Let K be a field and let F = <mrow class="MJX-TeXAtom-ORD"> P Q </mf

Savanah Boone 2022-07-01 Answered
Let K be a field and let F = P Q K ( X ) be a rational fraction, for simplicity we denote also by F the rational function associated to the rational fraction F. It is clear that if P and Q are both even or both odd polynomial functions, then F is an even rational function and we have also that if one of the two polynomial functions is even and the other is odd, then F must be odd. Now is the converse true, I mean do we have that if F is an even rational function then necessarely both P and Q are both odd or both even and that if F is odd then necesserely one is odd and the other is even ?
I think it is true, but i don't know how to prove it, suppose that F is even then
P ( x ) Q ( x ) = P ( x ) Q ( x )
Hence
P ( x ) Q ( x ) = P ( x ) Q ( x )
But how to go from here?
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Answers (2)

Jordin Church
Answered 2022-07-02 Author has 11 answers
You may suppose that P and Q have no commun divisors in K [ x ]. Your equality P ( x ) Q ( x ) = P ( x ) Q ( x ) show then that as P ( x ) divide P ( x ) Q ( x ) and is prime to Q ( x ), P ( x ) must divide P ( x ). As they have the same degree, there exists c such that P ( x ) = c P ( x ). Replacing x by x, we get c 2 = 1, hence c = 1 or c = 1 and we are done.
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Grimanijd
Answered 2022-07-03 Author has 4 answers
Take Q = x + x 2 , P = x Q. Then F = x, and F is odd, but Q isn't odd or even.
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