Find a basis for $\mathbb{Q}(\sqrt{2}+\sqrt{5})$ over $\mathbb{Q}(\sqrt{5})$. 
Find a basis for $\mathbb{Q}(\sqrt{2}+\sqrt{5})$ over $\mathbb{Q}(\sqrt{5})$.

I used the following facts:


*

*$\mathbb{Q}(\sqrt{2}+\sqrt{5}) = \mathbb{Q}(\sqrt{2},\sqrt{5})$, so a tower of fields is $\mathbb{Q}(\sqrt{5})\subset \mathbb{Q}(\sqrt{2}) \subset \mathbb{Q}(\sqrt{2},\sqrt{5})$.

*Minimal polynomial for $\sqrt{2}$ over $\mathbb{Q}(\sqrt{5})$ is $x^{2}-2$, so $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}(\sqrt{5}]=2$.

*Simple extension of $\mathbb{Q}(\sqrt{2})$ is $\mathbb{Q}(\sqrt{2})(\sqrt{5})$.

*Minimal polynomial for $\sqrt{5}$ over $\mathbb{Q}(\sqrt{2})$ is $x^{2}-5$. Then $[\mathbb{Q}\sqrt{2},\sqrt{5}):\mathbb{Q}(\sqrt{2})]=2$.


Therefore, $[\mathbb{Q}(\sqrt{2}+\sqrt{5}):\mathbb{Q}(\sqrt{5})]=4$ and a basis is $\left \{ 1,\sqrt{2},\sqrt{5},\sqrt{10}\right \}$.
Are these facts valid though?
 A: There is no such tower of subfields as $\mathbb{Q}(\sqrt{5})\subset\mathbb{Q}(\sqrt{2})\subset\mathbb{Q}(\sqrt{2},\sqrt{5})$, because the leftmost inclusion doesn't hold.
You should instead consider $\mathbb{Q}\subset\mathbb{Q}(\sqrt{5})\subset\mathbb{Q}(\sqrt{2},\sqrt{5})$. However, you claim that the minimal polynomial of $\sqrt{2}$ over $\mathbb{Q}(\sqrt{5})$ is $x^2-2$ without proving it. Once you prove it, together with the fact that $x^2-5$ is the minimal polynomial of $\sqrt{5}$ over $\mathbb{Q}$, you get
$$
[\mathbb{Q}(\sqrt{2},\sqrt{5}):\mathbb{Q}]=
[\mathbb{Q}(\sqrt{2},\sqrt{5}):\mathbb{Q}(\sqrt{5})]
[\mathbb{Q}(\sqrt{5}):\mathbb{Q}]=2\cdot2=4
$$
So your proof jumps over two facts:


*

*$\mathbb{Q}(\sqrt{2}+\sqrt{5})=\mathbb{Q}(\sqrt{2},\sqrt{5})$;

*$x^2-2$ is irreducible over $\mathbb{Q}(\sqrt{5})$.


Both facts can be easily proved.
Clearly $\mathbb{Q}(\sqrt{2}+\sqrt{5})\subset\mathbb{Q}(\sqrt{2},\sqrt{5})$. On the other hand
$$
\frac{1}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{3}\in\mathbb{Q}(\sqrt{2}+\sqrt{5})
$$
so also $\sqrt{5}-\sqrt{2}\in\mathbb{Q}(\sqrt{2}+\sqrt{5})$. Hence $\sqrt{2}$ and $\sqrt{5}$ both belong to $\mathbb{Q}(\sqrt{2}+\sqrt{5})$. This proves the reverse inclusion.
The polynomial $x^2-2$ has no root in $\mathbb{Q}(\sqrt{5})$, hence it is irreducible (having degree $2$). Indeed, if
$$
(a+b\sqrt{5}\,)^2=2
$$
we obtain $a^2+5b^2=2$ and $2ab=0$. This is a contradiction.
Therefore $\{1,\sqrt{2}\}$ is a basis for $\mathbb{Q}(\sqrt{2},\sqrt{5})$ over $\mathbb{Q}(\sqrt{5})$, which in turn has basis $\{1,\sqrt{5}\}$ over $\mathbb{Q}$.
The standard proof of the dimension theorem tells us that
$$
\{1,\sqrt{2},\sqrt{5},\sqrt{2}\sqrt{5}\}=\{1,\sqrt{2},\sqrt{5},\sqrt{10}\}
$$
is a basis for $\mathbb{Q}(\sqrt{2},\sqrt{5})$ over $\mathbb{Q}$.
A: Clearly $\mathbb{Q}(\sqrt2 + \sqrt5) \subseteq \mathbb{Q}(\sqrt2, \sqrt5)$. Since $\mathbb{Q}(\sqrt2 + \sqrt5)$ is a field, the inverse of $\sqrt2 + \sqrt5$ must be in $\mathbb{Q}(\sqrt2 + \sqrt5)$ i.e.
$$\frac{1}{\sqrt2 + \sqrt5} = \frac{\sqrt5 - \sqrt2}{3} \in \mathbb{Q}(\sqrt2 + \sqrt5)$$
Then $\sqrt5 = \frac{3}{2}(\frac{\sqrt5 - \sqrt2}{3}) + \frac{1}{2}(\sqrt2 + \sqrt5) \in \mathbb{Q}(\sqrt2 + \sqrt5)$ and similarly you can check that $\sqrt2 \in \mathbb{Q}(\sqrt2 + \sqrt5)$. Thus, $\mathbb{Q}(\sqrt2, \sqrt5) \subseteq \mathbb{Q}(\sqrt2 + \sqrt5)$ and hence $\mathbb{Q}(\sqrt2, \sqrt5) = \mathbb{Q}(\sqrt2 + \sqrt5)$.
Now you know that $\{ 1, \sqrt2 \}$ is a basis for $\mathbb{Q}(\sqrt2)$ over $\mathbb{Q}$ and $\{ 1, \sqrt5 \}$ is a basis for $\mathbb{Q}(\sqrt5)$ over $\mathbb{Q}$. Then, as you stated, $x^2 -2$ is the minimal polynomial for $\sqrt2$ over $\mathbb{Q}(\sqrt5)$ and hence $\mathbb{Q}(\sqrt2, \sqrt5)$ is an extension field of degree 2 over $\mathbb{Q}(\sqrt5)$. Hence a basis for $\mathbb{Q}(\sqrt2, \sqrt5)$ over $\mathbb{Q}(\sqrt5)$ is $\{ 1, \sqrt2 \}$ since $\sqrt2 \notin \mathbb{Q}(\sqrt5)$ and it is clear that it is also a basis for $\mathbb{Q}(\sqrt2 + \sqrt5)$ over $\mathbb{Q}(\sqrt5)$ since $\mathbb{Q}(\sqrt2, \sqrt5) = \mathbb{Q}(\sqrt2 + \sqrt5)$.
A: $Let$ $x=√2+√5$
$x-√5=√2$ $(x-√5)^2=(√2)^2$
$x^2-2√5x+5=2$
$x^2-2√5x+3=0$
This polynomial belongs to Q(√5) therefore degree of √2+√5 over Q(√5) is 2. That means basis for Q(√2+√5) over Q(√5) is {1, √2+√5 }.
Am I right?
