Proving $X^4 - 14X^2 + 9$ is irreducible

Question

I have to prove that $X^4 - 14X^2 + 9$ is the minimal polynomial of $\sqrt{2} + \sqrt{5}$ over $\mathbb{Q}$, and to do so I have to prove that it is irreducible over $\mathbb{Q}$. The only tools I have are Einstein's criterion, the mod-$p$ test, and the fact that translations of a polynomial maintain reducibility.

We can't apply Einstein's criterion since the constant term is $9 = 3^2$, and $3$ doesn't divide $14$. Neither the mod-2 nor the mod-7 tests work: in mod-2, $X^4 + 1$ decomposes as $(X^2 + 1)^2$, and in mod-7, $X^4 + 2$ decomposes as $(X^2 + X + 4)(X^2 +6X +4)$.

I could use Perron's irreducibility criterion, however I'm not allowed to use this result.

Answer 1

Let $d = [\mathbb{Q}(\sqrt{2} + \sqrt{5}) : \mathbb{Q}]$.

The fact that $\sqrt{2} + \sqrt{5}$ is a root of $X^4 - 14X^2 + 9$ implies that $d \leq 4$.

But one also has $d=[\mathbb{Q}(\sqrt{2} + \sqrt{5}) : \mathbb{Q}(\sqrt{2})] \times [\mathbb{Q}(\sqrt{2}) : \mathbb{Q}] = 2 \times [\mathbb{Q}(\sqrt{2} + \sqrt{5}) : \mathbb{Q}(\sqrt{2})] $, and because $ [\mathbb{Q}(\sqrt{2} + \sqrt{5}) : \mathbb{Q}(\sqrt{2})] \geq 2$, you get that $d \geq 4$.

So $d=4$, so $X^4 - 14X^2 + 9$ is the minimal polynomial of $\sqrt{2} + \sqrt{5}$.

Answer 2

$f=X^4-14X^2+9$ is not irreducible for all primes $p$, compare with this post:

Proving that $x^4 - 10x^2 + 1$ is not irreducible over $\mathbb{Z}_p$ for any prime $p$.

However, by the rational root test it has no rational root, and a decomposition $$ f=(X^2+aX+b)(X^2+cX+d) $$ is impossible by comparing coefficients and solving the equations over the integers. Hence $f$ is irreducible over $\Bbb Z$ and then also over $\Bbb Q$.