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I have the following situation: let $m,n$ be integers such that $m|n$ and let $\zeta_m$, $\zeta_n$ denote primitive $m$ and $n$th roots of unity. Then we have the inclusion of fields

$$\mathbb{Q}\subset \mathbb{Q}(\zeta_m) \subset \mathbb{Q}(\zeta_n)$$ Now suppose we also have primes (where $(p,n)=1$) $$(p)\subset \mathbb{Z}$$ and then $$\mathfrak{p}\subset \mathbb{Q}(\zeta_m)$$ lying over $(p)$ and $$\mathfrak{P}\subset \mathbb{Q}(\zeta_n)$$ lying over $\mathfrak{p}$.

I have a congruence in $\mathbb{Q}(\zeta_n)$ of the form $a\equiv b \pmod{\mathfrak{P}}$, where $a,b$ are actually elements of $\mathbb{Q}(\zeta_m)$.

What can I say about the congruence properties of $a,b$ in $\mathbb{Q}(\zeta_m)$? More importantly, if I take the trace or the norm down to $\mathbb{Q}$, can I say anything about their congruence properties there? Ideally I'd like a congruence of something in the integers.

Thanks!

Edit: Are there any assumptions that you can make that might give congruences mod a prime power?

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Sorry, my first answer was incomplete. For some $p$, you do get the desired congruence between traces, see my completed answer below. –  Alex B. Oct 5 '10 at 4:34
    
I'd actually suggest a p-adic metric formulation: equivalent but perhaps clearer in the end. –  Charles Matthews Oct 5 '10 at 7:43
    
I guess I am still curious about when can you say something further. For instance maybe when $a\equiv b \pmod{\mathfrak{B^{\sigma}}}$ for all $\sigma \in \textrm{Gal}(\mathbb{Q}(\zeta_n)\mathbb{Q}(\zeta_m))$. –  Jill Oct 5 '10 at 15:55
    
Thanks for all the great answers so far, also :) –  Jill Oct 5 '10 at 15:56
    
I've retagged this as "nt.number-theory" in line with the general principle that each question should have one tag corresponding to an Arxiv subject class. –  David Loeffler Oct 28 '10 at 11:39
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3 Answers

up vote 4 down vote accepted

Sure. $a\equiv b\pmod{\mathfrak{P}}$ just means $a-b\in\mathfrak{P}$. Taking norms to any subfield $K$ of $\mathbb{Q}(\zeta_n)$ (e.g., $\mathbb{Q}$ or $\mathbb{Q}(\zeta_m)$) gives you $N_{\mathbb{Q}(\zeta_n)/K}(a-b)\in N_{\mathbb{Q(\zeta_n)}/K}\mathfrak{P}.$

For $K=\mathbb{Q}$, the latter norm is just $p^f$ where $f$ is the order of $p\pmod{n}$.

For $K=\mathbb{Q}(\zeta_m)$, the former norm is $(a-b)^{\phi(n)/\phi(m)}$ and the latter is $\mathfrak{p}^{f'}$, where $f'$ is the easily-calculated relative residue degree.

This doesn't give you an explicit congruence between $a$ and $b$, but given Gerry's answer, that might have been too much to ask for anyway. On the other hand, if $\phi(n)/\phi(m)$ is small or (as in Alex's answer) if $p$ has few factors in $\mathbb{Q}(\zeta_m)$, you get something at least slightly non-stupid out.

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Of course, $N(a-b)\neq N(a)-N(b)$, so this is a looser notion of "congruence of something in the integers" than in Gerry's answer. –  Cam McLeman Oct 5 '10 at 4:16
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Let $m=3$, $n=6$, $p=7$. The prime over 7 is $3\pm\zeta_3$. Let $a=1$, $b=4+\zeta_3$, so $a\equiv b\pmod{3+\zeta_3}$. The trace of $a$ is 2, the trace of $b$ is $4+\zeta_3+4+\zeta_3^2=7$, so there's no congruence (mod 7) there. The norm of $a$ is 1, the norm of $b$ is $(4+\zeta_3)(4+\zeta_3^2)=16-4+1=13$, so there's no congruence (mod 7) there, either. I think this shows that, in general, there's no congruence in the integers.

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I don't think your factorization of 7 is correct. In particular, $N(3-\zeta)=(3-\zeta)(3-\zeta^2)=9+3+1=13$. –  Cam McLeman Oct 5 '10 at 4:19
    
(but good answer). –  Cam McLeman Oct 5 '10 at 4:34
    
Oops, I meant $3+\zeta_3$ and $3+\zeta_3^2$, whose product is 7. –  Gerry Myerson Oct 5 '10 at 6:05
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Edit: sometimes you do get the congruence you want for traces. See corrected answer:

to say that $a\equiv b \; ({\rm mod}\;{\mathfrak P})$ is equivalent to saying $a-b \in {\mathfrak P}$, so that ${\rm Tr}(a) - {\rm Tr}(b) \in \sum_{\sigma\in G}{\mathfrak P}^\sigma$, where $G$ is the Galois group of $\mathbb{Q}(\zeta_m)/\mathbb{Q}$. If $p$ splits in $\mathbb{Q}(\zeta_m)$, then the sum is just the whole ring of integers, since ${\mathfrak P}$ is prime, hence maximal. So in this case, this doesn't give you any information. If on the other hand ${\mathfrak P}$ is the unique prime above $p$, then you get the desired statement that ${\rm Tr}(a-b)\in {\mathfrak P} \cap \mathbb{Q}$, so ${\rm Tr}(a)\equiv {\rm Tr}(b)\; {\rm mod}\; p$, as required.

As for the norm, it is true that ${\rm Norm}(a-b)$ is in $(p)$, but since the norm is not linear, I wouldn't expect this to tell you anything about ${\rm Norm}(a) - {\rm Norm}(b)$.

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