Suppose that
\(\alpha=\sqrt[3]{2}\in\R\) and
\(\beta=\omega\alpha\in\C\text{,}\) where
\(\omega\) is a primitive cube root of unity. Suppose that
\(K=\Q(\alpha)\) and that
\(L=K(\beta)\text{.}\) We show that
\([L:\Q]=6\text{.}\)
First of all note that
\(\beta\not\in\R\) and hence
\(\beta\not\in K\text{.}\) Furthermore,
\(x^3-2\) is the minimal polynomial of
\(\alpha\) over
\(\Q\) and hence
\([K:\Q]=3\text{,}\) and
\(\{1,\alpha,\alpha^2\}\) is a
\(\Q\) basis of
\(K\text{.}\) The minimal polynomial of
\(\beta\) over
\(K\) is
\(x^2+\alpha x+\alpha^2\text{,}\) which is irreducible over
\(K\) since it has no roots in
\(K\text{.}\) Therefore,
\([L:K]=2\text{,}\) and
\(\{1,\beta\}\) is a
\(K\) basis of
\(L\text{.}\)
We claim that \(\{1,\alpha,\alpha^2,\beta,\alpha\beta,\alpha^2\beta\}\) is a \(\Q\) basis of \(L\text{.}\) Let \(x\in L\text{.}\) There exists a unique \(a_0,a_1\in K\) such that \(x=a_0+a_1\beta\text{.}\) Furthermore, there exists a unique \(b_{0i}\) and \(b_{1i}\text{,}\) for \(0\leq i\leq 2\) such that \(a_0=\sum_{i=0}^2b_{0i}\alpha^i\) and \(a_1=\sum_{i=0}^2b_{1i}\alpha^i\text{.}\) In particular, we can write \(x=\sum_{i=0}^2b_{0i}\alpha^i+\sum_{i=0}^2b_{1i}\alpha^i\beta\text{.}\) This shows that \(x\) can be expressed as a linear combination of the elements in \(\{1,\alpha,\alpha^2,\beta,\alpha\beta,\alpha^2\beta\}\text{.}\) Hence, this set spans \(L\) over \(\Q\text{.}\) To show that it is linearly independent, suppose that there are \(c_0,c_1,c_2,d_0,d_1,d_2\in\Q\) such that
\begin{equation*}
\sum_{i=0}^2 c_i\alpha^i+\sum_{i=0}^2d_i\alpha^i\beta=0.
\end{equation*}
As \(\{1,\beta\}\) is a basis of \(L\) over \(K\text{,}\) we get that
\begin{equation*}
\sum_{i=0}^2 c_i\alpha^i=0\in K\quad\text{ and }\quad\sum_{i=0}^2d_i\alpha^i=0\in K.
\end{equation*}
Using the fact that \(\{1,\alpha,\alpha^2\}\) is a basis of \(K\) over \(\Q\text{,}\) we get that \(c_i=0\) and \(d_i=0\) for all \(i\text{.}\) This shows that \([L:\Q]=6\text{.}\)
