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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 $$\mathfrak{p}\subset \mathbb{Z}(\zeta_m)$$ lying over $(p)$ and $$\mathfrak{P}\subset \mathbb{Z}(\zeta_n)$$ 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 $\mathbb{Q}(\zeta_m)$? Also, if I take the trace or the norm down to $\mathbb{Q}$, can I say anything about their congruence properties there?


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1 Answer 1

If $\mathcal{A}_m$ is the ring of integers of $\mathbb{Q}(\zeta_m)$, then $\mathfrak{B} \cap \mathcal{A}_m = \mathfrak{p}$.

You say $a \equiv b \pmod{\mathfrak{B}}$ where $a,b \in \mathbb{Q}(\zeta_m)$. Does this mean $a,b$ are algebraic integers (in $\mathcal{A}_m$)? If not, it must mean that there is some denominator $d \in \mathcal{A}_m$, $d \notin \mathfrak{B}$ such that $d(a-b) \in \mathfrak{B}$. (Just like we could say $\frac{2}{3} \equiv \frac{7}{3} \pmod{5}$.)

Either way, the above fact gives us $a \equiv b \pmod{\mathfrak{p}}$.

I don't think you can relate Norm(a) to Norm(b) or Trace(a) to Trace(b) modulo (p). Consider $1 \equiv 2i \pmod{(1-2i)}$ in the Gaussian integers $\mathbb{Z}[i]$. Taking norm or trace gives us nothing modulo (5).

We can, however, use $\mathfrak{p} \cap \mathbb{Z} = (p)$ to say that ${\rm Norm}(a-b) \in (p)$, if $a,b$ are algebraic integers. If, however, we have some denominator $d \in \mathfrak{p}'$ where $\mathfrak{p}'$ is a conjugate ideal to $\mathfrak{p}$, then the norm will give us powers of $p$ in the numerator and denominator, so we have to be more careful.

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