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seen Aug 5 at 8:32

Interested in understanding the Inverse Galois Problem.


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awarded  Curious
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awarded  Yearling
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awarded  Popular Question
Jan
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accepted Analog of Newton's theorem for symmetric polynomials
Jan
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awarded  Popular Question
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awarded  Yearling
Oct
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comment Let $x$ , $y$ be the basis of a free abelian group of rank 2, prove that $2x+3y$ and $ x-y$ generate a free subgroup
Exactly! That was the whole point of this question. Thanks for confirming it.
Oct
12
comment Let $x$ , $y$ be the basis of a free abelian group of rank 2, prove that $2x+3y$ and $ x-y$ generate a free subgroup
Yes. Thats what I wanted to confirm for myself. Thanks!
Oct
12
revised Let $x$ , $y$ be the basis of a free abelian group of rank 2, prove that $2x+3y$ and $ x-y$ generate a free subgroup
added 152 characters in body
Oct
12
comment Let $x$ , $y$ be the basis of a free abelian group of rank 2, prove that $2x+3y$ and $ x-y$ generate a free subgroup
Ok, apologies for an unclear question. The question verbatim from Munkres is this: "If $G$ is free abelian with basis {x,y}, show that {2x+3y,x-y} is also a basis for $G$"
Oct
12
comment Let $x$ , $y$ be the basis of a free abelian group of rank 2, prove that $2x+3y$ and $ x-y$ generate a free subgroup
Yes that's what I wrote in the question but will they 'generate' the group <x,y> ? I think no. But the question in Munkres was to prove that <2x+3y> and <x-y> are also a basis for '<x,y>'.
Oct
12
comment Let $x$ , $y$ be the basis of a free abelian group of rank 2, prove that $2x+3y$ and $ x-y$ generate a free subgroup
@Thomas Andrews that was clear but I am saying that $2x+3y$ and $x-y$ doest span <$x$,$y$>
Oct
12
awarded  Custodian
Oct
12
reviewed Approve suggested edit on Let $x$ , $y$ be the basis of a free abelian group of rank 2, prove that $2x+3y$ and $ x-y$ generate a free subgroup
Oct
12
asked Let $x$ , $y$ be the basis of a free abelian group of rank 2, prove that $2x+3y$ and $ x-y$ generate a free subgroup
Aug
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comment Analog of Newton's theorem for symmetric polynomials
By a field rational over $\mathbb{Q}$ I mean a field of the form $\mathbb{Q}(t_1,...,t_n)$ where $t_1,...,t_n$ are algebraically independent.
Aug
24
comment Analog of Newton's theorem for symmetric polynomials
..contd. In fact it gives an explicit polynomial for $G$ (when we have enough information). So if I know the analog of newton's theorem for $C_p$ then I can realize $C_p$ and might also be able to produce an explicit polynomial for it. The reason I chose $n$ to be prime is that, Lenstra showed that the fixed field of $C_8$ is not rational over $\mathbb{Q}$ (Reference: Serre's Topics in Galois theory).
Aug
24
comment Analog of Newton's theorem for symmetric polynomials
Qiaochu Yuan Thanks for your answer. I really meant $\mathbb{Q}$ not $\mathbb{C}$ , though it is interesting to note how it really matters and I am wondering how it does not matter in the case $S_{n}$. The motivation behind this question is Noether's formulation of inverse Galois problem( see 'Introduction' in Serre's Topics in Galois thory). It says that when a permutation group $G$ acts on $\mathbb{Q}(x_1,..x_n)$ and if its fixed field is rational over $\mathbb{Q}$ then via Hilbert's irreducibility theorem we can realize $G$ over $\mathbb{Q}$.
Aug
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revised Analog of Newton's theorem for symmetric polynomials
edited body
Aug
23
asked Analog of Newton's theorem for symmetric polynomials