Minimal polynomial of $\sqrt[10]{5}$ over $\mathbb{Q}(e^{2\pi i/10})$

Question

Determine the minimal polynomial of $\sqrt[10]{5}$ over $\mathbb{Q}(e^{2\pi i/10})$.

Progress so far: 1) Its degree must divide $10$ (since the extension $\mathbb{Q}(\sqrt[10]{5},e^{2\pi i/10})/\mathbb{Q}(e^{2\pi i/10})$ is a Kummer extension and it is well known that such have a cyclic Galois group of order dividing $10$.

2 ) If $\omega = e^{2\pi i/10}$, then $\omega^4 - \omega^3 + \omega^2 - \omega + 1 = 0$, so $(\omega + \omega^{-1})^2 - (\omega + \omega^{-1}) - 1 = 0$ and with $\omega + \omega^{-1} = 2\cos(\frac{\pi}{5}) > 0$ we obtain $(\sqrt[10]{5})^5 = \sqrt{5} = 2(\omega + \omega^{-1}) - 1$, so there is an annihilating polynomial of degree $5$.

Any help appreciated, especially if not too much theory is involved!

Answer 1

The easiest way to see the answer to this is to use the well-known result that for any field $ F $ and for a prime $ p $, the polynomial $ X^p - a $ is irreducible in $ F[X] $ if and only if $ a $ is not a $ p $-th power in $ F $.

The cyclotomic field $ \mathbf Q(\zeta_{10}) $ contains $ \sqrt{5} $ and can't contain any root of $ 5 $ of a higher degree because the extension over $ \mathbf Q $ is abelian and such a root would give rise to a subextension that is not Galois, which would result in a contradiction. Therefore $ \sqrt 5 $ has no roots of any nontrivial degree in $ \mathbf Q(\zeta_{10}) $ and by the above result this implies that the polynomial $ X^5 - \sqrt{5} $ is irreducible in $ \mathbf Q(\zeta_{10})[X] $, making it the minimal polynomial.

Answer 2

Let $$a=5^{1/10},b=e^{2\pi i/10},c=a^5=\sqrt{5}$$ You have already shown that $b$ is a root of $x^4-x^3+x^2-x+1\in\mathbb {Q} [x] $ and $$b+b^{-1}=2\cos(\pi/5)=\frac{1+c}{2}$$ This means that $c\in\mathbb {Q} (b) $ and $a$ is a root of $x^5-c\in\mathbb {Q} (b)[x] $.

So $[\mathbb {Q} (a,b) :\mathbb{Q} (b)] \leq 5$. On the other hand $[\mathbb {Q} (a) :\mathbb {Q}] =10$ and $[\mathbb{Q} (b) :\mathbb {Q}] =4$. Using these we can see that $$[\mathbb {Q} (a, b) :\mathbb{Q} (a)] =\frac{[\mathbb {Q} (a, b) :\mathbb {Q} (b)] [\mathbb {Q} (b) :\mathbb {Q}]} {[\mathbb {Q} (a) :\mathbb {Q}]} \leq 2$$ Next note that $\mathbb{Q} (a) \subseteq \mathbb {R} $ and $b\notin\mathbb {R} $ and hence $[\mathbb{Q} (a, b) :\mathbb{Q} (a)] >1$ and then we have $[\mathbb{Q} (a, b) :\mathbb {Q} (a)] =2$.

It now follows that $[\mathbb {Q} (a, b) :\mathbb {Q} (b)] =5$ and hence $x^5-c\in\mathbb {Q} (b) [x] $ is the desired minimal polynomial for $a$.