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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 $K$. Let $p \ne l$ be a prime number. Let $f$ be the order of $p$ modulo $l$, i.e. the smallest positive integer such that $p^f \equiv 1$ (mod $p$). Let $P$ be a prime ideal of $A$ lying over $p$.

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

Proposition Let $\alpha \in A$. Then there exist rational integers $a_0, ..., a_{f-1}$ such that $\alpha \equiv a_0 + a_1\zeta + ... + a_{f-1}\zeta^{f-1}$ (mod $P$). Here $a_0, ..., a_{f-1}$ are uniquely determined mod $p$.

Greg Martin
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Makoto Kato
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1 Answers1

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By my answer to this question, the degree of $P$ is $f$. Let $\omega$ be the image of $\zeta$ by the canonical homomorphism $\mathbb{Z}[\zeta] \rightarrow \mathbb{Z}[\zeta]/P$. Clearly $\mathbb{Z}[\zeta]/P = (\mathbb{Z}/p\mathbb{Z})(\omega)$. Hence every element $x \in \mathbb{Z}[\zeta]/P$ can be uniquely written as $x = a_0 + a_1\omega + \cdots + a_{f-1} \omega^{f-1}$, where $a_i \in \mathbb{Z}/p\mathbb{Z}$. This completes the proof.

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Makoto Kato
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