# On irreducible polynomial

Prove that the polynomial $f(X)=X^5-9X^3+15X+6$ is irreducible over $\mathbb{Q}\left(\sqrt{2},\sqrt{3}\right)$

Apply Eisenstein's Irreducibility Criterion with $p=3$ we see that $f$ is irreducible over $\mathbb{Q}$. Can conclude that $f$ is irreducible over $\mathbb{Q}\left(\sqrt{2},\sqrt{3}\right)$?

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No, you can't${}$. – Git Gud Nov 17 '13 at 12:56
What is $\mathbb{Q}(\sqrt2,\sqrt3)$ again? Is that the polynomial ring over the rationals modulo $\sqrt2$ and $sqrt3$? – Caleb Jares Nov 17 '13 at 18:19

You can't directly conclude that $f$ is irreducible over $\mathbb{Q}(\sqrt{2},\sqrt{3})$ from its irreducibility over $\mathbb{Q}$. But since the degree of $f$ is $5$, you can conclude that for any zero $\alpha$ of $f$,

$$[\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha):\mathbb{Q}] = [\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha):\mathbb{Q}(\alpha)]\cdot [\mathbb{Q}(\alpha):\mathbb{Q}]$$

is a multiple of $5$. On the other hand,

\begin{align} [\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha):\mathbb{Q}] &= [\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha):\mathbb{Q}(\sqrt{2},\sqrt{3})]\cdot [\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]\\ &= 4[\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha):\mathbb{Q}(\sqrt{2},\sqrt{3})], \end{align}

so

$$5 \mid [\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha):\mathbb{Q}(\sqrt{2},\sqrt{3})].$$

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I assume that this is an extended hint (rolls eyes). +1 for that! – Jyrki Lahtonen Nov 17 '13 at 13:14
I don't get why $[\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha):\mathbb{Q}]$ is a multiple of $5$. – Git Gud Nov 17 '13 at 13:15
@GitGud What's $[\mathbb{Q}(\alpha) : \mathbb{Q}]$? – Daniel Fischer Nov 17 '13 at 13:16
The degree of the minimal polynomial of $\alpha$ over $\Bbb Q$, which I don't know what it is. Edit: nevermind, I do after all. Another edit: thanks. – Git Gud Nov 17 '13 at 13:16
@Daniel Fischer: Thank you so much for your support – chuyenvien94 Nov 17 '13 at 13:36

It is not hard to see that $[{\mathbb Q}(\sqrt{2},\sqrt{3}):{\mathbb Q}]=4$. Since $4$ and $5$ are coprime we are done, because of the classical result proved here.

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Thank you very much. – chuyenvien94 Nov 17 '13 at 14:45