Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\mathcal{O}$ be the ring of integers of a number field, let $\mathfrak{p}$ be a prime ideal of $\mathcal{O}$, and let $\alpha$ be an element of $\mathcal{O}$. Is it always possible to find an integer $a$ such that $\alpha \equiv a \pmod{\mathfrak{p}}$? If so, how can one find it?

Thanks.

EDIT: Assume that the inertial degree of $\mathfrak{p}$ is 1. Gerry Myerson has shown that the answer to my question is no if this is not assumed.

EDIT: If the inertial degree of $\mathfrak{p}$ is 1, then $\mathcal{O} / \mathfrak{p} = \mathbb{Z} / p\mathbb{Z}$, and the question is trivial.

share|improve this question

1 Answer 1

up vote 5 down vote accepted

Not always possible. In the Gaussian integers, let $I=(3)$. There is no integer congruent to $i$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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