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The easy part is algebraic manipulation:

$x - \left( {\sqrt 5 + \sqrt 6 } \right) = 0 \Rightarrow {x^2} = 5 + 6 + 2\sqrt {30} \Rightarrow {x^2} - 11 = 2\sqrt {30} \Rightarrow p\left( x \right) = {x^4} - 22{x^2} + 1 = 0$. Solutions to this equation are ${x_{1,2,3,4}} = \pm \sqrt 5 \pm \sqrt 6 $ and these are all candidates for the conjugates. However, not all of them are necessarily conjugates of ${\sqrt 5 + \sqrt 6 }$.

The hard part is proving that $p$ is the minimal polynomial of ${\sqrt 5 + \sqrt 6 }$ over $\mathbb{Q}$. Eisenstein criterion doesn't give irreducibility when taking $p\left( {x + a} \right)$.

Even though $p$ has no rational zeroes, that still doesn't mean that it is irreducible over $\mathbb{Q}$.

One method which I can think of would be to calculate each of 7 divisors of $p$ and show that they are not in $\mathbb{Q}\left[ X \right]$. However, that would be impractical in the next problem which has a polynomial of degree 31.

How to see that all the candidates are indeed the conjugates of ${\sqrt 5 + \sqrt 6 }$ ?

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4 Answers 4

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You can show by a direct calculation that $\mathbb{Q}(\sqrt{5}+\sqrt{6})=\mathbb{Q}(\sqrt{5},\sqrt{6})$. The latter is the splitting field of $(x^2-5)(x^2-6)$ over $\mathbb{Q}$, hence is a Galois extension of $\mathbb{Q}$. You need to show that this field has four automorphisms, which by the fundamental theorem is the same as showing it has degree $4$ as an extension of $\mathbb{Q}$. But $[\mathbb{Q}(\sqrt{5},\sqrt{6}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{5},\sqrt{6}):\mathbb{Q}(\sqrt{5})]\cdot[\mathbb{Q}(\sqrt{5}):\mathbb{Q}]$, and it's not hard to see that both factors are $2$.

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For completeness, here's a proof without resorting to Galois theory:

By Gauss' lemma, it suffices to show that $p(x)$ is irreducible in $\mathbb{Z}[X]$. It has no integer zeroes, so if it factors it must be into two quadratics as: $$p(x)=(x^2+ax+b)(x^2+cx+d)=x^4+x^3(a+c)+x^2(b+d+ac)+x(ad+bc)+bd$$

Thus we get: $$c=-a$$ $$bd=1$$ $$a(d-b)=0$$ $$b+d-a^2=-22$$ It follows immediately that $b=d=\pm1$, from which it follows that $a^2=22\pm2$, contradicting that $a \in \mathbb{Z}$.

This proof also immediately generalises to show that $x^4-kx^2+1$ is reducible in $\mathbb{Q}[X]$ iff one of $k\pm2$ is a perfect square.

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One way to proceed would be to show that the extension field $\mathbb{Q}(\sqrt{5} + \sqrt{6}) / \mathbb{Q}$ has degree $4$. If you can do this, then since you've already found a degree $4$ polynomial for the primitive element $\alpha = \sqrt{5} + \sqrt{6}$, then that polynomial must be the minimum polynomial so the conjugates would be the ones you listed.

To prove that the extension has degree $4$, you can try proving that actually $\mathbb{Q}(\sqrt{5} + \sqrt{6}) = \mathbb{Q}(\sqrt{6}, \sqrt{5})$ and this is not hard to do.

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The direct method is to show that $\mathbb Q(\sqrt 5+\sqrt 6) \cong \mathbb Q(\sqrt 5)(\sqrt 6)$, then show explicit automorphisms of the second field that sends $\sqrt 5+\sqrt 6$ to its conjugates.

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