Question

Show that$\mathit{G}\mathit{a}\mathit{l}{}_{\mathbf{\mathbb{Q}}}\mathbf{\mathbb{Q}}\mathbf{|}\mathbf{\left(}\sqrt[\mathbf{4}]{\mathbf{2}}\mathbf{\right)}\mathbf{\ne }\mathbf{<}\mathbf{i}\mathbf{>}$.

Short answer

It is shown that $\mathit{G}\mathit{a}\mathit{l}{}_{\mathbf{\mathbb{Q}}}\mathbf{\mathbb{Q}}\mathbf{|}\mathbf{\left(}\sqrt[\mathbf{4}]{\mathbf{2}}\mathbf{\right)}\mathbf{\ne }\mathbf{<}\mathbf{i}\mathbf{>}$.

Step-by-step solution

Definition of field extension

A${\mathit{n}}^{\mathbf{t}\mathbf{h}}$ root of unity may be complex roots of numbers whose${\mathit{n}}^{\mathbf{th}}$ power is 1 and it is a root of the polynomial ${\mathit{x}}^{\mathbf{n}}\mathbf{-}\mathbf{1}$. The${\mathit{n}}^{\mathbf{th}}$ roots of unity form cyclic group of order$\mathit{n}$ under multiplication. A field extension is called a cycle extension if its Galois group is a cyclic group, and the Galois group of every finite field is finite and cyclic.

Showing that Galℚℚ|(24)≠<i> .

Consider the Galois group and prove that $\mathit{G}\mathit{a}\mathit{l}{}_{\mathbf{\mathbb{Q}}}\mathbf{\mathbb{Q}}\mathbf{|}\mathbf{\left(}\sqrt[\mathbf{4}]{\mathbf{2}}\mathbf{\right)}\mathbf{\ne }\mathbf{<}\mathbf{i}\mathbf{>}$.

Now, an${\mathit{n}}^{\mathbf{t}\mathbf{h}}$ root of unity may be thought of as a complex number whose${\mathit{n}}^{\mathbf{th}}$ power is 1. And it is a root of the polynomial ${\mathit{x}}^{\mathbf{n}}\mathbf{}\mathbf{-}\mathbf{1}$. So, the given Galois group$Gal{}_{\mathrm{\mathbb{Q}}}\mathrm{\mathbb{Q}}|\left(\sqrt[4]{2}\right)\ne <\mathrm{i}>$ has not a root of unit and it may think a complex root whose power is${\mathbf{4}}^{\mathbf{t}\mathbf{h}}$ of element 2. Since, the${\mathbf{4}}^{\mathbf{th}}$ root of this is not a unity and it is not formed a cyclic group of order${\mathbf{4}}^{\mathbf{th}}$ under multiplication.

Hence, the field extension is same as the cyclic extension and if it is a Galois group. It is cyclic but at the given group the Galois group is not the root of unity so it is not a cyclic group.