Primary ideal

An ideal $I$ is primary when $ab\in I$ and $a\notin I$ imply $b^n\in I$ for some positive integer $n$.

If a prime ideal $P=(r)$ is generated by a non-zero-divisor, its $P$-primary ideals are exactly $(r^k)$ for $k\ge 1$, in any commutative ring. To prove $(r^k)$ is primary, if $ab\in (r^k)$ and $a\notin (r)$, the prime ideal condition gives $b=r{b}_1$; cancelling $r$ and inducting gives $b\in (r^k)$. Also $\sqrt{(r^k)}=(r)$. Conversely, for a $P$-primary ideal $Q$, choose the least $k$ with $r^k\in Q$. If $x\in Q\setminus (r^k)$, factor $x=r^{j}u$ as far as possible with $0\le j<k$. Then $u\notin (r)$, so the primary ideal condition forces $r^j\in Q$, a contradiction. Thus $Q=(r^k)$. The powers are strictly decreasing: equality $(r^k)=(r^{k+1})$ would let cancellation give $1=rc$. This also proves uniqueness of $k$.

In a Noetherian ring, an ideal $Q$ is $\mathfrak{m}$-primary for a maximal ideal $\mathfrak{m}$ if and only if ${\mathfrak{m}}^n\subseteq Q\subseteq \mathfrak{m}$ for some $n\ge 1$. For necessity, choose generators $x_1,\dots ,x_d$ of $\mathfrak{m}$ and exponents $e_i\ge 1$ with $x_i^{e_i}\in Q$. Every product of $1+{\sum }_i(e_i-1)$ generators lies in $Q$. If $\mathfrak{m}=0$, take $n=1$. Conversely the inclusions imply $\sqrt{Q}=\mathfrak{m}$, so $R/Q$ is a local ring with maximal ideal $\mathfrak{m}/Q$. Every element outside this maximal ideal is a unit, while every element inside is nilpotent. Therefore every zero divisor is nilpotent, which is the primary ideal condition. The reverse implication works for any commutative ring.

In the Noetherian local ring $A=k[[u,v]]$ with maximal ideal $\mathfrak{m}=(u,v)$, the embedding dimension is two but ${\mathfrak{m}}^r$ needs $r+1$ generators for every $r\ge 1$. The degree-$r$ monomials $u^r,u^{r-1}v,\dots ,v^r$ generate ${\mathfrak{m}}^r$, and their images form a basis of ${\mathfrak{m}}^r/{\mathfrak{m}}^{r+1}$, so no smaller generating list exists. These ideals are $\mathfrak{m}$-primary: if $ab\in {\mathfrak{m}}^r$ and $a\notin {\mathfrak{m}}^r$, then $b$ cannot be a unit; hence $b\in \mathfrak{m}$ and $b^r\in {\mathfrak{m}}^r$. Consequently embedding dimension bounds the minimum over all maximal-primary ideals, because the maximal ideal itself is eligible, but does not bound the number of generators of each such ideal.

The radical $\sqrt{I}=\{x:x^n\in I\text{ for some }n\ge 1\}$ of a primary ideal is prime. In a number ring this forces a nonzero primary ideal to have only one prime in its unique factorization.

A minimal primary decomposition of an ideal $I$ is an irredundant finite intersection $I={\bigcap }_i{Q}_i$ of primary ideals whose radicals are pairwise distinct.

For a finite minimal primary decomposition $I={\bigcap }_i{Q}_i$, these are the prime ideals $\sqrt{Q_i}$. This terminology applies without a Noetherian hypothesis when the decomposition exists. Its invariance follows from the intrinsic description as those prime ideals of the form $\sqrt{(I:x)}$, where $x\notin I$ and $(I:x)$ is a colon ideal. Indeed, $(I:x)={\bigcap }_i(Q_i:x)$, and each factor is either the whole ring or a primary ideal with radical of an ideal $\sqrt{Q_i}$. A finite intersection of prime ideals can be prime only if it equals one of them: otherwise multiplying one element outside the intersection from each factor contradicts the prime ideal condition. Conversely, irredundancy supplies $x\in {\bigcap }_{j\ne i}{Q}_j\setminus Q_i$, giving $\sqrt{(I:x)}=\sqrt{Q_i}$. In a Noetherian ring these agree with the associated primes of a module $R/I$. For arbitrary rings, the annihilator definition of an associated prime of a module must not be substituted without a further hypothesis.

The distinct radicals in a minimal primary decomposition are exactly the associated primes of a module $R/I$, and hence are invariant. The injection into the sum of primary quotients proves one inclusion; an element supported on one component, supplied by irredundancy, proves the other. This does not imply uniqueness of embedded primary components.

A component in a minimal primary decomposition is embedded if its radical of an ideal strictly contains another radical occurring in that decomposition. Such components need not be unique. For example,in $k[x,y]$ gives different $(x,y)$-primary ideals as embedded components, with the same isolated component $(x)$.

An isolated prime of a minimal primary decomposition is a minimal member of the set of radicals of its primary components.

In a Noetherian ring, the primary component belonging to each isolated prime $\mathfrak{p}$ is uniquely determined byEmbedded primary components need not be unique.