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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.

ACB
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Anakhand
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    Maybe you can prove that $\mathbb{Q}(\sqrt{2} + \sqrt{5})$ has degree $4$ over $\mathbb{Q}$ ? – TheSilverDoe Feb 22 '22 at 18:58
  • Alt. hint: the polynomial factors over reals as the product of $,x \pm \sqrt{2} \pm \sqrt{5},$, and the product of any two of those factors is not a rational polynomial. – dxiv Feb 22 '22 at 22:07

2 Answers2

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$f=X^4-14X^2+9$ 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 $\Bbb Q$.

Dietrich Burde
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    One can slightly speed up the comparison of a hypothetical quadratic factorization by observing that $f$ is invariant under the reflection $X \mapsto -X$, so that reflection must either fix both factors, in which case $a = c = 0$, or interchange them, in which case $c = -a, d = b$. The former case amounts to factoring a quadratic in $u := X^2$, but the quadratic has discriminant $160$, which is a nonsquare. The latter amounts to checking whether there are integers $a, b$ such that $a^2 - 2 b = 14$ and $b^2 = 9$. – Travis Willse Feb 22 '22 at 19:24
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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 $\sqrt{2} + \sqrt{5}$.

TheSilverDoe
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