<s>[INST] Show $\mathbb{Q}( \sqrt{5},\sqrt{7} ) = \mathbb{Q}( \sqrt{5} + \sqrt{7} )$ The problem :

*

*find the minimal polynomial of $\sqrt{5} + \sqrt{7}$

*what is the degree of $ [ \mathbb{Q}( \sqrt{5},\sqrt{7} ) : \mathbb{Q} ] $

*conclude $\mathbb{Q}( \sqrt{5},\sqrt{7} ) = \mathbb{Q}( \sqrt{5} + \sqrt{7} )$
My question and my works
I found 1., I found that the minimal polynomial of $\sqrt{5} + \sqrt{7}$ is $P : X \mapsto X^4 - 24X^2 + 4 $
I know that $\deg P = [ \mathbb{Q}( \sqrt{5} + \sqrt{7} ): \mathbb{Q} ]$ where $P$ is the min. polynomial of $\sqrt{5} + \sqrt{7}$ over $\mathbb{Q}$. But, how can i calculate $ [ \mathbb{Q}( \sqrt{5},\sqrt{7} ) : \mathbb{Q} ] $ with 1. ?
I found 3., by showing each inclusion, without using 2...
Thanks you
 [/INST] Once you know that $[ \mathbb{Q}( \sqrt{5},\sqrt{7} ) : \mathbb{Q} ]=4$, with basis $\{1,\sqrt{5},\sqrt{7},\sqrt{35}\}$, you can proceed as follows, without finding the minimal polynomial of $\sqrt{5}+\sqrt{7}$.
Let $\alpha=\sqrt{5}+\sqrt{7}$. Then
$$
\begin{pmatrix} 1 \\ \alpha \\ \alpha^2 \\ \alpha^3 \end{pmatrix}
=
\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 12 & 0 & 0 & 2 \\ 0 & 26 & 22 & 0
 \end{pmatrix}
\begin{pmatrix} 1 \\ \sqrt{5} \\ \sqrt{7} \\ \sqrt{35} \end{pmatrix}
$$
The matrix has nonzero determinant and so is invertible. Therefore, $\{1,\alpha,\alpha^2,\alpha^3\}$ is also a basis and so generates the same space, that is, $\mathbb{Q}( \sqrt{5},\sqrt{7} ) = \mathbb{Q}( \sqrt{5} + \sqrt{7} )$.
</s>