# Jesko Hüttenhain

less info
reputation
625
bio website nullteilerfrei.de location Germany age 28 member for 2 years, 6 months seen yesterday profile views 580

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

 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. 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. 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. 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 $I(\pi^{-1}(Y))=I(\pi^{-1}(Y)\cup\{ \mathbf 0 \})=I(Y)$. Oct29 comment Simple example of an ample line bundle that is not very ample Very helpful indeed, thanks for giving all these examples. Oct29 comment Simple example of an ample line bundle that is not very ample Fascinating. This is presenting to be exactly the learning experience I had hoped it would be. I think I will have a close look at all of those ;). Oct29 comment Simple example of an ample line bundle that is not very ample Hum, now I am confused. You wrote "Sections of K define a 2:1 cover, so K is globally generated but not ample." I think you simply mistyped then, because if that means it's a finite cover of degree $2$, then it is ample but not very ample. And that'd be what I want. In (2), you say it isn't globally generated and I thought that being globally generated is necessary for being ample. Oct29 comment Simple example of an ample line bundle that is not very ample Unfortunately, it seems like only (3) really meets my requirements, because in the other cases the bundle isn't ample. I will have a look later. Oct18 comment transversal intersection Hartshorne Lemma V.1.2 p.358 Errr. yes of course. Let me fix that. Oct17 comment When does one have $f_\ast(Im(a))=Im(f_\ast a)$? Ah, jesus, I mixed up $F$ and $E$. I'll remove my comment. Oct17 comment Rick Miranda exercise complete intersection curve. Prove it and find genus. Dear Georges, that means a lot coming from you! And apparently, I am really bad at calculating. Oct16 comment Rick Miranda exercise complete intersection curve. Prove it and find genus. Anyway, my answer has a few more technical details and a different approach to determining the genus, maybe it will still be useful. Oct12 comment Let $R$ a commutative ring and let $a\in R$. What does $aR$ mean? Well, $aR=R$ for fields only because for every $r\in R$, you can write $r=a(a^{-1}r)$. If $a$ is not a unit, then this fails. I am very certain of the above "because", the one in my answer might be a bit bold. Still, the equality $aR=R$ holds if and only if $a$ is a unit. Oct9 comment negative multiple of ample line bundle has no global section For $X$ nonsingular and projective: A global section $s\in\mathcal L(X)$, the divisor of zeros $Z(s)$ is an effective divisor $D$ such that $\mathcal L \cong \mathcal O(D)$. This is II, Proposition 7.7 in Hartshorne. Hence, in your case, this would imply that $mH$ is effective. Then, $\mathcal O(-mH)$ is the ideal sheaf of $mH$. Since it is not globally generated, it can not be ample, but it is supposed to be because $H$ is. The assumptions are much stronger than yours, but maybe it helps. Oct9 comment negative multiple of ample line bundle has no global section @xyzzyz: Don't you only get that $X$ is quasi-projective? If you assume $X$ to be proper over $k$, then this implies projective, but not in general. Oct8 comment What is the codimension of matrices of rank $r$ as a manifold? Let $M_r$ be the matrices of rank \$