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?

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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.

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).

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.