# Jesko Hüttenhain

less info
reputation
630
bio website nullteilerfrei.de location Germany age 28 member for 2 years, 10 months seen 19 hours ago profile views 679

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

# 623 Actions

 Jan17 awarded Popular Question Jan14 comment $f_\ast \mathcal{O}_X$ locally free iff $X$ Cohen-Macaulay Well sure, the converse also holds. It is essentially the same argument. You might need that finite morphisms are affine. Jan13 comment $f_\ast \mathcal{O}_X$ locally free iff $X$ Cohen-Macaulay It is indeed. You can reformulate the first statement to "$R$ is CM $\Leftrightarrow$ All localizations at maximal ideals of $R$ are CM $\Leftrightarrow$ All localizations at prime ideals of $R$ are CM". Hence, even if $y$ corresponds only to a point and not a closed point, localization at $y$ will be CM. Jan13 reviewed Approve suggested edit on How to prove that $n^7\equiv n^3\mod40,\forall n\in\mathbb{Z}$ Jan13 answered $f_\ast \mathcal{O}_X$ locally free iff $X$ Cohen-Macaulay Jan11 revised An example of a $P$-primary ideal $I$ satisfying $I^2 = IP$ formatting cleanup Jan10 awarded Custodian Jan10 reviewed Close Tell a sequence in which there comes three consecutive even numbers after three odd numbers indefinitely? Jan10 reviewed Approve suggested edit on Let $U$ and $V$ be any two open sets with $U\cap V=\emptyset$ and $F\subset U\cup V$. Show that there is an integer $n$ with $F_n\subset U\cup V$. Jan10 comment P/B is isomorphic to the projective line $\mathbb{P}^1$ I am confused. I thought Borel subgroups of an algebraic group are the minimal parabolics. Wikipedia agrees with me. That would mean $P/B$ is trivial. Also, what do you mean by "containing a root"? A root is usually a character of a maximal torus in $B$, not an element of the group. Jan10 reviewed Approve suggested edit on How to calculate a complicated geometrical series? Jan9 comment Criterion for etaleness How much additional constraints are you willing to empose on your rings $A$ and $B$? Can they be Noetherian by any chance? Or even finitely generated $\Bbbk$-algebras over an algebraically closed field? Alright, I might be asking for a lot there. But still, anything beyond commutative? Jan7 revised Coplanar codition slightly fixed and unified the notation Jan7 comment Why $\mathcal{O}_{\mathbb{P}^n}(1)$ is a line bundle? I once wrote a little blog post about why vector bundles and locally free sheaves are the same. It might help. Dec29 comment Abstract nonsense proof of snake lemma @MarkS. is correct. I primarily want to understand the reasoning in Borceux' book. I hope you don't take my downvote as a personal offense, but this is precisely what I did not want. 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