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Let $\alpha$ be a complex number.

Suppose there exists a a monic polynomial $f(x) \in \mathbb{Z}[x]$ such that $f(\alpha)=0$. Show that there exists a monic polynomial $g(x) \in \mathbb{Z}[x]$ such that $g(\alpha ^2)=0$.

I tried to compute the elementary symmetric polynomials in order to put them in the coefficients, but it became increasingly non-practical, and I had no way to guarantee that the numbers are integers.

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    $\begingroup$ You probably want to exclude $f=0$; and $g=0$. $\endgroup$ – Marc van Leeuwen Apr 8 '14 at 13:31
  • $\begingroup$ Included more information, thank you. $\endgroup$ – Aloizio Macedo Apr 8 '14 at 13:35
  • $\begingroup$ But "monic" is much stronger than "nonzero". $\endgroup$ – Marc van Leeuwen Apr 8 '14 at 13:45
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    $\begingroup$ The result can be computed as resultant, $g(y)=Res_x(f(x),y-x^2)$. If the properties of the resultant are known, the claims follow directly. $\endgroup$ – Dr. Lutz Lehmann Apr 9 '14 at 5:35
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Hint: What's the relationship between the coefficients of $\prod_{i=1}^n (x-r_i)$ and $\prod_{i=1}^n (x+r_i)$?

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    $\begingroup$ This root-squaring procedure is the basis of the Dandelin-Gräffe method. -- More generally, if $q=\exp(i\,2\pi/d)$, then $f(x)f(qx)...f(q^{d-1}x)$ is a polynomial $g(x^d)$ that only contains powers of $x^d$. $\endgroup$ – Dr. Lutz Lehmann Apr 9 '14 at 5:32

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