Question

Find the minimal polynomial of the given element over Q .

(a)$\sqrt{\mathbf{1}\mathbf{+}\sqrt{\mathbf{5}}}$

(b)$\sqrt{\mathbf{3}\mathbf{i}}\mathbf{+}\sqrt{\mathbf{2}}$

Short answer

(a)$x^4-2x^2-4=0$is minimal polynomial.

(b) localid="1660199331121" $x^4+2x^2+25=0$is the minimal polynomial.

Step-by-step solution

Definition of minimal polynomial.

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

Step-2: Showing that the minimal polynomial is x4-2x2-4=0 .

(a)

Let $x=\sqrt{1+\sqrt{5}}$. Now solve for x .

$x^2={\left(\sqrt{1+\sqrt{5}}\right)}^2$

$x^2=1+\sqrt{5}$

$x^2-1=\sqrt{5}$

Now again square both side

$$\begin{gathered} {\left(x^2-1\right)}^2={\left(\sqrt{5}\right)}^2 \\ {x}^4+1-2x^2=5 \\ {x}^4-2x^2-4=0 \end{gathered}$$

Thus the minimal polynomial is $x^4-2x^2-4=0$.

Showing that the minimal polynomial is x4+2x2+25=0.

(b)

Let $x=\sqrt{3i}+\sqrt{2}$. Now solve for x .

$x=\sqrt{3i}+\sqrt{2}$

$x^2-2\sqrt{2}x=-5$

$x^2+5=2\sqrt{2}x$

Now square both side again.

${\left(x^2+5\right)}^2={\left(2\sqrt{2}x\right)}^2$

localid="1660199236267" $x^4+25+10x^2=8x^2$

$x^4+2x^2+25=0$

The required polynomial is $x^4+2x^2+25=0$.