tea.mathoverflow.net - Discussion Feed (A post)

awllower comments on "A post" (13170)

awllower — Tue, 08 Feb 2011 20:17:50 -0800

Well, I didn't have enough reputation points at that time.
And thank you for telling me how the transformation works.
In any case, thank you.

voloch comments on "A post" (13169)

voloch — Tue, 08 Feb 2011 20:14:31 -0800

This is not an appropriate discussion for meta. This could have been asked in the comments of the post in question.

Ryan Reich comments on "A post" (13168)

Ryan Reich — Tue, 08 Feb 2011 19:27:20 -0800

The coefficient of $x^n$ is $a^{n/(n - 1)}$, while that of $x$ is $a * a^{1/(n - 1)} = a^{n/(n - 1)}$, so in canceling one, you cancel both.

Valerio Talamanca comments on "A post" (13167)

Valerio Talamanca — Tue, 08 Feb 2011 19:27:13 -0800

in fact they are both 1, since the transformation proposed has the effect of making the leading coefficent and the coefficent of x equal.

awllower comments on "A post" (13166)

awllower — Tue, 08 Feb 2011 19:01:51 -0800

But he required that both the leading coefficient and the coefficient of x are 1, didn't him?

Gerry Myerson comments on "A post" (13150)

Gerry Myerson — Tue, 08 Feb 2011 04:08:30 -0800

It makes the leading coefficient not 1, but then you just divide by that coefficient to get a polynomial of the desired form, no?

awllower comments on "A post" (13147)

awllower — Tue, 08 Feb 2011 02:36:36 -0800

I have read a post in MO
http://mathoverflow.net/questions/48855/galois-theory-generalization-of-abels-theorem-better-version
and found a somewhat strange answer, i.e. the only answer.
The answer says that this transformation can change those two polynomials, but why is it acceptable?? It makes the leading coefficient not 1, doesn't it?
Please inform me, thank you.