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

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
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
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?
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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.
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?
added 141 characters in body; added 6 characters in body; added 44 characters in body
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?