3
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ What is $\mathbb{Q}(\sqrt2,\sqrt3)$ again? Is that the polynomial ring over the rationals modulo $\sqrt2$ and $sqrt3$? $\endgroup$ Nov 17 '13 at 18:19
11
$\begingroup$

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

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

$\endgroup$
0
1
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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