Timeline for Can an algebraic number on the unit circle have a conjugate with absolute value different from 1?
Current License: CC BY-SA 3.0
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| Apr 10, 2015 at 21:52 | comment | added | YCor | OK (you said this after your edit :). We have $f_k(t)=\prod_i (t-x_i^k)$, all $x_i$ are algebraic integers and hence so are the coefficients of $f_k$. So my remark is just a (partly) distinct argument, where instead of using basic Galois theory, I use that the ring of invariants of $\mathbf{Z}[t_1,\dots,t_n]$ under the symmetric group $S_n$ is generated (as a ring) by the elementary symmetric polynomials in $t_1\dots,t_n$, and I apply this to coefficients of the polynomial $\prod (t-t_n)$, and then substitute $t_i$ to $x_i$. | |
| Apr 10, 2015 at 16:40 | comment | added | user6976 | As I said, the coefficients are algebraic integers and stable under the action of the Galois group. Hence these coefficients are rational integers. What's wrong? | |
| Apr 9, 2015 at 9:42 | comment | added | YCor | I know what Vieta's formulas are: they describe the coefficients of a monic polynomials in terms of elementary symmetric polynomials on the roots (which have integer coefficients). I refer to something else, namely the fact that any symmetric polynomial can be described as a polynomials (with integer coefficients) on the elementary symmetric polynomials. | |
| Apr 9, 2015 at 9:06 | comment | added | user6976 | @YCor: Do you know what Vieta's formulas are, en.wikipedia.org/wiki/Vieta%27s_formulas ? I have edited the answer making it more accessible. | |
| Apr 7, 2015 at 16:59 | history | edited | user6976 | CC BY-SA 3.0 |
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| Apr 7, 2015 at 12:04 | comment | added | YCor | It has to be said that the coefficients of $f_k$ are symmetric functions of the roots of $f$, and therefore belong to the subring generated by the elementary symmetric polynomials of $x_1\dots x_n$, that is, because $f$ is monic, the subring of $\mathbf{Z}[x_1,\dots,x_n]$ generated by coefficients of $f$. In particular the coefficients of $f_k$ are integers. | |
| Sep 14, 2010 at 15:18 | history | answered | user6976 | CC BY-SA 2.5 |