TITLE: Ring of algebraic integers in a quadratic extension of a cyclotomic field
QUESTION [8 upvotes]: Hello,
I have a question which arose when trying to classify orders of certain algebras.
We know that if $K=\mathbb{Q}(\zeta)$ is any cyclotomic field, and $\zeta$ is an $n$-th root of unity (for some number $n$), then the ring of algebraic integers in $K$ is exactly $\mathcal{O}_K=\mathbb{Z}[\zeta]$.
Consider now the following quadratic extension $L=K(t)$ where $t$ satisfies the equation $$t^2 = \omega(1-\xi)$$ where $\xi$ is a $p$-th root of unity, where $p|n$ is some odd prime number, and $\omega$ is some unit in $\mathcal{O}_K=\mathbb{Z}[\zeta]$
I would like to ask the following question about the ring of integers $\mathcal{O}_L$ of $L$: 
does $\mathcal{O}_L$ contains elements of the form $X=\frac{1}{2}(a+bt)$ with $a,b\in \mathcal{O}_K$ and such that $2\nmid a$ and $2\nmid b$? The trace of such an element is $a\in\mathcal{O}_K$, but its determinant is $\frac{1}{4}(a^2-\omega(1-\xi)b^2)$, and I do not know if there are $a$ and $b$ in $\mathcal{O}_K - 2\mathcal{O}_K$ for which we will get an integral expression.
I will appreciate your help,
Thanks,
Udi.

REPLY [4 votes]: UPDATE: Previous argument was flawed. Here is what  can salvage. 
I can show there is no solution with $n$ an odd prime, or with $n$ odd and $\omega$ cyclotomic. 
Let $\sigma$ denote complex conjugation. 
Let $\zeta$ be a primitive $n$-th root of unity for $n$ odd and let $K = \mathbb{Q}(\zeta)$. 
Lemma: Let $n$ be an odd prime and let $u$ be any unit of $K$, or let $n$ be odd and let $u$ be a cyclotomic unit of $K$. We have $\sigma(u)/u = \zeta^k$ for some integer $k$.
Proof: First, note that $\sigma(u)/u$ is an algebraic integer and all its Galois conjugates have norm $1$. So, by a result of Kronecker, it is a root of unity, and must be of the form $\pm \zeta^k$. Our goal is to show that the minus sign is impossible.
If $u$ is a cyclotomic unit, then it is a product of terms of the form $(1-\zeta^a)/(1-\zeta)$ and an explicit computation shows that the sign is positive.
Now, suppose that $n$ is an odd prime. So, suppose that $\sigma(u)/u = - \zeta^k$. Since $k$ is only determined modulo the odd number $n$, we may assume that $k$ is even. Replacing $u$ by $\zeta^{-k/2} u$, we have $\sigma(u)/u = -1$. 
But, for any algebraic integer $v$ in $K$, we have $\sigma(v) \equiv v \mod 1-\zeta$. So $\sigma(u) \equiv u \mod 1- \zeta$ and (since $u$ is a unit) we have $\sigma(u)/u \equiv 1 \mod 1-\zeta$. Putting these together, we deduce that $1 \equiv -1 \mod 1-\zeta$. Since $n$ is an odd prime, $1-\zeta$ is a prime which does not divide $2$, a contradiction. $\square$
The error in the earlier version was forgetting that $1-\zeta$ can itself be a unit when $n$ is not prime. (In fact, this occur whenever $n$ is a square free non-prime.) And this unit, of course, violates the lemma. Not sure whether the original statement might still be true in these cases. 
Now suppose that we have a solution to 
$$\omega \equiv  1-\zeta^m \mod 4$$
for $m$ a proper divisor of $n$ and $\omega$ a unit. (I am using Franz's rephrasing.)
We hit both sides with $u \mapsto \sigma(u)/u$. By the lemma, we have $\sigma(\omega)/\omega = \zeta^j$ for some $j$. Also, $\sigma(1-\zeta^m)/(1-\zeta^m) = - \zeta^{-m}$.
So
$$\zeta^j \equiv - \zeta^k \mod 4$$
This equation is not true (using again that $n$ is odd), so we have a contradiction. $\square$