Q:

Minimal Polynomial of $\mathbb{Q}(\sqrt a +\sqrt b)$

I want to find the Minimal Polynomial of $\mathbb{Q}(\sqrt a +\sqrt b)$ over $\mathbb{Q}$. I don't really know where to start, I've seen solutions to specific cases but I'm not sure about the general case.

A:

If $a=b$, it's easy, so I'll assume $a\ne b$ and also that neither $\sqrt{a}$ nor $\sqrt{b}$ is rational (or it would be easy as well).
Set $r=\sqrt{a}+\sqrt{b}$; then $(r-\sqrt{a})^2=b$, which becomes
$$
r^2+a-b=2r\sqrt{a}
$$
Squaring again,
$$
r^4+2r^2(a-b)+(a-b)^2=4r^2a
$$
and so $r$ is a root of
$$
X^4-2(a+b)X^2+(a-b)^2
$$
Now note that
$$
\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{1}{a-b}(\sqrt{a}-\sqrt{b})
$$
which implies $\sqrt{a}-\sqrt{b}\in\mathbb{Q}(r)$ and, therefore, $\sqrt{a}\in\mathbb{Q}(r)$ and $\sqrt{b}\in\mathbb{Q}(r)$.
In particular, $\mathbb{Q}(r)=\mathbb{Q}(\sqrt{a},\sqrt{b})$.
Can you apply the dimension formula to find out when the degree of $\mathbb{Q}(r)$ is $2$ or $4$?

