Timeline for Vanishing of permanent of a Vandermonde matrix [Edited]

Current License: CC BY-SA 3.0

12 events

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Jun 11, 2015 at 13:21	comment	added	Vesselin Dimitrov		If this is of any help, for each fixed $n$ there is an effective procedure to determine all root of unity solutions to $A(x_1,\ldots,x_n) = 0$, valid for an arbitrary polynomial $A$. See, for example, Bombieri and Zannier's paper "Algebraic points on subvarieties of $\mathbb{G}_m^n$."
Jun 11, 2015 at 13:17	history	edited	Mostafa - Free Palestine	CC BY-SA 3.0	added 116 characters in body
Jun 11, 2015 at 13:11	history	edited	Mostafa - Free Palestine	CC BY-SA 3.0	added 116 characters in body
Jun 11, 2015 at 9:27	comment	added	Mostafa - Free Palestine		@FedorPetrov Good question! in fact I didn't assumed that $x_i$'s generate $\mu_N$ and so the condition $n<N$ was redundant and I removed it. So in particular you can ask your question when all of $x_i$'s are $p$-roots of 1 and $p$ dose not divide $n$.
Jun 11, 2015 at 9:19	history	edited	Mostafa - Free Palestine	CC BY-SA 3.0	added 8 characters in body
Jun 11, 2015 at 9:09	comment	added	Fedor Petrov		Say, if $N=pq$ for primes $p,q$, could not it appear that $A$ is linear combination of (sum of $p$-roots of 1) and (sum of $q$-roots of 1)? If not, what is the reason? This specific question is rather combinatorial.
Jun 11, 2015 at 8:11	history	edited	Mostafa - Free Palestine	CC BY-SA 3.0	added 189 characters in body; edited title
Jun 11, 2015 at 8:03	history	edited	Mostafa - Free Palestine	CC BY-SA 3.0	added 189 characters in body; edited title
Jun 10, 2015 at 15:41	answer	added	Vesselin Dimitrov		timeline score: 5
Jun 10, 2015 at 14:28	history	edited	Mostafa - Free Palestine	CC BY-SA 3.0	added 38 characters in body
Jun 10, 2015 at 13:01	history	edited	Mostafa - Free Palestine	CC BY-SA 3.0	deleted 1 character in body
Jun 10, 2015 at 12:48	history	asked	Mostafa - Free Palestine	CC BY-SA 3.0