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comment cartier duality is a contravariant equivalence?
I think that it's probably easier to just check it over $k^\text{sep}$ and make sure that the obvious isomorphism $(\mathbb{Z}/n\mathbb{Z})^{\vee\vee}\cong\mathbb{Z}/n\mathbb{Z}$ preserves Galois action and that the obvious map isomorphism $\text{Hom}(\mathbb{Z}/n\mathbb{Z},\mathbb{Z}/m\mathbb{Z})\to \text{Hom}((\mathbb{Z}/m\mathbb{Z})^\vee,(\mathbb{Z}/n\mathbb{Z}^\vee)$ sends Gal-equivariant things to Gal equivariant things. Of course, if you're not working over $\text{Spec}(k)$ but just some connected $T$ just replace Gal with the fundamental group.
Feb
1
awarded  Guru
Jan
29
awarded  Nice Answer
Dec
14
awarded  Good Answer
Nov
15
comment Is it possible to upgrade the statement that an unramified finite map is a “cover” to the statement that a ??? map is a fiber bundle?
Usually the analogue in algebraic geometry, as far as I know, is the statement that if $f:X\to Y$ is smooth proper then $R^f_\ast \mathcal{F}$ is lisse is $\mathcal{F}$ is lisse (equivalently, higher pushforwards of LCC sheaves are LCC). This is one of the main applications, at least cohomologically, of Ehresmann's theorem. PS, this is known as 'smooth proper base change' in a different light.
Nov
13
comment Example of a normal variety (or scheme) which is not locally factorial?
Does locally factorial mean 'covered by open factorial schemes', or the local rings are factorial?
Nov
12
comment Condition s.t. all closed set with dimension d in $\mathbb P^n$ intersects a closed set X
I don't know what Theorem 1.7.2 of Hartshorne says, but this is just the fact that to eat up all of the irrelevant ideal you need the sum of the heights to be $n+1$.
Nov
10
comment Cohomology and normalization of a curve
The map $\pi$ is finite.
Nov
8
comment Why do we care whether a functor is representable?
@Hoot It's a two way street. If schemes weren't easier then functors, then everything which is true of schemes would be obviously true of algebraic spaces.
Nov
8
comment Why do we care whether a functor is representable?
@Hoot While this is true in some situations, I don't think it's true always. For example, while smoothness is easy to encode in the functor, notions like 'finite type' are not (I guess you can say that finite presentation means commutes with filtered colimits?). And, basic things like Noetherian become very annoying to phrase in that language. And, even then, even when it's nicer to write down a definition functorially it's often easier to intuit this definition geometrically (smoothness maybe being a rare counterexample).
Nov
8
comment Why do we care whether a functor is representable?
Your second statement is a little misleading. It was actually the study of $X_0(N)$, which is never representable, which constituted most of Mazur's proof.
Nov
6
comment If a complex Lie group has the structure of an algebraic group, is this structure unique?
I can of course write something up, but perhaps a reference would be more useful. I learned this fact originally from Brian Conrad's book Reductive Group Schemes. In particular, it's proposition D.2.1 here.
Nov
5
comment If a complex Lie group has the structure of an algebraic group, is this structure unique?
Interesting question. The answer is yes if $G$ is reductive. I don't know if you are interested in non-reductive groups. The important thing being that you can prove it for semisimple groups by looking at their Lie algebras, and then extend to general reductive by working over extensions by tori. In fact, you get that every homomorphism is algebraic (note that this isn't a contradiction with your example since $\mathbb{C}$ is unipotent).
Oct
30
comment What is “Field with One Element”?
click here. Another good reference for ideas is this article. Of course, someone else might be able to say something more reasonable.
Oct
30
comment What is “Field with One Element”?
@HaiderAtrah I mean, not to be snarky, but what precisely do you even mean by that? What is your definition of an 'algebraic structure'? It does not exist as a scheme in any reasonable sense. It may exist, from what I've heard, in the context of 'generalized rings', but these objects were constructed largely to contain this the theory $\mathbb{F}_1$. The oft cited paper is this one (where the notion was developed). Another good thing to look at if you want to get an idea for the possible uses of $\mathbb{F}_1$ this article of Conne's is nice
Oct
30
comment What is “Field with One Element”?
For your consideration: youtu.be/x95hJ6F87fw?t=59m59s
Oct
28
comment What is the correct generalization of degree of a divisor to the number field case?
@Dorebell As Qiaochu mentioned, the correct thing to think about here is Arakelov geometry. In particular, the that one would want to do on a 'compactification' of $\mathrm{Spec}(\mathcal{O}_K)$ in general is intersection theory (defining degree being a basic part of this). This was, in fact, one of the original goals of Arakelov geometry—to attempt to prove something like, say, the Mordell conjecture similar to how one proves the Weil conjectures (for curves): by looking at analogues of $\Delta\cap\Gamma_\mathrm{Frob}$.
Oct
27
comment Failure of flatness in an integer ring
I mean, that is certainly one approach, although I do not see how to use it by looking at that one prime. I don't know if you read my comment above, but I thought about just asserting the constancy of fiber size to argue that flatness (+finiteness, I suppose) should imply our map is an isomorphism, since it's birational.
Oct
27
comment Failure of flatness in an integer ring
I'm confused. I don't think he's making the claim that $R'_\mathfrak{p}$ is local, is he? I mean, this isn't true in any generality that I can see applying this problem.
Oct
27
comment Failure of flatness in an integer ring
He's trying to give a proof of your second paragraph. He's assuming that $R'/R$ is flat. Then, since it's finite type (since, I assume, we're working with something like number rings) this implies that $R'/R$ is projective. I think he then meant to say that $(R')_\mathfrak{p}$ is free over $R_\mathfrak{p}$. But, as I said in my above comment, or you mentioned here, it's just true that normalization maps are not isomorphisms unless they are isomorphisms—insert Serre $S_1$ and $R_2$ statements here.