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Let $F = \{ x^{4}+1,x^{5}+2\}$ in $\mathbb{Q}\left[ x \right]$ and $L$ its splitting field over $\mathbb{Q}$. I'm asked to explicitly find $L$ as an algebraic extension of $\mathbb{Q}$ and compute its degree $\left[ L:\mathbb{Q} \right]$.

My work:

Let $\alpha$ be a fourth root of -1, $\beta$ be a primitive fifth root of -1, then the splitting field $L$ can be written as (by looking at the roots of the two polynomials):

$$L=\mathbb{Q}(\alpha,\beta,\sqrt[5]{2})$$

But I'm struggling to characterise the intermediate extensions in order to compute the extension degree:

  • Clearly $\mathbb{Q}(\sqrt[5]{2})\subsetneq \mathbb{Q}(\alpha,\sqrt[5]{2})$ and $\mathbb{Q}(\sqrt[5]{2})\subsetneq \mathbb{Q}(\beta,\sqrt[5]{2})$, as $\mathbb{Q}(\sqrt[5]{2})\subseteq \mathbb{R}$
  • Then $\left[\mathbb{Q}(\sqrt[5]{2}):\mathbb{Q} \right] = 5$ as $x^5 + 2$ is irreducible in $\mathbb{Q}\left[ x \right]$ by Eisenstein criterion

But I do't know how to proceed (maybe something related to the fact that $L$ by being a splitting field it's a normal extension of $\mathbb{Q}$). Thanks for your help.

user26857
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iki
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    Since $\mathbb{Q}(\alpha,\beta)$ has subextensions of degree $5$ and $4$, which generate it (namely $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\beta)$), the degree is at most $5\times 4=20$, and must be a multiple of both $5$ and $4$, so the extension has degree exactly $20$. So the question is whether $x^5+2$ factors over this field in some way or not. – Arturo Magidin Oct 19 '23 at 18:37
  • Why is $\mathbb{Q}(\alpha)$ of degree 5? Isn’t it of degree at most 4? @ArturoMagidin – iki Oct 20 '23 at 09:34
  • $\alpha$ is degree $4$, $\beta$ is degree $5$. Poor phrasing. – Arturo Magidin Oct 20 '23 at 13:07
  • But then again the minimal polynomial for $\beta$ over $\mathbb{Q}$ has degree at most 4 @ArturoMagidin – iki Oct 20 '23 at 15:01
  • Oh, right... You can get $20$ by using $\sqrt[5]{2}$ and either $\alpha$ or $\beta$, in that case. – Arturo Magidin Oct 20 '23 at 15:38
  • But yes know that we know that there’s a subextension of degree 20 how can we continue? @ArturoMagidin – iki Oct 20 '23 at 15:51
  • You really should specify that $\beta$ is a primitive $5$th root of unity, not "a fifth root of $-1$" (that would be $-1$, for example). That's what you need to get a splitting field of $x^5+2$. So check the irreducibility fo $x^4+1$ over $\mathbb{Q}(\beta,\sqrt[5]{2})$ and proceed. I'm not going to solve this for you. – Arturo Magidin Oct 20 '23 at 16:09
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    Take $\alpha$ a primitive 8th root unity, $\beta$ a primitive 5th root of unity. and add $\alpha$, $\beta$ and $\sqrt[5]{-2}$ to $\mathbb Q$ in order to get $L$. Now notice that $[\mathbb Q(\alpha,\beta):\mathbb Q]=16$. (See https://math.stackexchange.com/questions/2838521.) Then $[L:\mathbb Q]=5\cdot 16=80$. – user26857 Oct 21 '23 at 07:41
  • @user26857 the link is broken on my end – iki Oct 21 '23 at 08:28
  • https://math.stackexchange.com/questions/2838521 – user26857 Oct 21 '23 at 08:31

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