# Finding $a \in \mathbb{Z}$ such that $\alpha \equiv a \pmod{\mathfrak{p}}$

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.

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