# Ring of integers of a CM field

Let $L$ be a totally real number field such that $\zeta_p+\zeta_p^{-1} \in L$, where $\zeta_p$ is a primitive $p$-th root of unity ($p$ is an odd prime number). Let $F = L(\zeta_p)$. Then $F/L$ is a CM extension.

Write $O_F$ for the ring of integers of $F$ and $O_L$ for the ring of integers of $L$. I wonder if it is true that the ring $O_F$ is can be generated by a single element, so is of the form $O_L[\alpha]$.

If the ramification index of $p$ in $L$ is equal to $(p-1)/2$ (for all primes above p), then $O_F = O_L[\zeta_p]$, but in general this is false. For instance you can take $p$ congruent to $3$ modulo $4$, and $L = {\mathbb Q}(\sqrt p, \zeta_p + \zeta_p^{-1})$. In this case $O_F = O_L[i]$ but $O_F \neq O_L[\zeta_p]$.

In the comments a proof for $O_F \neq O_L[\zeta_p]$ was requested, but since it didn't fit, I put it here. There should be a simpler way to argue, but here is something: Suppose $O_F = O_L[\zeta_p]$ for a contradiction. The different $D_F$ of $F/{\mathbb Q}$ is equal to the product of the different $D_L$ of $L/{\mathbb Q}$ with the relative different $D_{F/L}$. Since by assumption $O_F = O_L[\zeta_p]$, I compute $D_{F/L} = (\zeta_p - \zeta_p^{-1})$. Thus the discriminant of $F$ is equal to the product of the norm of the $O_F$-ideal generated by $D_L$ down to ${\mathbb Q}$ times the norm of $(\zeta_p - \zeta_p^{-1})$ down to ${\mathbb Q}$. We have $N_{F/{\mathbb Q}}(\zeta_p - \zeta_p^{-1}) = p^2$, which is coprime to $2$. One can also see that $2$ is unramified in $L/{\mathbb Q}$, and thus $2$ does not divide the norm of $D_L$ down to ${\mathbb Q}$. The conclusion is that $2$ is coprime to the discriminant of $F/{\mathbb Q}$, but that is a contradiction since $2$ ramifies in ${\mathbb Q}(i) \subset F$.

• Can you prove your very last claim? Aug 30 '17 at 9:13
• I used Prop 17 (p. 68) from Lang's ANT book. This states that if E and M are two number fields which are (a) linearly disjoint and (b) have coprime discriminant, then $O_{EM}$ is the compositum of $O_E$ and $O_M$. So it suffices to apply this lemma to $M = {\mathbb Q}(i)$ and $E = {\mathbb Q}(\zeta_p+\zeta_p^{-1})$.
– mnr
Aug 30 '17 at 9:36
• This proves the next to last claim. How do you prove that $\zeta_p$ does not generate the ring of integers? Aug 30 '17 at 11:07
• I added a sketch in the question (it didn't fit here in the comments).
– mnr
Aug 30 '17 at 14:12
• The extension $L/F$ is only ramified at the infinite primes, hence $D_{F/L} = (1)$. This follows e.g. from $F = L(i) = L(\zeta_p)$. I also doubt that $O_F = O_L[i]$ since clearly $(i + \sqrt{p})/2$ is integral for $p \equiv 3 \bmod 4$. As a rule, relative integral bases are a mess. Aug 30 '17 at 14:46