Ring of algebraic integers in a quadratic extension of a cyclotomic field

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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. $\square$