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Sep
7
comment 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
comment 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
comment 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
revised Philosophy behind Mochizuki's work on the ABC conjecture
deleted 4 characters in body; deleted 21 characters in body
Sep
7
comment 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
comment Philosophy behind Mochizuki's work on the ABC conjecture
Correction: "an enthusiastic report". Sorry, Jordan!
Sep
7
asked Philosophy behind Mochizuki's work on the ABC conjecture
Aug
29
awarded  Popular Question
Jul
6
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Jul
2
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May
16
awarded  Nice Question
May
16
comment What is the intuition for $\mathbb{Q}^{ab}$ having cohomological dimension $1$?
Carnahan: can you expatiate a little more about your comment?
May
16
comment 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 $Gal(\mathbb{Q}^{ab}/\mathbb{Q})$.
May
15
asked What is the intuition for $\mathbb{Q}^{ab}$ having cohomological dimension $1$?
May
1
awarded  Yearling
Mar
10
awarded  Popular Question
Feb
12
accepted Are there n polynomials for which all intersection multiplicities are at least m?
Feb
12
comment Are there n polynomials for which all intersection multiplicities are at least m?
Yes, you're right!
Feb
12
comment 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
comment 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 $1$.