Proving $\mathbb{Q}(\sqrt{2},\sqrt{5})=\mathbb{Q}(\sqrt{2}+\sqrt{5})$ using matrices I am trying to show that $\mathbb{Q}(\sqrt{2},\sqrt{5})=\mathbb{Q}(\sqrt{2}+\sqrt{5})$. I have calculated the minimal polynomial $x^4-14x^2+9$ and I know that its basis as a $\mathbb{Q}$-vector space is $\{1, 
\sqrt{2}, \sqrt{5}, \sqrt{10}\}$. But I'm not sure how to construct an invertible matrix to express the basis as rational polynomials. I'm thinking of Fly by Night's comment here
 A: As Mariano notes in the comments, there's a much quicker argument using degrees, but here's the matrix argument. (A nice bonus of this argument is that you can explicitly express the basis elements for $\mathbb{Q}(\sqrt{2}, \sqrt{5})$ in terms of powers of $\alpha$.)
Let $\alpha = \sqrt{2} + \sqrt{5}$. Note that $\alpha \in \mathbb{Q}(\sqrt{2}, \sqrt{5}),$ so immediately we have $\mathbb{Q}(\alpha) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{5})$. Suppose there exists an invertible rational $4 \times 4$ matrix $A$ such that
$$\begin{pmatrix}\alpha \\ \alpha^{2} \\ \alpha^{3} \\ \alpha^{4}\end{pmatrix} = A \begin{pmatrix}1 \\ \sqrt{2} \\ \sqrt{5} \\ \sqrt{10} \end{pmatrix}.$$
Then, we have
$$\begin{pmatrix}1 \\ \sqrt{2} \\ \sqrt{5} \\ \sqrt{10} \end{pmatrix} = A^{-1}\begin{pmatrix}\alpha \\ \alpha^{2} \\ \alpha^{3} \\ \alpha^{4}\end{pmatrix},$$ which would prove that each of the basis elements for $\mathbb{Q}(\sqrt{2}, \sqrt{5})$ can be expressed as a $\mathbb{Q}$-linear combination of powers of $\alpha$, and thus we must have $\mathbb{Q}(\sqrt{2}, \sqrt{5}) \subseteq \mathbb{Q}(\alpha)$. Since we already have inclusion the other way, we would then be done.
So, we just need to find such an $A$. Well, there's only one real candidate for what $A$ can be: the $i$th row of $A$ must be $\begin{pmatrix}a & b  & c & d \end{pmatrix}$, where $a + b\sqrt{2} + c\sqrt{5} + d\sqrt{10}$ is the unique way to write $\alpha^{i}$ as a $\mathbb{Q}$-linear combination of the basis elements $\{1, \sqrt{2}, \sqrt{5}, \sqrt{10}\}.$ (This is unique because $\alpha^{i} \in \mathbb{Q}(\sqrt{2}, \sqrt{5}).$)
Let's compute:
$$\alpha = 0 \cdot 1 + 1 \cdot \sqrt{2} + 1 \cdot \sqrt{5} + 0 \cdot \sqrt{10},$$
$$\alpha^{2} = 7 + 2\sqrt{10} = 7 \cdot 1 + 0 \cdot \sqrt{2} + 0 \cdot \sqrt{5} + 2 \cdot \sqrt{10},$$
$$\alpha^{3} = 17\sqrt{2} + 11\sqrt{5} = 0 \cdot 1 + 17 \cdot \sqrt{2} + 11 \cdot \sqrt{5} + 0 \cdot \sqrt{10},$$
$$\alpha^{4} = 89 + 28\sqrt{10} = 89 \cdot 1 + 0 \cdot \sqrt{2} + 0 \cdot \sqrt{5} + 28 \cdot \sqrt{10}.$$
So, our candidate for $A$ is
$$A = \begin{pmatrix} 0 & 1 & 1 &0 \\ 7 & 0 & 0 & 2 \\ 0 & 17 & 11 & 0\\ 89 & 0 & 0 &28 \end{pmatrix}.$$ A quick computation shows that this matrix has nonzero determinant, hence is invertible. At this point, we are already done, but we can actually invert $A$ to explicitly express the basis elements for $\mathbb{Q}(\sqrt{2}, \sqrt{5})$ in terms of powers of $\alpha$,
A more laborious computation (just kidding, I used a calculator) finds the inverse to be
$$A^{-1} = \begin{pmatrix} 0 & \frac{14}{9} & 0 & - \frac{1}{9} \\ -\frac{11}{6} & 0 & \frac{1}{6} & 0 \\ \frac{17}{6} & 0 & -\frac{1}{6} & 0\\ 0 & -\frac{89}{18} & 0 & \frac{7}{18} \end{pmatrix}.$$
Since we have
$$\begin{pmatrix}1 \\ \sqrt{2} \\ \sqrt{5} \\ \sqrt{10} \end{pmatrix} = A^{-1}\begin{pmatrix}\alpha \\ \alpha^{2} \\ \alpha^{3} \\ \alpha^{4}\end{pmatrix},$$ it follows that
$$1 =\frac{14}{9} \alpha^{2} - \frac{1}{9} \alpha^{4},$$
(Look! It's the minimal polynomial!)
$$\sqrt{2} = -\frac{11}{6}\alpha  + \frac{1}{6} \alpha^{3},$$
$$\sqrt{5} = \frac{17}{6}\alpha - \frac{1}{6} \alpha^{3},$$
$$\sqrt{10} = -\frac{89}{18}\alpha^{2} + \frac{7}{18}\alpha^{4}.$$
