Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\mathbb{Q}\left(\sqrt{2},\sqrt{3}\right)$?

share|cite|improve this question
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
up vote 11 down vote accepted

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,

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


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

share|cite|improve this answer
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.

share|cite|improve this answer
Thank you very much. – chuyenvien94 Nov 17 '13 at 14:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.