finding minimal polynomial for $\frac{\theta^5}{5}$ when minimal polynomial for $\theta$ is $x^3-2$

Question

suppose $f(x)= x^3-2$ is the minimal polynomial for $\theta$. How can I find a minimal polynomial for $a = \frac{\theta^5}{5}$

My idea was to substitute $u = \sqrt[5]{5 \cdot x} $

and then plug it into $f(x)$ and reduce it to a minimal polynomial.

While $f(u)$ would have $a$ as a root, it may not be in $\mathbb{Q}[x]$.

So my question is, is there any method to find a polynomial or even better a minimal polynomial?

A "better" minimal polynomial? There is only one up to a multiple by a degree zero polynomial (=a number different from zero) — DonAntonio, Jan 12 '17 at 18:27
Sorry, maybe it is a punctuation error. I meant ..a polynomial, or even better, a minimal polynomial. — user315567, Jan 12 '17 at 18:32

score 1 · Accepted Answer · answered Jan 12 '17 at 18:33

1

So we have that

$$\;\theta\in\left\{\,\sqrt[3]2,\,\sqrt[3]2\,\omega\,,\,\,\sqrt[3]2\,\omega^2\,\right\}\;,\;\;\omega:=e^{2\pi i/3}\implies\frac{\theta^5}5\in\left\{\,\frac{2^{5/3}}5\,,\,\,\frac{2^{5/3}\omega^2}5\,,\,\,\frac{2^{5/3}\omega}5\,\right\}$$

and now:

$$r=\frac{\theta^5}5\implies125r^3=2^5=32\implies g(x):=x^3-\frac{32}{125}\in\Bbb Q[x]$$

makes the cut.

answered Jan 12 '17 at 18:33

DonAntonio

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but it looks like I can use this method only, if I can separate $\theta$. what if $\theta$ is a root of $x^3-x-2$? – user315567 Jan 12 '17 at 18:54
@user315567 Then things get way tougher... – DonAntonio Jan 12 '17 at 18:56

finding minimal polynomial for $\frac{\theta^5}{5}$ when minimal polynomial for $\theta$ is $x^3-2$

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