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Find the the splitting field of $X^5-2$ over $\mathbb{Q}$ and find it's degree.

My approach: The roots of $X^5-2$ are $\{\sqrt[5]{2},\sqrt[5]{2}\omega,\sqrt[5]{2}\omega^2, \sqrt[5]{2}\omega^3, \sqrt[5]{2}\omega^4\}$ where $\omega=e^{2\pi i/5}$.

It's quite easy to show that splitting field of $X^5-2$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt[5]{2},\omega)$.

Let's find the value of $[\mathbb{Q}(\sqrt[5]{2},\omega):\mathbb{Q}]$.

By tower's Theorem $[\mathbb{Q}(\sqrt[5]{2},\omega):\mathbb{Q}]=[\mathbb{Q}(\sqrt[5]{2},\omega):\mathbb{Q}(\sqrt[5]{2})][\mathbb{Q}(\sqrt[5]{2}):\mathbb{Q}]$ and it's obvious that $[\mathbb{Q}(\sqrt[5]{2}):\mathbb{Q}]=5$.

$\omega$ is the root of polynomial $X^4+X^3+X^2+X+1$ which shows that $[\mathbb{Q}(\sqrt[5]{2},\omega):\mathbb{Q}(\sqrt[5]{2})]\leq 4$.

How to show that polynomial $X^4+X^3+X^2+X+1$ is irreducible over $\mathbb{Q}(\sqrt[5]{2})$?

I was trying in that way: since $\omega \notin \mathbb{Q}(\sqrt[5]{2})$ then it factors as a product of quadratic polynomials $$X^4+X^3+X^2+X+1=(X^2+AX+B)(X^2+CX+D),$$ where $A,B,C,D\in \mathbb{Q}(\sqrt[5]{2})$.

How to get contradiction?

I would be very thankful if anyone can show how to complete this reasoning?

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    $\begingroup$ $x^{p-1}+x^{p-2}+\cdots+x+1$ is irreducible over $\mathbb{Q}$ for all primes $p$: do the shift $x=y+1$ and apply Eisenstein. $\endgroup$ – Arturo Magidin Feb 20 at 19:35
  • $\begingroup$ @Arturo Magidin, i know this fact. I have to show irreducibility over $\mathbb{Q}(\sqrt[5]{2})$. $\endgroup$ – ZFR Feb 20 at 19:38
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    $\begingroup$ No, you don't. You have a subextension of degree $5$, and a subextension of degree $4$. Their compositum must be of degree $20$. (as a consequence you can deduce the polynomial is irreducible over $\mathbb{Q}(\sqrt[5]{2})$, but you don't need to prove it ex nihilo....) $\endgroup$ – Arturo Magidin Feb 20 at 19:41
  • $\begingroup$ @ArturoMagidin Do you know why it's obvious that the degree of the extension $\Bbb{Q}(\sqrt[5]{2},\omega)$ is at most $20$? This is the only part I don't understand. $\endgroup$ – user193319 Apr 22 at 19:13
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    $\begingroup$ @user193319: Because the degree of $\sqrt[5]{2}$ is $5$, and the degree of $\omega$ is $4$ over $\mathbb{Q}$, and hence at most $4$ over $\mathbb{Q}(\sqrt[5]{2})$. Then Dedekind's Product Theorem tells you that $[\mathbb{Q}(\sqrt[5]{2},\omega):\mathbb{Q}] = [\mathbb{Q}(\sqrt[5]{2})(\omega):\mathbb{Q}(\sqrt[5]{2})][\mathbb{Q}(\sqrt[5]{2}):\mathbb{Q}]\leq (4)(5) = 20$. $\endgroup$ – Arturo Magidin Apr 22 at 21:07
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It has degree $20$. It has subfields $\Bbb Q(\sqrt[5]2)$ of degree $5$ (Eisenstein) and $\Bbb Q(\omega)$ of degree $4$ (cyclotomy, or Eisenstein again). So its degree is a multiple of $4$ and of $5$, so is s multiple of $20$. But clearly the degree is at most $20$ also.

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  • $\begingroup$ Indeed, great answer! +1 $\endgroup$ – ZFR Feb 21 at 16:48
  • $\begingroup$ Sorry, but do you have any good answer on this math.stackexchange.com/questions/3121481/…? $\endgroup$ – ZFR Feb 21 at 16:48
  • $\begingroup$ Why is the degree at most $20$? $\endgroup$ – user193319 Apr 21 at 20:44

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