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This is a review problem for an algebra exam:

Let $\alpha, \beta$ be distinct roots of the polynomial $f(x)=x^5 - 20 \in \mathbb Q[x]$, and let $K$ be the splitting field of $f(x)$.

  1. Show that $K= \mathbb Q(\alpha, \beta)$.
  2. What is the degree of the extension $K/ \mathbb Q$?
  3. Show that $K \neq \mathbb Q(\alpha + \beta)$.

Work so far. I know that the roots of $f(x)$ are $\zeta^k \sqrt[5]{20}$ where $k=0,1,2,3,4$ and $\zeta$ is a primitive 5th root of unity. Therefore $K=\mathbb Q(\zeta, \sqrt[5]{20})$.

For (1), I can show this by considering the quotient $\alpha / \beta$.

For (2), I can show that the degree is $20$ by using the minimal polynomials of $\zeta$ and $\sqrt[5]{20}$ and then using the tower property of field extensions.

However, I'm stuck on (3). Of course $\mathbb Q(\alpha + \beta) \subseteq K$, so I just need to show that $K$ is not contained in $\mathbb Q(\alpha + \beta)$. For this it would suffice to show that $\zeta$ or $\sqrt[5]{20}$ are not contained in $\mathbb Q(\alpha + \beta)$. I'm not sure how to show this directly, since I'm not sure how to write down an arbitrary explicit element of $\mathbb Q(\alpha+ \beta)$ (i.e., as a $\mathbb Q$-linear combination of powers of $\alpha+\beta$). I also considered showing that $\mathbb Q(\alpha + \beta)/\mathbb Q$ has a different degree than $K/\mathbb Q$, but I'm not sure how to find the minimal polynomial of $\alpha + \beta$ or otherwise calculate this degree.

Question. Is showing the degrees are different the right strategy for (3)? If so, how can I calculate the degree of $\mathbb Q(\alpha + \beta)/ \mathbb Q$?

algebra learner
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  • You may assume, that $\alpha = \sqrt[5]{20}$ and that $\beta = \zeta \alpha$, since any nontrivial 5th root of unity is primitive. You could then try to calculate the minimal polynomial of $\alpha+\beta$, maybe there is a better way to do it, but that's something you can work with. – Julian Quast Apr 28 '25 at 17:51
  • See also this post for the splitting field of $x^n-a$, for $n=5$ and $a=20$. The degree over $\Bbb Q$ is $n\phi(n)=5\cdot 4=20$, as you have found. – Dietrich Burde Apr 28 '25 at 18:14
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    $(\sqrt[5]20+\zeta\sqrt[5]{20})^5$ is in $\mathbb Q(\zeta)\cap\mathbb R$, which has degree $2$ over $\mathbb Q$ – J. W. Tanner Apr 28 '25 at 18:30

1 Answers1

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Combining the very helpful comments, we can take, for example, $\alpha = \sqrt[5]{20}$ and $\beta = \zeta\alpha$. Then from J.W. Tanner's comment we can show $(\sqrt[5]{20} + \zeta \sqrt[5]{20})^5 = (\alpha + \beta)^5 \in \mathbb{Q}(\zeta)\cap \mathbb{R}$.

So $\alpha+\beta$ has degree at most 5 over $\mathbb{Q}(\zeta)\cap \mathbb{R}$, and since $\mathbb{Q}(\zeta)\cap \mathbb{R}$ has degree 2 over $\mathbb{Q}$, by the tower property of field extensions we have that $\mathbb{Q}(\alpha+\beta)$ has at most degree $5\cdot 2 = 10$ over $\mathbb{Q}$. But $[\mathbb{Q}(\alpha, \beta): \mathbb{Q}] = 20$, as you've shown, so these two field extensions cannot be equal.

(To see that $(\alpha+\beta)^5 \in \mathbb{Q}(\zeta)\cap \mathbb{R}$, we have $(\sqrt[5]{20} + \zeta \sqrt[5]{20})^5 = (\sqrt[5]{20}(1+\zeta))^5 = 20(1+\zeta)^5$. Then expanding with the binomial formula we then get $$(1+\zeta)^5 = 1+5\zeta+10\zeta^2+10\zeta^3+5\zeta^4+\zeta^5$$ and conjugating gives us the same element, just with the orders of the elements reversed. This seems like quite a convoluted way to do this. Anyone have a better way?)

yms
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