Find the minimal polynomial of $\sqrt2 + \sqrt3 $ over $\mathbb Q$ I have no idea how to do this.
To find the minimal polynomial of say $\sqrt2 + \sqrt3$, we need to find the monic polynomial $p \in \mathbb Q$ (correct if I am wrong but monic polynomial is when the coefficient of the highest degree term is $1$) of the smallest possible degree such that $\sqrt2 + \sqrt3$ is a root of $p$.
If we let $u=\sqrt2 + \sqrt3$ then $u ^2 = 5+ 2 \sqrt6 \iff u^2 - 5 = 2 \sqrt6 $, then $(u^2 - 5)^2=24 \iff u^4 -10u^2 +1=0$
All I did was keep squaring until all of the irrational terms go away. But what next? Am I doing this correctly and what do we do next if I am? 
 A: You know that the minimal polynomial for $\sqrt3$ over $\Bbb Q$ is $X^2-3$, and we’ll believe that this is still the minimal polynomial for $\sqrt3$ over $\Bbb Q(\sqrt2\,)$. This means that the polynomial for $\sqrt3+\sqrt2$ over $\Bbb Q(\sqrt2\,)$ is $(X-\sqrt2\,)^2-3$. Expand this out, and multiply it by its “conjugate” (replacing $\sqrt2$ by $-\sqrt2\,$) and get a $\Bbb Q$-polynomial. That’s it.
A: First, show that $\sqrt{3}$ is not in the quadratic extension generated by $\sqrt{2}.$ That means that the degree of the extension is at least $4.$ But you have found a polynomial of degree $4,$ so it must be minimal.
A: Suppose $\;\sqrt2\in\Bbb Q(\sqrt3)\;$ , then there exist $\;a,b\in\Bbb Q\;$ such that
$$\sqrt2=a+b\sqrt3\implies 2=a^2+3b^2+2ab\sqrt3\implies\sqrt3\in\Bbb Q\;,\;\;\text{contradiction}$$
Thus, $\;x^2-2\;$ must be irreducible in $\;\Bbb Q(\sqrt3)[x]\;$ , so that $\;\sqrt2+\sqrt3\;$ must belong to an extension of $\;\Bbb Q\;$ of at least degree $\;4\;$ . Since you already found a rational polynomial of degree four which vanishes at $\;\sqrt2+\sqrt3\;$ you finished, as then this must be an irreducible polynomial (otherwise this sum of square roots would belong to an extension of degree less than four).
