<s>[INST] How to formally prove that we cannot find a polynomial in $\textbf Z[x]$ with degree $2$ with such a root? I am trying to find the kernel of the map from $\textbf Z[x]$ to $\textbf C$. The map is evaluating at $\sqrt 2 + \sqrt 3$.
A solution says that we cannot find polynomials of degree $2$ or $3$ that has such a root. So it skips the procedure of trying degree $2$ and $3$. And the final solution is the ideal in $\textbf Z[x]$ generated by $$x^4 − 10x^2 + 1 = (x − \sqrt 2 − \sqrt 3)(x − \sqrt 2 + \sqrt 3)(x + \sqrt 2 − \sqrt 3)(x + \sqrt 2 + \sqrt 3)$$
A solution says there are $\sqrt 6$'s in $(\sqrt 2 + \sqrt 3)^2$ and $(\sqrt 2 + \sqrt 3)^3$. So we cannot find polynomials with root $\sqrt 2 + \sqrt 3$ of degree $2$ or $3$. And that is what I am confused about: How does it imply the fact?
Thanks in advance!
 [/INST] Proceed by contradiction. 
Assume that there exist integer $a,b,c$ such that
$$ax^2+bx+c=0$$
with $x=\sqrt{2}+\sqrt{3}$.
$$a(5+2\sqrt{6})+b(\sqrt{2}+\sqrt{3})+c=0$$
Note that $1,\sqrt{2},\sqrt{3},$ and $\sqrt{6}$ are linearly independent, which means that there exist no integers $n_1,n_2,n_3,n_4$, not all $0$, such that
$$n_1+n_2\sqrt{2}+n_3\sqrt{3}+n_4\sqrt{6}$$
(A proof of this fact can be found here.) However, this means that
$$(5a+c)+b\sqrt{2}+c\sqrt{3}+(2a)\sqrt{6} \neq 0$$
which is a contradiction. 
</s>