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So I've been reading that this statement is true.

Take, for example, $\mathbb{Q}(\sqrt[3]{2}, \omega)$ where $\omega$ is cube root of unity.

This is the splitting field of $x^3 - 2$. It has degree 6. It is also a finite separable field extension. But if it were simple, then it would be generated by some $\alpha$ and this $\alpha$ would have degree 6 minimal polynomial? How would one find this $\alpha$, if possible? Also, in general, how would one find a basis for this field extension?

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The proof of the multiplicativity formula for extension degrees tells you that $E = \mathbb{Q}(\beta, \omega)$ (where $\beta = \sqrt[3]{2}$) has a basis over $\Bbb{Q}$ given by $1, \beta, \beta^{2}, \beta, \omega \beta, \omega \beta^{2}$.

This is because $1, \beta, \beta^{2}$ is a basis of $\Bbb{Q}(\beta)$ over $\Bbb{Q}$, and $1, \omega$ is a basis of $E = (\mathbb{Q}(\alpha)) (\omega)$ over $\mathbb{Q}(\alpha)$.

To find $\alpha$, you may use the Galois group, which maps $\beta \mapsto \omega^{k} \beta$, for $k = 0, 1, 2$, and $\omega \mapsto \omega^{j}$ for $j = 1, 2$.

Now consider $\alpha = \beta + \omega$, and note that it has $6$ distinct images under the Galois group. (It suffices to see that no element of the Galois group but the identity fixes $\alpha$.)

Thus the minimal polynomial of $\alpha$ over $\Bbb{Q}$ has degree $6$, and $E = \Bbb{Q}(\alpha)$. This yields another basis $\alpha^{j}$, for $0 \le j < 6$ for $E = \Bbb{Q}(\alpha)$ over $\Bbb{Q}$.

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    $\begingroup$ As a note: in the general case, the general theory says that every finite separable extension is simple, but the proof is not constructive. It is generally a hard problem to actually find a generator (even for finite fields). $\endgroup$ – Ittay Weiss Apr 27 '13 at 8:08

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