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visits | member for | 4 years, 4 months |
seen | Jan 3 at 9:16 | |
stats | profile views | 3,279 |
I'm interested in many things, but algebraic geometry I love.
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 12 |
asked | Which groups are quotients of symmetric groups? |
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 |
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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 |
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?
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Feb 12 |
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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 |
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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. |
Feb 12 |
comment |
Are there n polynomials for which all intersection multiplicities are at least m?
@Mahdi: I don't see why that would solve it. It would just change the intersection number over $x=a$, but not over other values of $x$. |
Feb 12 |
revised |
Are there n polynomials for which all intersection multiplicities are at least m?
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Feb 12 |
comment |
Are there n polynomials for which all intersection multiplicities are at least m?
Hmmm, let me clarify in the body of the question. |
Feb 12 |
asked | Are there n polynomials for which all intersection multiplicities are at least m? |
Feb 4 |
answered | Why should the anabelian geometry conjectures be true? |
Jan 25 |
asked | What is the obstruction for a local set of models of a curve to come from a global model? |
Dec 25 |
awarded | Popular Question |