Find th eminimal polynomial of sqrt2+sqrt3 over QQ

Abstract algebra
Find th eminimal polynomial of $$\displaystyle\sqrt{{2}}+\sqrt{{3}}$$ over $$\displaystyle\mathbb{Q}$$

2020-11-11
To find the smallest degree polynomial f(x) with coefficients in Q, such that $$\displaystyle{f{{\left(\sqrt{{2}}+\sqrt{{3}}\right)}}}={0}.$$
The impo
rtant point is that the coefficients should all be in Q. For example, $$\displaystyle{x}-{\left(\sqrt{{2}}+\sqrt{{3}}\right)}$$ is not allowed, as its coefficients are not rational numbers.
Now, $$\displaystyle\sqrt{{3}}\notin\mathbb{Q}\sqrt{{2}}.$$
So, we expect the minimal polynomial of $$\displaystyle\sqrt{{2}}+\sqrt{{3}}$$ to have degree $$\displaystyle{4}{\left(,'\sqrt{{2}}+\sqrt{{3}}\in\mathbb{Q}{\left(\sqrt{{2}}\right)}\cdot{\left(\sqrt{{3}}\right)}\right.}$$, a quadrantic extension of a quadrantic extension of $$\displaystyle\mathbb{Q}$$.
We proceed to find this minimal polynomial by repeated squaring (to clear radicals and obtain rational coefficients)
$$\displaystyle{x}=\sqrt{{2}}+\sqrt{{3}}\Rightarrow{x}-\sqrt{{2}}=\sqrt{{3}},$$ squaring
$$\displaystyle{\left({x}-\sqrt{{2}}\right)}^{{2}}={3}\Rightarrow{x}^{{2}}-{2}\sqrt{{2}}{x}+{2}={3}$$, rearrange
$$\displaystyle{x}^{{2}}-{1}={2}\sqrt{{2}}{x}$$, square again
$$\displaystyle{\left({x}^{{2}}-{1}\right)}^{{2}}={8}{x}^{{2}}\Rightarrow{x}^{{4}}-{2}{x}^{{2}}+{1}={8}{x}^{{2}}$$, rearrange
Minimal polynomial is $$\displaystyle{x}^{{4}}-{10}{x}^{{2}}+{1}$$