tea.mathoverflow.net - Discussion Feed (A post)2019-10-21T14:24:14-07:00http://mathoverflow.tqft.net/
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awllower comments on "A post" (13170)http://mathoverflow.tqft.net/discussion/948/a-post/?Focus=13170#Comment_131702011-02-08T20:17:50-08:002019-10-21T14:24:14-07:00awllowerhttp://mathoverflow.tqft.net/account/498/
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.
And thank you for telling me how the transformation works. In any case, thank you.]]>
voloch comments on "A post" (13169)http://mathoverflow.tqft.net/discussion/948/a-post/?Focus=13169#Comment_131692011-02-08T20:14:31-08:002019-10-21T14:24:14-07:00volochhttp://mathoverflow.tqft.net/account/211/
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)http://mathoverflow.tqft.net/discussion/948/a-post/?Focus=13168#Comment_131682011-02-08T19:27:20-08:002019-10-21T14:24:14-07:00Ryan Reichhttp://mathoverflow.tqft.net/account/434/
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.
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.
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Valerio Talamanca comments on "A post" (13167)http://mathoverflow.tqft.net/discussion/948/a-post/?Focus=13167#Comment_131672011-02-08T19:27:13-08:002019-10-21T14:24:14-07:00Valerio Talamancahttp://mathoverflow.tqft.net/account/492/
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)http://mathoverflow.tqft.net/discussion/948/a-post/?Focus=13166#Comment_131662011-02-08T19:01:51-08:002019-10-21T14:24:14-07:00awllowerhttp://mathoverflow.tqft.net/account/498/
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)http://mathoverflow.tqft.net/discussion/948/a-post/?Focus=13150#Comment_131502011-02-08T04:08:30-08:002019-10-21T14:24:14-07:00Gerry Myersonhttp://mathoverflow.tqft.net/account/370/
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)http://mathoverflow.tqft.net/discussion/948/a-post/?Focus=13147#Comment_131472011-02-08T02:36:36-08:002019-10-21T14:24:14-07:00awllowerhttp://mathoverflow.tqft.net/account/498/
I have read a post in MOhttp://mathoverflow.net/questions/48855/galois-theory-generalization-of-abels-theorem-better-versionand found a somewhat strange answer, i.e. the only answer.The answer says ...
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.]]>