[[Prime ideal]]
# Prime order of an ideal

Let $R$ be a [[commutative ring]], $\mathfrak{p}$ be a prime ideal, and $\mathfrak{a}$ be an ideal.
Then $\opn{ord}_{\mathfrak{p}}(\mathfrak{a})$ is the largest $m \in \mathbb{N}_{0}$ such that $\mathfrak{p}^m \mid \mathfrak{a}$, #m/def/ring 
see [[product ideal]].[^2022]
For $\alpha \in R$, we also write $\opn{ord}_{\mathfrak{p}}(\alpha) := \opn{ord}_{\mathfrak{p}}(\langle \alpha \rangle)$.

  [^2022]: 2022\. [[Sources/@bakerAlgebraicNumberTheory2022|Algebraic number theory course notes]], p. 26

## Properties

1. $\opn{ord}_{\mathfrak{p}}(\mathfrak{a}\mathfrak{b}) \geq \opn{ord}_{\mathfrak{p}}(\mathfrak{a}) + \opn{ord}_{\mathfrak{p}}(\mathfrak{b})$, which becomes an equality if $R$ admits [[Unique factorization of ideals|UFI]].

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#state/tidy | #lang/en | #SemBr