Question

Prove that the given element is algebraic over .

(a)$\mathbf{3}\mathbf{+}\mathbf{5}\mathit{i}$

(b)$\sqrt{\mathbf{i}\mathbf{-}\sqrt{\mathbf{2}}}$

(C)$\mathbf{1}\mathbf{+}\sqrt[\mathbf{3}]{\mathbf{2}}$

Short answer

(a) $3+5i$ is algebraic over Q.

(b) $\sqrt{\mathbf{i}\mathbf{-}\sqrt{\mathbf{2}}}$is algebraic over Q.

(c) $1+\sqrt[3]{2}$ is algebraic over Q.

Step-by-step solution

Definition of algebraic extension.

Let K an extension field of F. Then K is said to be algebraic extension of F if every element of K is algebraic over F.

Showing that   is algebraic over  .

(a)

Consider $3+5i$ over Q. Let $x=3+5i$ . The simplify for x .

$x-3=5i$

${\left(x-3\right)}^2={\left(5i\right)}^2$

$x^2+9-6x=-25$

$x^2-6x+34=0$

Verify $3+5i$ is a root of role="math" localid="1657945135295" $x^2-6x+34=0$as follows.

$$\begin{gathered} {\left(3+5i\right)}^2-6\left(3+5i\right)+34=9-25+30i-18-30i+34 \\ =43-43+30i-30i \\ =0 \end{gathered}$$

Since 3+5isatisfy the equation $x^2-6x+34=0$. Therefore $3+5i$ is algebraic overQ.

Step-3: Showing that i-2 is algebraic over  Q.

(b)

Let over $x=\sqrt{i-\sqrt{2}}$ over Q. Simplify for x.

$x^2={\left(\sqrt{i-\sqrt{2}}\right)}^2$

$x^2=\left(\sqrt{i-\sqrt{2}}\right)$

$x^2=\sqrt{2}=i$

Again square both side

$$\begin{gathered} {\left(x^2+\sqrt{2}\right)}^2={\left(i\right)}^2 \\ {x}^4+2-2\sqrt{2}{x}^2=-1 \\ {x}^4+3-2\sqrt{2}{x}^2=0 \\ {x}^4+3=2\sqrt{2}{x}^3 \end{gathered}$$

Since the coefficient does not belong Q. Therefore square both side again

$$\begin{gathered} {\left(x^4+3\right)}^2={\left(2\sqrt{2}{x}^3\right)}^2 \\ {x}^8+9+6x^4=8x^4 \\ {x}^8-2x^4+9=0 \end{gathered}$$

Now check $x=\sqrt{i-\sqrt{2}}$ is a root of the equation$x^8-2x^4+9=0$ as it satisfy the equation. Therefore $x=\sqrt{i-\sqrt{2}}$is algebraic overQ.

Step-4: Showing that  1+23 is algebraic over Q.

Let $x=1\sqrt[3]{2}$. Now simplify this for x.

localid="1657946126499" $$\begin{gathered} x^2={\left(1\sqrt[3]{2}\right)}^2 \\ {x}^3-1-3x^2+3x=2 \\ {x}^3-3x^2+3x-3=0 \end{gathered}$$

Now the value $x=1+\sqrt[3]{2}$ is a root of the equation $x^3-3x^2+3x-3=0$ as it satisfy the equation. Therefore $x=1+\sqrt[3]{2}$is algebraic over Q