Fixed ring $\mathbb{Z}[\zeta]^\tau$ is the ring of integers of the fixed field $\mathbb{Q}(\zeta)^\tau$

Question

Let $p$ be a prime number and $\zeta=\exp(2\pi i/p)$ the primitive $p^{\rm th}$ root of unity. We know $\mathbb{Q}(\zeta)$ is a cyclotomic field with ring of integers $\mathbb{Z}[\zeta]$. We also know $Aut_\mathbb{Q}(\mathbb{Q}(\zeta))$ is cyclic of order $p-1$, generated by $\sigma$, say.

Now, if $e|p-1$, then there exists one (and only one) intermediate field, and this corresponds to the subgroup $<\sigma^{(p-1)/e}>$ (let's write $\tau=\sigma^{(p-1)/e}$ for ease). This field is the fixed field of $\tau$, which we write as $\mathbb{Q}(\zeta)^\tau$ and $|\mathbb{Q}(\zeta)^\tau/\mathbb{Q}|=|<\sigma>/<\sigma^{(p-1)/e}>|=(p-1)/e$.

It is my understanding that $\mathbb{Q}^\tau=\mathbb{Q}(\eta)$, for some $\eta$ which is a finite sum of powers of $\zeta$. In other words, $\mathbb{Q}^\tau$ is a number field. I am wondering, is it true that $\mathbb{Z}[\zeta]^\tau$ is the ring of integers of $\mathbb{Q}(\zeta)^\tau$?

My thinking is that it must be, since $\alpha\in\mathbb{Q}(\zeta)^\tau\subset\mathbb{Q}(\zeta)$ is an algebraic integer if and only if there exists a monic polynomial $f(x)\in\mathbb{Z}[x]$ such that $f(\alpha)=0$. So $\alpha\in\mathbb{Z}[\zeta]^\tau$. Is this enough?

Answer 1

Yes. This is correct.

An integer of this intermediate field is:

• also an integer of the bigger field $\Bbb{Q}(\zeta)$, and
• fixed under $\tau$.