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seen Dec 27 '13 at 2:36

Apr
28
comment Why is $O_K\otimes \mathbb{Z}_p\cong \oplus_{\mathfrak{p}|p}O_{K,\mathfrak{p}}$?
(Some would say that this is the problem with my generation.)
Apr
28
comment Why is $O_K\otimes \mathbb{Z}_p\cong \oplus_{\mathfrak{p}|p}O_{K,\mathfrak{p}}$?
Thank you! I didn't think of actually constructing the isomorphism...
Feb
12
comment Is $\bar{\mathbb{Q}}(x)\cap \mathbb{Q}((x))=\mathbb{Q}(x)$? [unsolved (even though we earlier thought it was)]
It is true that some details are missing (giving a consistent way to extend $Gal(\mathbb{Q})$), but they are clear to me, and it is the most elegant of the answers.
Feb
12
comment Is $\bar{\mathbb{Q}}(x)\cap \mathbb{Q}((x))=\mathbb{Q}(x)$? [unsolved (even though we earlier thought it was)]
The field of Laurent series. $\mathbb{Q}((x))=Quot(\mathbb{Q}[[x]])$, the quotient field of the ring of formal power series with coefficients in $\mathbb{Q}$.
Jan
23
comment For a $G$-module $A$ is there a maximal subgroup $H$ of $G$ such that the image $H^2(G,A)\rightarrow H^2(H,A)=0$?
Just out of curiosity, what would you have said under the assumption that $G$ is finite?
Jan
23
comment For a $G$-module $A$ is there a maximal subgroup $H$ of $G$ such that the image $H^2(G,A)\rightarrow H^2(H,A)=0$?
No, not necessarily. Although in the cases that I'm thinking of A is finite.
Dec
28
comment Interpretations of the first cohomology group
Aha! You're quite correct. The problem is that the "automorphism" I gave is not a homomorphism in general. This means that the action is not transitive, but is free. And the action is free and transitive if we restrict to the set of group-theoretic sections (which may be an empty set, but is not empty iff it's a semi-direct product). Good, I'm happy with this conclusion!
Dec
28
comment Interpretations of the first cohomology group
P.S. You can find "torsor" on wiki. The idea is that $H^1(X,G)$, where $X$ is a "space" (e.g. scheme, manifold, topological space, ...) and $G$ is a "group object" (e.g. group scheme, group variety, finite group, ...), classifies $G$-torsors over $X$. Think of this naively as principal $G$-bundles over a topological space $X$.
Dec
28
comment Interpretations of the first cohomology group
Aha... So what you're saying is that I can't just choose any group extension in this interpretation - I have to work with the trivial extension (the semi-direct product). I see... Can you tell me where I was wrong with my argument in the question? It seemed like I defined an action of $H^1(G,A)$ on the set-theoretic sections of any given extension (not nec. the semi-direct product extension) which is free and transitive. Where did I go wrong?
Dec
27
comment Interpretations of the first cohomology group
Ah! I will correct this.
Dec
27
comment Interpretations of the first cohomology group
$E$ is some (any) specific group extension of $G$ by $A$. As for your second comment, I'm not at all sure what you mean. The first and second group cohomologies are sometimes trivial, but usually not.
Oct
30
comment How can one deduce quadratic reciprocity from Hilbert reciprocity?
Thanks! I thought it might be something with $a=p$ and $b=q$. I can't quite convince myself of all of the facts above... Is there a proof of these identities in Cassels?
Oct
29
comment How could $Y(1)$ be the quotient of the upper half plane?
I can't think of points where it is branched, which would make it a covering map...
Sep
13
comment A few questions about $\mathbb{Q}$-models of modular curves (curves given by congruence groups)
The cover is defined, but it is not nec. Galois.
Sep
5
comment In what sense is Taniyama-Shimura the $n=2$ case of Langlands?
Every time I reread it I retain more. Thanks again for the answer!
Sep
4
comment A few questions about $\mathbb{Q}$-models of modular curves (curves given by congruence groups)
Are you saying $X(N)\rightarrow \mathbb{P}^1$ is always defined together with the group action over $\mathbb{Q}$? This surprises me. I thought $PSL_2(\mathbb{F}_p)$ is not known (for a general prime $p$) to be realized as a Galois group over $\mathbb{Q}$. I think Shih proved it was just for some of the primes.... Are you sure you didn't mean defined without the group action over $\mathbb{Q}$? That would make more sense. (P.S. Thanks a lot for the answer)
Sep
3
comment Can one write a PhD thesis before going into a PhD progam?
You can peruse my questions and see that I, too, am advanced for my age. But just read some theses. Much of the point of theses is that they are in depth. You need to be familiar with most of the literature on the topic (most of which you will probably not be able to read until years from now) and you have to make sure that you thought about all the ways to think about the phenomena discussed in the thesis. A thesis is meant to be work that takes years to do, and prepares you for independent research.
Sep
3
comment Can one write a PhD thesis before going into a PhD progam?
Unless you are one of the 10 most mathematically talented people in the world AND started doing professional math in your early teens, it is completely unrealistic.
Sep
3
comment Is it true that $\forall G\leq Aut(\mathbb{P}^1)=PGL_2(\mathbb{C})$ the map $\mathbb{P}^1\rightarrow \mathbb{P}^1/G$ is defined over $\mathbb{Q}$?
I'm not aware of this literature, and I only vaguely remember hearing about the McKay correspondence. Are you saying it is true?
Sep
1
comment What is $\operatorname{Aut}(\operatorname{PSL}_2(\mathbb{F}_q))$?
Cool! Thanks...