Let p,q in ZZ be district primes. Prove that QQ(sqrtp, sqrtq)=QQ(sqrtp+sqrtq), and that QQ sube QQ (sqrtp+sqrtq) is a degree 4 extension.

p,q distinct prime numbers
Let ${\displaystyle K=\mathbb{Q}(\sqrt{p},\sqrt{q}),L=\mathbb{Q}(\sqrt{p}+\sqrt{q})}$
Claim:
1) $K=L$
2) ${\displaystyle deg\left(\frac{L}{\mathbb{Q}}\right)=4}$
1) Proof that K is contained in L.
p,q distinct prime numbers
Let ${\displaystyle K=\mathbb{Q}(\sqrt{p},\sqrt{q}),L=\mathbb{Q}(\sqrt{p}+\sqrt{q})}$
Claim:
1) $K=L$
Proof:
L sube K. Follow from ${\displaystyle \sqrt{p},\sqrt{q}\in K\Rightarrow \sqrt{p}+\sqrt{q}\in K}$
So, ${\displaystyle L=Q(\sqrt{p}+\sqrt{q})\subseteq K}$. So, proved.
2) Proof that L is contained in K. Note that i) abd ii) imply that $K=L$, as required ${\displaystyle K\subseteq L}$
Now, as p and q are distinct primes, the conjugates of ${\displaystyle \sqrt{p}+\sqrt{q}}$ are -${\displaystyle \sqrt{p}+\sqrt{q},\sqrt{p}-\sqrt{q},-\sqrt{p},-\sqrt{i}L.}$
So ${\displaystyle (\sqrt{p}+\sqrt{q})+(\sqrt{p}-\sqrt{q})=2\sqrt{p}\in L\Rightarrow \sqrt{p}\in L}$
Similarly, ${\displaystyle \sqrt{q}\in L.}$
So, $K=QQ(\sqrt{p},\sqrt{q})$ sube L, thus proved rArr $K=L$
3) Proof ofdepends on claim , whose is proved first
${\displaystyle deg\left(\frac{L}{\mathbb{Q}}\right)=4}$
Proof: Form $K=L$.
But ${\displaystyle K=Q(\sqrt{p}.\sqrt{q})=Q\left(\sqrt{p}\right)\left(\sqrt{q}\right)}$
${\displaystyle \sqrt{q}\notin Q\left(\sqrt{p}\right)}$
Otherwise, ${\displaystyle \sqrt{q}=a+b\sqrt{p}}$, for some ${\displaystyle a,b\in \mathbb{Q}}$
Square both sides to get ${\displaystyle q=a^2+b^{2}p+2ab\cdot \sqrt{p}\Rightarrow \sqrt{p}=\frac{q-a^2+b^{2}p}{2a\cdot b}\in \mathbb{Q}}$
which is absurd, as p is prime ${\displaystyle \Rightarrow \sqrt{p}\in \mathbb{Q}}$
Completing the proof that deg $(L)=deg(K)=4.$
${\displaystyle deg\left(\frac{L}{\mathbb{Q}}\right)=deg\left(\frac{K}{\mathbb{Q}}\right)=deg\left(\frac{Q\left(\sqrt{p}\right)}{\mathbb{Q}}\right)\cdot deg\left(Q\frac{\sqrt{p}\sqrt{q}}{\mathbb{Q}}\left(\sqrt{p}\right)\right)=2\cdot 2=4}$