Can You Prove the Equality of Field Theories with Different Prime Numbers?

[ElDavidas]

Homework Statement

Show $Q(\sqrt{p},\sqrt{q}) = Q(\sqrt{p} + \sqrt{q})$

Homework Equations

$p$ and $q$ are two different prime numbers

The Attempt at a Solution

I can show $\sqrt{p} + \sqrt{q} \in Q(\sqrt{p},\sqrt{p})$

I have trouble with the other direction though, i.e $\sqrt{p},\sqrt{p} \in Q(\sqrt{p} + \sqrt{q})$.

So far I've let $\alpha = \sqrt{p} + \sqrt{q}$

and found the powers $\alpha^2 = p + q + 2 \sqrt{p}\sqrt{q}$ and $\alpha^3 = (p + 3q )\sqrt{p} + (3p + q)\sqrt{q}$

Not sure what to do now though.

 

[matt grime]

So you know that
$\sqrt{p} + \sqrt{q}$

and

$(p + 3q )\sqrt{p} + (3p + q)\sqrt{q}$

are in the field. That is all you need. HINT: think simlutaneous linear equations.