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Apr
12
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Apr
10
comment Rank 1 Azumaya algebra
What does constant rank $1$ mean? That etale locally on $X$, $A$ is $\mathrm{Mat}_1(\mathcal{O}_U)=\mathcal{O}_U$? The map $\mathrm{Br}(X)\to \mathrm{Br}'(X)$ is injective (it's just the limit of the inclusions $H^1(X,\mathrm{PGL}_n)/H^1(X,\mathrm{GL}_n)$), so the image is trivial if and only if your original algebra is trivial!
Mar
15
comment A simple question about rational Hodge conjecture
I'm not quite sure, but perhaps it's because we have Lefschetz $(1,1)$?
Mar
12
comment Why are finite group schemes usually assumed flat?
important, I feel like, often times, just masks the true nature of theorems/notions. So, in case I am understanding you, if you assume that everything in sight is Noetherian, do you feel satisfactorily about the definitions?
Mar
12
comment Why are finite group schemes usually assumed flat?
I'm confused by your question. I think you're saying that you understand why you want to assume locally free of finite, constant rank, right? Because then you can define the order of your finite group scheme. So, you're asking why not just say locally free of finite constant rank? Probably because, as you observed, for finite morphisms to Noetherian things, this is the same thing (assuming connected base). So, I think the answer to both, and I mean this sincerely: don't pay attention to Noetherian hypotheses. Wondering why you need this or that to be Noetherian, while theoretically
Mar
12
comment $\mathbb{G}_a$ or $\mathbb{G}_m$ as subgroups of Affine Algebraic Groups
If you're working over $\bar{k}$, then this is equivalent to "does every affine algebraic group contain an integral smooth subgroup of dimension $1$". Somehow, this seems even harder though!
Mar
12
comment For a $k$-morphism $X \to Y$ to be determined by $X(\overline{k}) \to Y(\overline{k})$, does $X$ really have to be _geometrically_ reduced?
@KReiser No butting in. I couldn't have said it better myself :) You should ping c_c_chaos though so he knows you responded.
Mar
11
revised For a $k$-morphism $X \to Y$ to be determined by $X(\overline{k}) \to Y(\overline{k})$, does $X$ really have to be _geometrically_ reduced?
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Mar
11
comment Hartshorne's Algebraic geometry Chapter III ex. 9.10
Very nice example Georges! I guess what you wrote is fairly intuitive. You have a family of conics which, fiber by fiber are of course trivial (they have a zero), but to create a global trivialization you'd have to choose a family of rational points which, as you mentioned there is none.
Mar
11
comment Pushforward of a constant sheaf in the etale topology
@user113969 Isn't that a bit of overkill? The pushfoward is $\mathbb{Z}/n\mathbb{Z}$ since $\mathbb{A}^1$ is open-dense in $\mathbb{P}^1$. No?
Mar
11
comment Pull-back line bundle under morphism of degree $d$
@miguels You asked why "$(\ast)$" was true, which is that $f_\ast\mathcal{O}_X=\mathcal{O}_X\oplus \mathcal{O}(-1)^{\oplus(d-1)}$, this is why I wrote that answer. What about the answer do you not understand?
Mar
11
answered For a $k$-morphism $X \to Y$ to be determined by $X(\overline{k}) \to Y(\overline{k})$, does $X$ really have to be _geometrically_ reduced?
Mar
11
comment coordinate ring of quasi projective varieties are regular?
Interesting question! Usually these "do things affine locally hold imply global sections hold" are deceivingly tricky (e.g. finitely type over $k$). My only observation at this point, which may not be helpful, is to use the sheaf condition to understand $\mathcal{O}_X(X)$ as a subring of the finite product of regular rings, which are certainly regular. I don't think there are any nice conditions that say "a subring of the form ___ of a regular ring is regular" so this might not be helpful, but perhaps you can do something with it.
Mar
11
comment Can a birational morphism surject from an affine to a projective variety?
@JeskoHüttenhain No problem. I think there is probably an easier way with messing around with normalization maps, but this is intuitively nice for me (even though it requires a bit more in the way of justification): such a map clearly doesn't exist for curves, and any higher-dimensional example would produce an example with curves.
Mar
11
revised Can a birational morphism surject from an affine to a projective variety?
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Mar
11
comment Hartshorne's Algebraic geometry Chapter III ex. 9.10
Here's a hint. By considering $\mathbb{P}(\mathcal{E}):=\mathbf{Proj}(\mathrm{Sym}^\bullet(\mathcal{E}))$, you should be able to reduce this to a question about finding a coherent sheaf $\mathbb{A}^2$ which is not trivial on any neighborhood of the origin.
Mar
10
comment Can a birational morphism surject from an affine to a projective variety?
@Ben This is figured out. I'll write it up later. By normalizing Y you may assume it's normal then use Bertini to find the desired curve.
Mar
10
comment Can a birational morphism surject from an affine to a projective variety?
@Ben Yeah, I agree. When I have time tonight/tomorrow I'll think of how this can, if it can be, fixed. It's also worth mentioning, in case there is some obvious counterexample to the general case ($C$ intersecting good locus densely but no component of pullback intersecting good locus densely surjects) that we can always just pick a different $C$. So, we could just not worry about generalities by thinking if we can pick such a $C$ that does our job. PS, as you noted, and as is clear in the above proof, we may as well assume that $C$ is irreducible to begin with.
Mar
10
comment Can a birational morphism surject from an affine to a projective variety?
For any readers, the correct statement, which eluded me last night, is that the base change of birational by dominant is birational. I still think this issue might be fixable in this special case, but I just wanted to point out how to fix the incorrect statement in general.
Mar
10
revised Can a birational morphism surject from an affine to a projective variety?
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