Intermediate field extensions of $\mathbb{Q}(\zeta_3)/\mathbb{Q}$

I'm trying to determine the intermediate field extensions of $\mathbb{Q}(\zeta_3)/\mathbb{Q}$. The minimal polynomial of $\zeta_3$ is $x^3+1$, which has roots $\zeta_3, \zeta_3^2$ and $-1$. Therefore, the automorphisms are given by sending $\zeta_3 \mapsto \zeta_3$, $\zeta_3 \mapsto \zeta_3^2$ and $\zeta_3 \mapsto -1$. The first automorphism trivially fixes the entire field $\mathbb{Q}(\zeta_3)$, thus we need to do nothing more for that. For the second automorphism, we consider that $\zeta_3 + \zeta_3^2$ is invariant under this automorphism. Therefore, we find that $(\zeta_3+\zeta_3^2)=-1 \implies$ this automorphism fixes $\mathbb{Q}$. Thus, we need only consider that $\zeta_3 -1$ is invariant under the third automorphism. But we find that $(\zeta_3 - 1)^2 = \zeta_3 - 2\zeta_3 + 1=-\zeta_3+1 = \frac{3}{2}-\frac{3\sqrt{3}}{2}i$... What do I do here?

Is the intermediate extension $\mathbb{Q}(\sqrt{3}i)$?

• If you are talking about $x^{3}+1$ then the roots are $-1,\xi_{3}$ and $-\xi_{3}^{2}$ not $\xi_{3}^{2}$, where $\xi_{3}=e^{\frac{i\pi}{3}}=\frac{1}{2}+\frac{\sqrt3}{2}i.$ And $x^{3}+1=(x+1)(x^{2}-x+1)$ is reducible. Please also see comments below by @carmichael561 – user188634 May 25 '16 at 5:09

The minimal polynomial of $\zeta_3$ is in fact $x^2+x+1$, since $x^3-1$ has $1$ as a root. Therefore $\mathbb{Q}(\zeta_3)$ is a quadratic extension of $\mathbb{Q}$, so there are no intermediate subfields.
• I think Jordan is talking about the root of $x^{3}+1$. – user188634 May 25 '16 at 5:07
• Then he should have said $\zeta_6$! And noted that $-1$ is a root! – user14972 May 25 '16 at 5:09
• @Yifan Wu: $\zeta_3$ generally means a primitive third root of unity. – carmichael561 May 25 '16 at 5:09
• Also, $-\zeta_3$ is a primitive $6$th root of unity, so my answer is still correct. – carmichael561 May 25 '16 at 5:11