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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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Please don't edit questions in such a way that render existing answers non-sense. –  Willie Wong Jul 25 '12 at 13:57
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Don't you think that 40+ questions in 2 days is enough for the community to think about? Why not leaning back and thinking about accepting some of given answers? –  draks ... Jul 25 '12 at 20:59
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Dear Makoto, You have been asking many questions about algebraic number theory, and in particular, the algebraic number theory of cyclotomic fields. These are very well-studied subjects, with many texts available. (E.g. the books on cyclotomic fields by Lang and Washington.) Have you considered looking at some of these books to answer your questions? Regards, –  Matt E Jul 26 '12 at 2:32
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I'd actually say that of the OP's many recent questions on cyclotomic fields, this one (in its edited form) is probably the most interesting, so I'm not sure why people are piling in to down-vote it. –  David Loeffler Jul 26 '12 at 8:46
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@Makoto: I can understand when you have difficulty understanding what I wrote, sometimes my use of language is a bit obtuse. But try to at least understand what you wrote yourself! I quote: positive statement "Is there a purely imaginary unit in A?" negative statement "Is the following proposition true? Proposition: there is no imaginary unit in A." –  Willie Wong Jul 26 '12 at 11:44

2 Answers 2

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. –  Willie Wong Jul 25 '12 at 13:58

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.]

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