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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.

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Are you asking if ${\mathcal O}_L$ contains at least one such number for every choice of $L$ or for just some choice of $L$? – KConrad Nov 25 '11 at 14:28
@KConrad: The answer is clearly no if n=2, 3, 4. The question is whether there exist cyclotomic fields that contain a unit congruent to 1-\xi modulo 4. Methinks that this is a legitimate question. – Franz Lemmermeyer Nov 25 '11 at 19:20
@Franz: or any other $n$ up to 15, but maybe less clearly and more a magma computation. – Dror Speiser Nov 25 '11 at 23:26
I understood that the result isn't true for some $L$, but I was trying to get Udi to clarify what it is exactly he's looking for: would just one such field $L$ be enough for him or is he looking for infinitely many? Perhaps if he would explain why he is asking this question it would be clearer what he really wants to know. – KConrad Nov 26 '11 at 2:20
well, I will be happy with one such $L$, I certainly do not need infinitely many. The reason I ask the question is the following: I am trying to classify Hopf orders of a certain Hopf algebra. I am trying to understand if orders exist over some cyclotomic ring. The question boils down to understanding if there are p,n, and omega such that such an algebraic integer exist. So I will be happy with just one example, or a prove that such an algebraic integer cannot exist. Thanks! – Udi Nov 27 '11 at 13:58

3 Answers 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. $\square$

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In the formulation of the question, the $\zeta$ on the right is not the same as the one in $\omega$, but a primitive pth root of unity (it is actually denoted $\xi$) so $1-\zeta$ is never a unit. – Dror Speiser Dec 19 '11 at 8:58
Right, but what I am saying is that $\sigma(u) \equiv u \mod 1-\zeta$, and it is the $n$-th root of unity for which that is true. – David Speyer Dec 19 '11 at 14:08
The same proof goes through for the more general congruence \omega = \alpha^2 (1-\zeta) mod 4 as long as n is odd. – Franz Lemmermeyer Dec 19 '11 at 17:24

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.

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Small magma computation shows that looking only at cyclotomic units doesn't work for all odd $n\le 100$. – Dror Speiser Dec 17 '11 at 13:28
If you post the code as an answer, perhaps others could modify it and possibly push it further? – Franz Lemmermeyer Dec 17 '11 at 14:23

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.

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I have written a few (~30) more lines so to actually get the units satisfying Franz's congruence. If anyone is interested feel free to email me. – Dror Speiser Dec 20 '11 at 22:32

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