# Dinesh

Chennai, India

Age: 22

Interested in understanding the Inverse Galois Problem.

 Sep 29 awarded Supporter Sep 27 awarded Teacher Apr 2 awarded Yearling Apr 2 awarded Yearling 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 subgroupExactly! 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 subgroupYes. 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 subgroupadded 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 subgroupOk, 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 subgroupYes that's what I wrote in the question but will they 'generate' the group ? I think no. But the question in Munkres was to prove that <2x+3y> and are also a basis for ''. 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 25 comment Analog of Newton's theorem for symmetric polynomialsBy 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 polynomialsQiaochu 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 23 revised Analog of Newton's theorem for symmetric polynomialsedited body Aug 23 asked Analog of Newton's theorem for symmetric polynomials Jun 8 awarded Constituent Jun 8 awarded Caucus