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 $4$?
A: Of course we assume that neither $a$ nor $b$ is a perfect square.
But also if $ab$ is a perfect square, we have simply $$\Bbb Q(\sqrt a+\sqrt b)=\Bbb Q(\sqrt a)\cong \Bbb Q[X]/(X^2-a).$$
In the remaining general case, the obvious candidates for conjugates of $\sqrt a+\sqrt b$ are $\sqrt a-\sqrt b$, $-\sqrt a+\sqrt b$, $-\sqrt a-\sqrt b$, so we try
$$(X-\sqrt a-\sqrt b)(X-\sqrt a+\sqrt b)(X+\sqrt a-\sqrt b)(X+\sqrt a+\sqrt b)\\
=((X-\sqrt a)^2-b)((X+\sqrt a)^2-b)\\
=(X^2-2\sqrt a X+a-b)(X^2+2\sqrt a X+a-b)\\
=(X^2+a-b)^2-4aX^2\\
=X^4+2(a-b)X^2+(a-b)^2-4aX^2\\
=X^4-2(a+b)X^2+(a-b)^2.  $$
Because all of its roots $\pm\sqrt a\pm \sqrt b$ are irrational, this cannot have a degree 1 factor.
For any candidate degree 2 factor, either the sum of its roots would be irrational (for example, $(\sqrt a+\sqrt b)+(\sqrt a-\sqrt b)=2\sqrt a$), or the  product of the roots is irrational (for example $(\sqrt a+\sqrt b)(-\sqrt a-\sqrt b)=-a-b-a\sqrt{ab}$, where we use that $ab$ is not a perfect square). Therefore the above polynomial is irreducible and
$$ \Bbb Q(\sqrt a +\sqrt b)\cong\Bbb Q[X]/(X^4-2(a+b)X^2+(a-b)^2).$$
