---
tag: V026AAV8E6
kind: definition
name: Primary ideal
msc: ["13A15"]
defines: TMTPWVETQK
uses: [0GYGHKX834, 0MYSA6MYNA, 437YRCNYRJ, 06HC6HK0Z8]
sources:
  - url: https://en.wikipedia.org/wiki/Primary_ideal
---

## Definition

Let $R$ be a [[commutative ring|0GYGHKX834]].
An [[ideal|0MYSA6MYNA]] $Q$ of $R$ is *primary* if these two conditions hold:

1. $Q \ne R$.
2. For all $a, b \in R$ with $ab \in Q$ and $a \notin Q$, there is an integer $n \ge 1$ with $b^n \in Q$.

## Remarks

The [[radical|437YRCNYRJ]] $P = \sqrt{Q}$ of a primary ideal is a [[prime ideal|06HC6HK0Z8]], and one says that $Q$ is $P$-primary.
Example: for a prime $p$ and $n \ge 1$, the ideal $(p^n)$ of $\Z$ is primary; if $p^n \mid ab$ and $p^n \nmid a$, then $p \mid b$, so $p^n \mid b^n$.
Non-example: $(6)$ in $\Z$ fails clause 2, because $2 \cdot 3 \in (6)$ and $2 \notin (6)$, but no power of $3$ is divisible by $6$.
Every prime ideal is primary, and every power of a maximal ideal is primary.
A power of a prime ideal need not be primary: in $R = k[x, y, z]/(xy - z^2)$ over a field $k$, with $P = (x, z)$, one has $xy = z^2 \in P^2$, $x \notin P^2$ and no power of $y$ in $P^2$.

## Explanations

### L1 human

A primary ideal contains a product only for a strong reason.
Either the first factor already lies in it, or some power of the second factor does.