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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]
let us continue this discussion in chat
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 $\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$
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]
@KCd Yes it is. I updated the tags.
Mar
17
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 tags
Mar
17
awarded  Student