bio | website | |
---|---|---|
location | ||
age | ||
visits | member for | 5 years |
seen | Jan 3 '14 at 9:16 | |
stats | profile views | 3,525 |
I'm interested in many things, but algebraic geometry I love.
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 |
awarded | Popular Question |
Jul 2 |
awarded | Popular Question |
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 $1$. |
Feb 12 |
comment |
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 $f_i-f_j$. |
Feb 12 |
revised |
Are there n polynomials for which all intersection multiplicities are at least m?
added 170 characters in body |
Feb 12 |
comment |
Are there n polynomials for which all intersection multiplicities are at least m?
You're right! So I guess this proves the $m=2$, $n=4$ case. I can't see how this would generalize, though... |
Feb 12 |
comment |
Are there n polynomials for which all intersection multiplicities are at least m?
Let's do a quick example: $0,x,1,1+x$ have the desired property at 0. If $m=2$, you're saying to multiply by $(x-1)^2$, say. $1(x-1)^2$ and $x(x-1)^2$ indeed intersect with multiplicity $2$ at $x=1$, but they would also intersect with multiplicity $1$ at some $x\neq 0,1$. So that's undesirable. |