Primary ideal explained

In mathematics, specifically commutative algebra, a proper ideal Q of a commutative ring A is said to be primary if whenever xy is an element of Q then x or yn is also an element of Q, for some n > 0. For example, in the ring of integers Z, (pn) is a primary ideal if p is a prime number.

The notion of primary ideals is important in commutative ring theory because every ideal of a Noetherian ring has a primary decomposition, that is, can be written as an intersection of finitely many primary ideals. This result is known as the Lasker–Noether theorem. Consequently,[1] an irreducible ideal of a Noetherian ring is primary.

Various methods of generalizing primary ideals to noncommutative rings exist,[2] but the topic is most often studied for commutative rings. Therefore, the rings in this article are assumed to be commutative rings with identity.

Examples and properties

ak{q}

is primary if, whenever

xy\inak{q}

, we have

x\inak{q}

or

y\inak{q}

or

x,y\in\sqrt{ak{q}}

. (Here

\sqrt{ak{q}}

denotes the radical of

ak{q}

.)

R=k[x,y,z]/(xy-z2)

,

ak{p}=(\overline{x},\overline{z})

, and

ak{q}=ak{p}2

, then

ak{p}

is prime and

\sqrt{ak{q}}=ak{p}

, but we have

\overline{x}\overline{y}={\overline{z}}2\inak{p}2=ak{q}

,

\overline{x}\not\inak{q}

, and

{\overline{y}}n\not\inak{q}

for all n > 0, so

ak{q}

is not primary. The primary decomposition of

ak{q}

is

(\overline{x})\cap({\overline{x}}2,\overline{x}\overline{z},\overline{y})

; here

(\overline{x})

is

ak{p}

-primary and

({\overline{x}}2,\overline{x}\overline{z},\overline{y})

is

(\overline{x},\overline{y},\overline{z})

-primary.

A\toAP

, the map from A to the localization of A at P, is the intersection of all P-primary ideals.[4]

ak{p}

-primary ideals is

ak{p}

-primary but an infinite product of

ak{p}

-primary ideals may not be

akp

-primary; since for example, in a Noetherian local ring with maximal ideal

akm

,

\capnak{m}n=0

(Krull intersection theorem) where each

ak{m}n

is

ak{m}

-primary, for example the infinite product of the maximal (and hence prime and hence primary) ideal

m=\langlex,y\rangle

of the local ring

K[x,y]/\langlex2,xy\rangle

yields the zero ideal, which in this case is not primary (because the zero divisor

y

is not nilpotent). In fact, in a Noetherian ring, a nonempty product of

ak{p}

-primary ideals

Qi

is

ak{p}

-primary if and only if there exists some integer

n>0

such that

ak{p}n\subset\capiQi

.

Footnotes

  1. To be precise, one usually uses this fact to prove the theorem.
  2. See the references to Chatters–Hajarnavis, Goldman, Gorton–Heatherly, and Lesieur–Croisot.
  3. For the proof of the second part see the article of Fuchs.
  4. Atiyah–Macdonald, Corollary 10.21

References

External links