Question

Which of the following are normal extension of Q.

$$\begin{gathered} \mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{}\mathbf{Q}\left(\sqrt{3}\right)\mathbf{.} \\ \left(b\right)\mathbf{}\mathbf{Q}\left(\sqrt[3]{3}\right)\mathbf{.} \\ \left(c\right)\mathbf{}\mathbf{Q}\left(\sqrt{5},i\right)\mathbf{.} \end{gathered}$$

Short answer

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{Q}\left(\sqrt{3}\right)\mathrm{is}\mathrm{normal}\mathrm{extension}\mathrm{of}\mathrm{Q}. \\ \left(\mathrm{b}\right)\mathrm{Q}\left(\sqrt[3]{3}\right)\mathrm{is}\mathrm{not}\mathrm{normal}\mathrm{extension}\mathrm{of}\mathrm{Q}. \\ \left(\mathrm{c}\right)\mathrm{Q}(\sqrt{5},\mathrm{i})\mathrm{is}\mathrm{normal}\mathrm{extension}\mathrm{of}\mathrm{Q}. \end{gathered}$$

Step-by-step solution

Definition of normal extension.

$$\begin{gathered} \mathrm{An}\mathrm{algebraic}\mathrm{extension}\mathrm{field}\mathrm{K}\mathrm{of}\mathrm{F}\mathrm{is}\mathrm{normal}\mathrm{provided}\mathrm{that}\mathrm{whenever}\mathrm{an} \\ \mathrm{irreducible}\mathrm{polynomial}\mathrm{in}\mathrm{F}\left[\mathrm{x}\right]\mathrm{has}\mathrm{one}\mathrm{root}\mathrm{in}\mathrm{K}\mathrm{then}\mathrm{it}\mathrm{split}\mathrm{over}\mathrm{K}. \end{gathered}$$

Step-2: Showing that Q(3) is normal extension of .

(a)

Consider a polynomial ${\mathrm{x}}^2-3\mathrm{over}\mathrm{Q}\left(\sqrt{3}\right).\mathrm{Noe}\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2-3$

Find the root of $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2-3$as follows,

$$\begin{gathered} f\left(x\right)=0 \\ {x}^2-3=0 \\ {x}^2=3 \\ x=\pm \sqrt{3} \end{gathered}$$

Since the root of $f\left(x\right)are\pm \sqrt{3}$and both of them are in $\mathrm{Q}\left(\sqrt{3}\right).\mathrm{So}\mathrm{Q}\left(\sqrt{3}\right)$is a normal extension of Q.

Showing that Q33 is not normal extension of Q.

(b)

Consider a polynomial ${\mathrm{x}}^3-3\mathrm{over}\mathrm{Q}\left(\sqrt[3]{3}\right).\mathrm{Let}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3-3.$

Now find the root of $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3-3$as follows,

$$\begin{gathered} f\left(x\right)=0 \\ {x}^3-3=0 \\ {x}^3=3 \\ x=\sqrt[3]{3},\sqrt[3]{3\omega },\sqrt[3]{3{\omega }^2} \end{gathered}$$

Since the root of $f\left(x\right)$are $\sqrt[3]{3},\sqrt[3]{3\omega },\sqrt[3]{3{\omega }^2}$and one of them is in $\mathrm{Q}\left(\sqrt[3]{3}\right)$that is $\left(\sqrt[3]{3}\right)$and rest two roots are not in$Q\left(\sqrt[3]{3}\right)$.

Hence $Q\left(\sqrt[3]{3}\right)$is not a normal extension of Q .

Step-4: Showing that Q(5,i)is normal extension over Q .

(c)

Consider a polynomial${\mathrm{x}}^2+5\mathrm{over}\mathrm{Q}\left(\sqrt{5},\mathrm{i}\right).\mathrm{Let}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2+5.$

First find the root of the polynomial $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2+5$as follows

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=0 \\ {\mathrm{x}}^2+5=0 \\ {\mathrm{x}}^2=-5 \\ \mathrm{x}=\pm \sqrt{5}\mathrm{i} \end{gathered}$$

Since the root of the polynomial $\mathrm{f}\left(\mathrm{x}\right)\mathrm{are}\pm \sqrt{5}\mathrm{i}$and both of them are in$\mathrm{Q}\left(\sqrt{5,}\mathrm{i}\right)$

Hence $Q\left(\sqrt{5},i\right)$is normal extension over Q.