Steven-Owen
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 Apr 1 accepted The field of rational functions in $n$ variables is a Galois extension of the field of symmetric rational polynomials in $n$ variables. Mar 31 awarded Peer Pressure Mar 31 comment The field of rational functions in $n$ variables is a Galois extension of the field of symmetric rational polynomials in $n$ variables. The intermediate field fixed by this is trivially the symmetric rational functions. So the extension is clearly Galois with Galois group $S_n$. Mar 31 comment The field of rational functions in $n$ variables is a Galois extension of the field of symmetric rational polynomials in $n$ variables. So it's trivial? By your lemma, given a finite group $G$ of automorphisms on a field K, then K is a Galois extension of $K^G$ and $G$ is a the Galois group of $K/K^G$. Since the by the symmetric functions theorem that says every symmetric polynomial with coefficients in a ring can be written as a polynomial in the elementary symmetric functions, the field generated by the elementary symmetric polynomials in $n$ variables is precisely the symmetric rational functions in $n$ variables. Now let $S_n$ act on the field of rational functions of $n$ variables by permuting the variables. Mar 31 revised The field of rational functions in $n$ variables is a Galois extension of the field of symmetric rational polynomials in $n$ variables. added 2 characters in body Mar 31 revised The field of rational functions in $n$ variables is a Galois extension of the field of symmetric rational polynomials in $n$ variables. deleted 209 characters in body Mar 31 revised The field of rational functions in $n$ variables is a Galois extension of the field of symmetric rational polynomials in $n$ variables. added 211 characters in body Mar 31 asked The field of rational functions in $n$ variables is a Galois extension of the field of symmetric rational polynomials in $n$ variables. Mar 21 awarded Commentator Mar 21 comment Let K/F be a finite extension, given a polynomial in K[x] find another so that their product is in F[x] This is exactly why I had put the ring theory tag on there as well. My course begins with rings and it seemed as though this problem could be solved using those methods, and here we have it. Thanks Darij! Mar 18 revised Let K/F be a finite extension, given a polynomial in K[x] find another so that their product is in F[x] edited title Mar 18 awarded Editor Mar 18 revised Finding the irreducible subrepresentations. added 167 characters in body Mar 18 comment Finding the irreducible subrepresentations. Do you understand what the question (the first paragraph) is asking (it's verbatim from my prof)? I can't seem to parse it into a question that makes sense. Mar 18 asked Finding the irreducible subrepresentations. Mar 17 awarded Supporter Mar 17 comment Let K/F be a finite extension, given a polynomial in K[x] find another so that their product is in F[x] Mar 17 comment How to prove $\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$ Apply this change to the integral on the right. You'll see that the change brought by the limits of integration and changing from dx to -du cancel out giving you the integral on the left with u instead of x. Mar 17 awarded Teacher Mar 17 answered How to prove $\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$