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.


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

  • $\begingroup$ Sorry, my first answer was incomplete. For some $p$, you do get the desired congruence between traces, see my completed answer below. $\endgroup$ – Alex B. Oct 5 '10 at 4:34
  • $\begingroup$ I'd actually suggest a p-adic metric formulation: equivalent but perhaps clearer in the end. $\endgroup$ – Charles Matthews Oct 5 '10 at 7:43
  • $\begingroup$ 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 $\sigma \in \textrm{Gal}(\mathbb{Q}(\zeta_n)\mathbb{Q}(\zeta_m))$. $\endgroup$ – Jill Oct 5 '10 at 15:55
  • $\begingroup$ 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. $\endgroup$ – David Loeffler Oct 28 '10 at 11:39

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.

  • $\begingroup$ 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. $\endgroup$ – Cam McLeman Oct 5 '10 at 4:16

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.

  • $\begingroup$ I don't think your factorization of 7 is correct. In particular, $N(3-\zeta)=(3-\zeta)(3-\zeta^2)=9+3+1=13$. $\endgroup$ – Cam McLeman Oct 5 '10 at 4:19
  • $\begingroup$ (but good answer). $\endgroup$ – Cam McLeman Oct 5 '10 at 4:34
  • $\begingroup$ Oops, I meant $3+\zeta_3$ and $3+\zeta_3^2$, whose product is 7. $\endgroup$ – Gerry Myerson Oct 5 '10 at 6:05

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 ${\rm Norm}(a) - {\rm Norm}(b)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.