is a finite normal subgroup of a reductive algebraic group central? Nov1 comment The cone over a projective variety @user42912: That's not quite correct, because $I(\mathbf 0)=(x_0,\ldots,x_n)$ is a homogeneous ideal, but of course it contains polynomials that are not homogeneous. You simply have $I(\pi^{-1}(Y))=I(\pi^{-1}(Y)\cup\{ \mathbf 0 \})=I(Y)$. Oct31 answered The cone over a projective variety Oct29 comment Simple example of an ample line bundle that is not very ample Very helpful indeed, thanks for giving all these examples.