Exercise about field extensions

Consider $a_1,\ldots,a_n\in \mathbb Z$.

i) Suppose $a_1,\ldots, a_n$ are pairwise relatively prime. I have to see by induction on n that $[\mathbb Q(\sqrt a_1,\ldots,\sqrt a_n):\mathbb Q]=2^n$

Once I proved the equality is true for $n=1$, I suppose it is true for $n-1$, so let's prove it for $n$:
Applying the tower law:
$[\mathbb Q(\sqrt a_1,\ldots,\sqrt a_n):\mathbb Q]=[\mathbb Q(\sqrt a_1,\ldots,\sqrt a_n):\mathbb Q(\sqrt a_1,\ldots,\sqrt a_{n-1})] [\mathbb Q(\sqrt a_1,\ldots,\sqrt a_{n-1}):\mathbb Q]$

By induction, $[\mathbb Q(\sqrt a_1,\ldots,\sqrt a_{n-1}):\mathbb Q]=2^{n-1}$

So we only have to see that $[\mathbb Q(\sqrt a_1,\ldots,\sqrt a_n):\mathbb Q(\sqrt a_1,...,\sqrt a_{n-1})]=2$

$[\mathbb Q(\sqrt a_1,\ldots,\sqrt a_n):\mathbb Q(\sqrt a_1,\ldots,\sqrt a_{n-1})]=[\mathbb Q(\sqrt a_1,\ldots,\sqrt a_{n-1})(a_n):\mathbb Q(\sqrt a_1,\ldots,\sqrt a_{n-1})]=deg(Irr(\sqrt a_n,\mathbb Q(\sqrt a_1,\ldots,\sqrt a_n))$

$Irr(\sqrt a_n,\mathbb Q(\sqrt a_1,\ldots,\sqrt a_{n-1})=X^2-a_n??$

I have to see that $\sqrt a_n \notin Q(\sqrt a_1\ldots,\sqrt a_{n-1})$. By contradiction,

if$\sqrt a_n \in Q(\sqrt a_1,\ldots,\sqrt a_{n-1}) \to \sqrt a_n= a+b\sqrt a_{n-1}$ where $a,b\in Q(\sqrt a_1,\ldots,\sqrt a_{n-1}$. How can I get to a contradiction???

ii) Consider $P$ the set of prime numbers and $F$ an extension of $\mathbb Q$: $F=\mathbb Q (\sqrt p, p\in P)$ Which is the degree of $F/\mathbb Q$? Is it finitely generated?

Could you help me with this problem please?

Thank you for your time and help.

Answer

I think that the most elegant proof of this question, which is a consequence of the fact that

Square roots of different square free positive integers are linearly independent over $\mathbb Q$,

can be found in:

http://www.thehcmr.org/issue2_1/mfp.pdf