Conjugates of ${\sqrt 5 + \sqrt 6 }$ over $\mathbb{Q}$ and irreducibility of $x^4 -22x^2+1$ (for minimal polynomial)

Question

The easy part is algebraic manipulation:

$$x - \left( {\sqrt 5 + \sqrt 6 } \right) = 0 \Rightarrow {x^2} = 5 + 6 + 2\sqrt {30} \Rightarrow {x^2} - 11 = 2\sqrt {30} \Rightarrow {x^4} - 22{x^2} + 1 = 0$$ 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 $p\left( x \right) = {x^4} - 22{x^2} + 1$ 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 ${\sqrt 5 + \sqrt 6 }$ ?

Answer 1

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 $2$.

Answer 2

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.

Answer 3

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.