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Jan
10
reviewed Approve 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 $F_n\subset U\cup V$.
Jan
10
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.
Jan
10
reviewed Approve How to calculate a complicated geometrical series?
Jan
9
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?
Jan
7
revised Coplanar codition
slightly fixed and unified the notation
Jan
7
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.
Dec
29
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.
Dec
29
accepted Any affine algebraic group is linear.
Dec
29
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.
Dec
6
awarded  Taxonomist
Dec
6
awarded  Good Question
Dec
2
accepted Infinite group with no maximal normal solvable subgroup
Dec
1
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.
Dec
1
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.
Dec
1
asked Infinite group with no maximal normal solvable subgroup
Nov
11
awarded  Benefactor
Nov
11
accepted Again: Ample and very ample line bundles
Nov
10
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.
Nov
9
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.
Nov
6
asked Again: Ample and very ample line bundles