Question

Find th eminimal polynomial of \(\displaystyle\sqrt{{2}}+\sqrt{{3}}\) over \(\displaystyle\mathbb{Q}\)

Answer 1

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 \(\displaystyle{x}^{{4}}-{10}{x}^{{2}}+{1}\)