Ring of algebraic integers in a quadratic extension of a cyclotomic field 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. 
 A: This is more of a longish comment explaining why I believe that the answer should be yes,
and that a confirmation should be within reach using current technology.
Let $K$ be the field of $n$-th roots of unity, and $\zeta_p$ a primitive $p$-th root of unity for some prime factor $p \mid n$. The question is whether there is a nonsquare unit $\omega \in {\mathcal O}_K^\times$ such that $\omega \equiv 1 - \zeta \bmod 4$.
I can see no reason why such units should not exist, even in the special case $n = p$. But one may have to look for a while before stumbling over an example. The related problem of finding a nonsquare unit $\omega \equiv 1 \bmod 4$, for example, is not solvable for $n = p < 29$ since the corresponding  cyclotomic fields have odd class number; ${\mathbb Q}(\zeta_{29})$, on the other hand, has a class group of type $(2,2,2)$ and a good chance of containing such a unit.
This can probably be verified by looking only at cyclotomic units, which are known explicitly.
In your case, you should look at products of units of the form
$$ \omega = (1+\xi)^{a_1}(1+\xi+\xi^2)^{a_2}(1+\xi+\xi^2+\xi^3)^{a_3} \cdots $$
with not all $a_j$ even, and check whether one of these lies in the residue class 
$1 - \xi^j \bmod 4$. Using sage or pari, this should actually be doable. Perhaps some linear algebra and the Chinese remainder theorem can be used to speed up the calculations.
Edit. I was so convinced that there would be a solution of the problem for a small $n$ that I did not do what I should have done: the problem in question is equivalent to
the congruence $\omega \equiv \alpha^2 (1 - \zeta) \bmod 4$ for some cyclotomic integer $\alpha$ coprime to $2$. I guess Dror's code can easily be adapted to the more general congruence.
A: As requested by Franz, here is the short Magma code looking for solutions in cyclotomic units:
test_n := function(n)
    K<z> := CyclotomicField(n);
    O := MaximalOrder(K);
    I := ideal<O|4>;
    R := quo<O|I>;
    G,p := MultiplicativeGroup(R);
    p1 := p^-1;
    H := sub<G|[p1(z)] cat [p1(c) : c in CyclotomicUnits(K)]>;
    return [p : p in PrimeDivisors(n) | p ne 2 and p1(1-z^(n div p)) in H];
end function;

I have run this for $n$ up to $171$, always returning an empty set. This is naive code and can be speeded up like Franz says.

Given David's answer, it is now interesting to look at the even case. Here there are in fact many solutions. For $n\le 100$ there are solutions when $n$ is 28, 56, 60, 92, with $p$ being 7, 7, 3, 23, respectively.
A: 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. $\square$
