# Splitting field of $X^5-2$ over $\mathbb{Q}$

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 $$A,B,C,D\in \mathbb{Q}(\sqrt[5]{2})$$.

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

• $x^{p-1}+x^{p-2}+\cdots+x+1$ is irreducible over $\mathbb{Q}$ for all primes $p$: do the shift $x=y+1$ and apply Eisenstein. – Arturo Magidin Feb 20 at 19:35
• @Arturo Magidin, i know this fact. I have to show irreducibility over $\mathbb{Q}(\sqrt[5]{2})$. – ZFR Feb 20 at 19:38
• 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....) – Arturo Magidin Feb 20 at 19:41
• @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. – user193319 Apr 22 at 19:13
• @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 $[\mathbb{Q}(\sqrt[5]{2},\omega):\mathbb{Q}] = [\mathbb{Q}(\sqrt[5]{2})(\omega):\mathbb{Q}(\sqrt[5]{2})][\mathbb{Q}(\sqrt[5]{2}):\mathbb{Q}]\leq (4)(5) = 20$. – Arturo Magidin Apr 22 at 21:07

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.
• Why is the degree at most $20$? – user193319 Apr 21 at 20:44