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accepted Analog of Newton's theorem for symmetric polynomials
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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 <$x$,$y$>
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reviewed Approve 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
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25
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.