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Let $l$ be an odd prime number and $\zeta$ be a primitive $l$-th root of unity in $\mathbb{C}$. Let $K = \mathbb{Q}(\zeta)$. Let $A$ be the ring of algebraic integers in $K$.

My question: Is there a purely imaginary unit in $A$?

EDIT New question: Is the following proposition true? If yes, how would you prove this?

Proposition There is no purely imaginary unit in $A$.

Related questions:

On a certain property of the different of an extension of an algebraic number field of a prime relative degree

Maximal real subfield of Q(ζ)

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    @WillieWong It's not that simple. If I didn't add the latter question, I doubt that Matt E would give a definitive answer and David Loeffler would think the question interesting. Anyway, you know I almost always ask the latter type questions. I don't think this would ever happen again. I apologize anyone who was confused by my edit.2012-07-26

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Not necessarily: if $\ell = 3$ then there are only six units in $A$ and none of them are totally imaginary.

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    Just to ping you so you know that the OP changed his question and so you may want to re-write your answer.2012-07-25
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The proposition is true:

Let $K^+$ denote the totally real subfield of $K$. Let $U$ be the group of units in $K$, and $U^+$ be the group of units in $K^+$. Dirichlet's Theorem implies that $U^+$ has finite index in $U$. I claim that in fact the index is odd. Indeed, suppose that $u$ is an element of whose image in $U/U^+$ is of exact order $2$.

Then $K = K^+(u) = K^+(\sqrt{u^2})$ is obtained by extracting the square root of a unit, and so is unramified over $K^+$ except possibly at primes lying over $2$. However, we know that $K/K^+$ is ramified at precisely the prime lying over $l$.

Consequently $U/U^+$ has odd order.

If $u \in U$ were purely imaginary, then $u \not\in U^+$, but $u^2 = - u \overline{u} = - |u|^2$ is an element of $U^+$, contradicting what we have just proved. Thus $U$ contains no purely imaginary elements.

[Hopefully this is correct; the previous argument I posted was nonsense.]