Splitting Field of $p(x) = (x^4-1)(x^3-2)$

Question

I found the splitting field of a polynomial $p(x)=(x^4-1)(x^3-2)$ to be $\mathbb{Q}(\sqrt[3]{2},\omega,i)$ where $ \omega=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$, but is it possible to simplify this field?

Answer 1

The primitive element theorem says every finite extension of $\mathbf Q$ can be generated by a single element (called a primitive element). For "random" examples, the sum typically works. And this is such a case: $\mathbf Q(\sqrt[3]{2},\omega,i) = \mathbf Q\mathbf(\sqrt[3]{2}+\omega+i)$, but the minimal polynomial over $\mathbf Q$ for $\sqrt[3]{2}+\omega+i$ is very ugly. Using a computer, here it is: $$ x^{12} + 6x^{11} + 27x^{10} + 72x^9 + 150x^8 + 258x^7 + 395x^6 + 348x^5 - 54x^4 - 558x^3 - 417x^2 + 96x + 277. $$ I do not regard this as simpler. The field is better viewed with the $3$ generators you gave.