Extension of $\mathbb{Q}(\sqrt{2})$ by $\sqrt{3}$ and then by $\sqrt{5}$ I am trying to write down the extension of $\mathbb{Q}(\sqrt{2})$ by $\sqrt{3}$ and then by $\sqrt{5}$
$[[\mathbb{Q}(\sqrt{2})](\sqrt{3})](\sqrt{5})$
Here is what I have done so far:
$[[\mathbb{Q}(\sqrt{2})](\sqrt{3})](\sqrt{5})=\left \{ (a+b\sqrt{2})+(c+d\sqrt{2})\sqrt{3}+[(e+f\sqrt{2})\sqrt{3}]\sqrt{5}: a, b, c, d, e, f\in \mathbb{Q} \right \}$
Can some one tell me that am I right? Thanks a lot!
 A: No you do not have it correct unfortunately. 
I will do it in two ways to illustrate.
First way: (your way)
$[[\mathbb{Q}(\sqrt{2})](\sqrt{3})](\sqrt{5})=\left \{ (a+b\sqrt{2})+(c+d\sqrt{2})\sqrt{3}+[(e+f\sqrt{2})+(g+h\sqrt{2})\sqrt{3}]\sqrt{5}: a, b, c, d, e, f\in \mathbb{Q} \right \}$
The reason for the above is that for the $sqrt{5}$ you should multiply it by the coefficients in your field that you just extended from i.e $[\mathbb{Q}(\sqrt2)](\sqrt3)$
To see that your set (which is not a field) is not the as mine. consider $\sqrt{10}$ you will not have that.
Second way: (easy way)
We first extend to $L = [\mathbb{Q}(\sqrt2)](\sqrt3) = \left \{ a + b\sqrt{2} + c \sqrt{3} + d \sqrt{6}: a, b, c, d\in \mathbb{Q} \right \}$ 
(as $\sqrt{3},\sqrt{2}$ must be in the field but the field is multiplicatively closed.
Then:
$L(\sqrt{5}) = [\mathbb{Q}(\sqrt{2})](\sqrt{3})](\sqrt{5})=\left \{ a + b\sqrt{5} : a, b \in L \right \} = \left \{ a + b\sqrt{2} + c \sqrt{3} + d \sqrt{6} + e\sqrt{5} + f\sqrt{10} + g\sqrt{15} + h\sqrt{30} : a, b \in L \right \}$
(You could do the second method using your way, or what assume is your way, but make sure you simplify at the end)
Tip: it also almost always helps to double check you are getting the correct degree by the tower law.
A: If $\mathbb{Q}[\sqrt{2}]$ is your base field, then make sure you extend based on that. For example, the extension $\mathbb{Q}[\sqrt{2}][\sqrt{3}]$ is given by:
$$\mathbb{Q}[\sqrt{2}][\sqrt{3}] = \{a + b\sqrt{3}:a,b \in\mathbb{Q}[\sqrt{2}] \}$$
Therefore, the extension $\mathbb{Q}[\sqrt{2}][\sqrt{3},\sqrt{5}]$ is given by:
$$\mathbb{Q}[\sqrt{2}][\sqrt{3},\sqrt{5}] = \{a + b\sqrt{3} + c\sqrt{5} + d\sqrt{15}:a,b,c,d \in\mathbb{Q}[\sqrt{2}]\}$$
As should be expected, you can form a basis of $\mathbb{Q}[\sqrt{2}][\sqrt{3},\sqrt{5}]$ with 4 elements from $\mathbb{Q}[\sqrt{2}]$, by the tower law.
