Question

Find the minimal polynomial of $\sqrt{\mathbf{2}}\mathbf{+}\stackrel{\mathbf{⏜}}{\mathbf{i}}$over $\mathit{Q}$and over R

Short answer

The minimal polynomial is $x^4-6x^2+9$over $Q$and $x^2-2ix-3=0$over R.

Step-by-step solution

Definition of monic polynomial.

The monic polynomial $p\left(x\right)$over a field $F$of least degree such that $\rho \left(\alpha \right)=0$an algebraic element $\alpha$ over a field $F$.

Showing that x4-6x2+9 is minimal polynomial over Q and x2-2ix-3=0 over R.

Let $\sqrt{2}+i$ over Qas$Q⊑Q\left(\sqrt{2}\right)⊑\left(\sqrt{2},i\right)$.

Thus $\left[Q\sqrt{2}:Q\right]=2$

Since $\sqrt{2}$is a root of polynomial of degree 2 over $Q\sqrt{2}$but it is not in $Q\sqrt{2}$. Thus

$Q\left[\sqrt{2}+i:Q\right]=4$

Thus the minimal polynomial of $\sqrt{2}+i$over $Q$should be of degree 4.

Now let $x=\sqrt{2}+i$

As,

$$\begin{gathered} \left(x-i\right)=\sqrt{2} \\ {\left(x-1\right)}^2={\sqrt{2}}^2 \\ {x}^2-1-2xi=2 \\ {x}^2-3=2xi \end{gathered}$$

Square again

$$\begin{gathered} {\left(x^2-3\right)}^2={\left(2xi\right)}^2 \\ {x}^4-6x^2+9=-4x^2 \\ {x}^4-2x^2+9=0 \end{gathered}$$

Hence the required minimal polynomial of $\sqrt{2}+i$over $Q$is $x^2-2x^2+9=0$.

Now let $x=\sqrt{2}+i$over $R$.

$$\begin{gathered} {\left(x-i\right)}^2={\sqrt{2}}^2 \\ {x}^2-1-2xi=2 \\ {x}^2-3=2xi \end{gathered}$$

Thus the required polynomial of $\sqrt{2}+i$over $R$is $x^2-2ix-3=0$.