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Let $\alpha$ be an algebraic integer and let $\mathcal{O}_{\mathbb{Q}(\alpha)}$ be the ring of integers of $\mathbb{Q}(\alpha)$.

Question: How to characterize an algebraic integer $\alpha$ such that $\alpha \mathcal{O}_{\mathbb{Q}(\alpha)}$ is a prime ideal of $\mathcal{O}_{\mathbb{Q}(\alpha)}$?

Example: if $\alpha \in \mathbb{Z}$, then $\mathcal{O}_{\mathbb{Q}(\alpha)} = \mathbb{Z}$, so that $\alpha \mathcal{O}_{\mathbb{Q}(\alpha)}$ is a prime ideal iff $\alpha$ is a prime number (up to a sign).

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    $\begingroup$ If I'm not mistaken, a necessary condition is that the norm of $\alpha$ must be a prime power, as if $\alpha\mathcal{O}_{\mathbb{Q}(\alpha)}$ is a prime ideal, then $\mathcal{O}_{\mathbb{Q}(\alpha)}/\alpha\mathcal{O}_{\mathbb{Q}(\alpha)}$ is a finite field with cardinality $|N_{\mathbb{Q}(\alpha)/\mathbb{Q}}(\alpha)|$. This condition is clearly not sufficient, as taking a prime power in $\mathbb{Z}$ shows. $\endgroup$
    – Alex Wertheim
    Commented Jul 6, 2020 at 3:10
  • $\begingroup$ @AlexWertheim: If I am not mistaken, $α$ is a unit in $\mathcal{O}_{\mathbb{Q}(\alpha)}$ when (by definition) its norm is one. $\endgroup$
    – Sebastien Palcoux
    Commented Jul 7, 2020 at 5:05
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    $\begingroup$ Trivially, $\alpha=0$ qualifies. As remarked, for $\alpha\ne0$ we must have $|N_{K/\mathbb{Q}}(\alpha)|=p^m$ with $p$ prime and $1\leq m\leq n=[K:\mathbb{Q}]$, where $K=\mathbb{Q}(\alpha)$. If $m=1$ then $\alpha\mathcal{O}_{K}$ is a prime ideal. If $m=n$, the ideal is prime iff $p$ is inert in $K$; in that case, $\alpha$ simply equals $p\beta$ for some unit $\beta\in\mathcal{O}_{K}$. Don't know whether much can be said in case $1<m<n$. $\endgroup$
    – Matthé van der Lee
    Commented Jul 8, 2020 at 12:03
  • $\begingroup$ If $m=n-1$, for $\alpha\mathcal{O}_{K}$ to be prime one would have $p\in\alpha\mathcal{O}_{K}$, hence $\beta=\alpha/p$ is integral (which can be verified easily enough) of norm $p$, hence $\beta\mathcal{O}_{K}$ is prime, and $p\mathcal{O}_{K}$ $=$ $\alpha\mathcal{O}_{K}\cdot\beta\mathcal{O}_{K}$, So $p$ would need to split as the product of a degree $n-1$ and a degree $1$ prime in $K$. But this is not a criterion that can be readily checked from the minimum polynomial of $\alpha$ alone - even if it factorizes as $x^{n-1}(x-a)$ in $\mathbb{F}_{p}[x]$. $\endgroup$
    – Matthé van der Lee
    Commented Jul 8, 2020 at 22:51
  • $\begingroup$ $\beta=p/\alpha$, of course. $\endgroup$
    – Matthé van der Lee
    Commented Jul 8, 2020 at 23:04

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