Find a field extension given a galois group $Z_{3} \times Z_{3} \times Z_{3}$

Question

So I want to find a field extension that has the galois group $Z_{3} \times Z_{3} \times Z_{3} $. Now if the 3's where changed to 2's then I guess for example $(x^2-2)(x^2-3)(x^2-5)$ would suffice but I don't see how any clever way to do it with $Z_{3}$. I tried a bit with cyclotomic extensions but came up empty handed. Any hints & help are appreciated, thanks!

Answer 1

If we take three primes $p\equiv1\pmod3$ then each cyclotomic field $\Bbb Q(\zeta_p)$ has a cyclic cubic subextension (generated by a Gaussian period). Take the compositum of these three fields.

To be specific, take $p=7$, $13$ and $19$. Then we can take $\Bbb Q(\gamma_7,\gamma_{13},\gamma_{19})$ where $$\gamma_7=\zeta_7+\zeta_7^6,$$ $$\gamma_{13}=\zeta_{13}+\zeta_{13}^5+\zeta_{13}^8+\zeta_{13}^{12}$$ and $$\gamma_{19}=\zeta_{19}+\zeta_{19}^7+\zeta_{19}^8+\zeta_{19}^{11} +\zeta_{19}^{12}+\zeta_{19}^{18}.$$

These Gaussian periods are $\gamma_p=\sum_{c\in C} \zeta_p^c$ where $c$ is the set of nonzero cubes modulo $p$.

Answer 2

First: Your example won't work. For example take $\theta=\sqrt[3]{2}.$

Then $$x^3-2=(x-\theta)(x^2+\theta x+\theta^2).$$

Since You can see that $(x^2+\theta x+\theta^2)$ has roots $$x=\theta\left(\dfrac{1}{2}\pm\dfrac{\sqrt{3}}{2}i\right).$$

Taking an extension over $\Bbb{Q}$ you find that the extension $\Bbb{Q}(\sqrt[3]{2})$ has a group of $\Bbb{Q}$ fixing automorphisms $G=\{1\}$ since the only place you can send $\sqrt[3]{2}$ is itself. So you need a bigger extension: $\Bbb{Q}\subset \Bbb{Q}(\sqrt[3]{2},\sqrt{3}i).$ Now, you have three places to send $\sqrt[3]{2},$ that is, any of the roots of the above polynomial, and that gives you a copy of $\Bbb{Z}/3\Bbb{Z}.$ However, you need to check that the automorphisms fixing $\Bbb{Q}$ don't commute (if you don't believe then use the complex conjugation automorphism), so you get a Galois group isomorphic to $S_3.$

Now go read Lord Shark the Unknown's answer.