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visits | member for | 4 years, 11 months |

seen | Jan 3 '14 at 9:16 | |

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I'm interested in many things, but algebraic geometry I love.

Dec 22 |
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What is the motivation for defining the conductor of an abelian variety?
Kestutis, is the description you gave here the motivation for defining a conductor of an abelian group? What (originally) was the purpose of this notion? |

Dec 22 |
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What is the motivation for defining the conductor of an abelian variety?
Kestutis, that sounds exactly like the kind of explanation I want. Where can I read more about this? (In particular I am not familiar with "semisimplification" and "Grothendieck's quasi-unipotence thm".) Noam, in that case what makes the definition of an abelian variety over a global field natural? |

Dec 22 |
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What is the motivation for defining the conductor of an abelian variety?
Ah, I see. Is it some limit of this procedure, or am I completely on the wrong track? |

Jun 23 |
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How can one interpret homology and Stokes' Theorem via derived categories?
Thanks, Denis! That sounds exactly like the kind of thing I'm looking for. Do you have a reference I can look at? |

Jun 23 |
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How can one interpret homology and Stokes' Theorem via derived categories?
The question is: what is a reference for a proof of Stokes' Theorem using derived categories? That sounds pretty well defined for me. If people prefer, they can give a proof of Stokes' Theorem using derived categories rather than to give a reference. Is that really not well-defined? |

Apr 3 |
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Is there a nice criterion for when the splitting fields of two irreducible polynomials are equal?
I'm looking for something with more insight, I'm afraid... I suppose if there is none then there is none. But I'm still holding out hope that someone will tell me that there is some paper that I should read about it, or that it somehow has to do with cohomology, or whatever insight might come this way... (BTW, the third question was a statement, not a question: "It is possible" rather than "Is it possible".) |

Jan 21 |
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Does the proof of GAGA use the axiom of choice?
*sorry, I meant: "the algebraic sheaf that induces a particular analytic sheaf". @nosr, that sounds more along the lines I was thinking of, but I can't say that I myself am familiar with all the details of the proof. It most certainly is false that GAGA is constructive, and there have been many papers trying to understand the relationship between the analytic side and the algebraic side better. |

Jan 21 |
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Does the proof of GAGA use the axiom of choice?
Duff is right. It is generally not well understood how to find the algebraic sheaf that induces a particular algebraic sheaf. It is very misleading to say that GAGA is obvious in any way. |

Oct 4 |
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Why is there no “regular etale fundamental group”?
unknown, it sounds like you have an example in mind. |

Sep 7 |
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Philosophy behind Mochizuki's work on the ABC conjecture
David, your comments are precisely the type of answer I'm looking for. It is okay that it's 20 years old. |

Sep 7 |
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Philosophy behind Mochizuki's work on the ABC conjecture
You are correct that I was inaccurate on that point, although I did know that it was Weil's idea. As for you argument for patience, I think you have misunderstood my question. I am not asking for a sketch of the methods, but only what those methods aim to achieve. An example of a good answer is David Speyer's comments. So saying "here is the rough argument in the function fields case, and so what we want is a number theoretic analogue of ____" is precisely the answer I was looking for. |

Sep 7 |
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Philosophy behind Mochizuki's work on the ABC conjecture
@quid: you're being stubborn. Is it not legitimate to ask questions about mathematics that is available but difficult to read and understand? @Kevin: thanks! |

Sep 7 |
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Philosophy behind Mochizuki's work on the ABC conjecture
@quid: the expositions I've seen (such as kurims.kyoto-u.ac.jp/~motizuki/2010-10-abstract.pdf) are mostly teasers to make people read more. My question is about the sketch underlying the proof of the ABC conjecture, which I don't see evident there. If you have an exposition that you would recommend, I suggest that you write it as an answer. |

Sep 7 |
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Philosophy behind Mochizuki's work on the ABC conjecture
Correction: "an enthusiastic report". Sorry, Jordan! |

May 16 |
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What is the intuition for $\mathbb{Q}^{ab}$ having cohomological dimension $1$?
Carnahan: can you expatiate a little more about your comment? |

May 16 |
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What is the intuition for $\mathbb{Q}^{ab}$ having cohomological dimension $1$?
Alex: $\mathbb{Q}^{ab}$ is indeed the maximal abelian extension of $\mathbb{Q}$, but its absolute Galois group is $Gal(\bar{\mathbb{Q}}/\mathbb{Q}^{ab})$ not $Gal(\mathbb{Q}^{ab}/\mathbb{Q})$. |

Feb 12 |
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Are there n polynomials for which all intersection multiplicities are at least m?
Yes, you're right! |

Feb 12 |
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Are there n polynomials for which all intersection multiplicities are at least m?
@Florian: $0$ and $(x-1)^2(x+1)$ intersect with multiplicity $1$ at $x=-1$. |

Feb 12 |
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Are there n polynomials for which all intersection multiplicities are at least m?
Look at the discussion above -- what you suggest will only make the condition hold at $x=a$, but not for other values of $x$. For example try multiplying $0,x,1,x+1$ (which satisfy the condition over $x=0$) by $(x-3)^2$ and see that there are values where some of these intersect with multiplicity $1$. |

Feb 12 |
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Are there n polynomials for which all intersection multiplicities are at least m?
@Will: I want that all of its nonzero roots will have multiplicity $\geq m$. Furthermore, I want that for every $i$ there will exist a unique $j$ such that $0$ is a root of $f_i-f_j$. |