# Primary decomposition

In mathematics, the Lasker–Noether theorem states that every Noetherian ring is a Lasker ring, which means that every ideal can be decomposed as an intersection, called primary decomposition, of finitely many primary ideals (which are related to, but not quite the same as, powers of prime ideals). The theorem was first proven by Emanuel Lasker (1905) for the special case of polynomial rings and convergent power series rings, and was proven in its full generality by Emmy Noether (1921).

The Lasker–Noether theorem is an extension of the fundamental theorem of arithmetic, and more generally the fundamental theorem of finitely generated abelian groups to all Noetherian rings. The Lasker–Noether theorem plays an important role in algebraic geometry, by asserting that every algebraic set may be uniquely decomposed into a finite union of irreducible components.

It has a straightforward extension to modules stating that every submodule of a finitely generated module over a Noetherian ring is a finite intersection of primary submodules. This contains the case for rings as a special case, considering the ring as a module over itself, so that ideals are submodules. This also generalizes the primary decomposition form of the structure theorem for finitely generated modules over a principal ideal domain, and for the special case of polynomial rings over a field, it generalizes the decomposition of an algebraic set into a finite union of (irreducible) varieties.

The first algorithm for computing primary decompositions for polynomial rings over a field of characteristic 0[Note 1] was published by Noether's student Grete Hermann (1926).[1][better source needed] The decomposition does not hold in general for non-commutative Noetherian rings. Noether gave an example of a non-commutative Noetherian ring with a right ideal that is not an intersection of primary ideals.

## Primary decomposition of an ideal

Let R be a Noetherian commutative ring. An ideal I of R is called primary if it is a proper ideal and for each pair of elements x and y in R such xy is in I, either x or some power of y is in I; equivalently, every zero-divisor in the quotient R/I is nilpotent. The radical of a primary ideal Q is a prime ideal and Q is said to be ${\displaystyle {\mathfrak {p}}}$ -primary for ${\displaystyle {\mathfrak {p}}={\sqrt {Q}}}$ .

Let I be an ideal in R. Then I has an irredundant primary decomposition into primary ideals:

${\displaystyle I=Q_{1}\cap \cdots \cap Q_{n}\ }$ .

Irredundancy means:

• Removing any of the ${\displaystyle Q_{i}}$  changes the intersection, i.e. for each i we have: ${\displaystyle \cap _{j\neq i}Q_{j}\not \subset Q_{i}}$ .
• The prime ideals ${\displaystyle {\sqrt {Q_{i}}}}$  are all distinct.

Moreover, this decomposition is unique in the two ways:

• The set ${\displaystyle \{{\sqrt {Q_{i}}}\mid i\}}$  is uniquely determined by I, and
• If ${\displaystyle {\mathfrak {p}}={\sqrt {Q_{i}}}}$  is a minimal element of the above set, then ${\displaystyle Q_{i}}$  is uniquely determined by ${\displaystyle I}$ ; in fact, ${\displaystyle Q_{i}}$  is the pre-image of ${\displaystyle IR_{\mathfrak {p}}}$  under the localization map ${\displaystyle R\to R_{\mathfrak {p}}}$ .

Primary ideals which correspond to non-minimal prime ideals over I are in general not unique (see an example below). For the existence of the decomposition, see #Primary decomposition from associated primes below.

The elements of ${\displaystyle \{{\sqrt {Q_{i}}}\mid i\}}$  are called the prime divisors of I or the primes belonging to I. In the language of module theory, as discussed below, the set ${\displaystyle \{{\sqrt {Q_{i}}}\mid i\}}$  is also the set of associated primes of the ${\displaystyle R}$ -module ${\displaystyle R/I}$ . Explicitly, that means that there exist elements ${\displaystyle g_{1},\dots ,g_{n}}$  in R such that

${\displaystyle {\sqrt {Q_{i}}}=\{f\in R\mid fg_{i}\in I\}.}$ [2]

By a way of shortcut, some authors call an associated prime of ${\displaystyle R/I}$  simply an associated prime of I (note this practice will conflict with the usage in the module theory).

• The minimal elements of ${\displaystyle \{{\sqrt {Q_{i}}}\mid i\}}$  are the same as the minimal prime ideals containing I and are called isolated primes.
• The non-minimal elements, on the other hand, are called the embedded primes.

In the case of the ring of integers ${\displaystyle \mathbb {Z} }$ , the Lasker–Noether theorem is equivalent to the fundamental theorem of arithmetic. If an integer n has prime factorization ${\displaystyle n=\pm p_{1}^{d_{1}}\cdots p_{r}^{d_{r}}}$ , then the primary decomposition of the ideal ${\displaystyle \langle n\rangle }$  generated by n in ${\displaystyle \mathbb {Z} }$ , is

${\displaystyle \langle n\rangle =\langle p_{1}^{d_{1}}\rangle \cap \cdots \cap \langle p_{r}^{d_{r}}\rangle .}$

Similarly, in a unique factorization domain, if an element has a prime factorization ${\displaystyle f=up_{1}^{d_{1}}\cdots p_{r}^{d_{r}},}$  where u is a unit, then the primary decomposition of the principal ideal generated by f is

${\displaystyle \langle f\rangle =\langle p_{1}^{d_{1}}\rangle \cap \cdots \cap \langle p_{r}^{d_{r}}\rangle .}$

### Examples

The examples of the section are designed for illustrating some properties of primary decompositions, which may appear as surprising or counter-intuitive. All examples are ideals in a polynomial ring over a field k.

#### Intersection vs. product

The primary decomposition in ${\displaystyle k[x,y,z]}$  of the ideal ${\displaystyle I=\langle x,yz\rangle }$  is

${\displaystyle I=\langle x,yz\rangle =\langle x,y\rangle \cap \langle x,z\rangle .}$

Because of the generator of degree one, I is not the product of two larger ideals. A similar example is given, in two indeterminates by

${\displaystyle I=\langle x,y(y+1)\rangle =\langle x,y\rangle \cap \langle x,y+1\rangle .}$

#### Primary vs. prime power

In ${\displaystyle k[x,y]}$ , the ideal ${\displaystyle \langle x,y^{2}\rangle }$  is a primary ideal that has ${\displaystyle \langle x,y\rangle }$  as associated prime. It is not a power of its associated prime.

#### Non-uniqueness and embedded prime

For every positive integer n, a primary decomposition in ${\displaystyle k[x,y]}$  of the ideal ${\displaystyle I=\langle x^{2},xy\rangle }$  is

${\displaystyle I=\langle x^{2},xy\rangle =\langle x\rangle \cap \langle x^{2},xy,y^{n}\rangle .}$

The associated primes are

${\displaystyle \langle x\rangle \subset \langle x,y\rangle .}$

Example: Let N = R = k[xy] for some field k, and let M be the ideal (xyy2). Then M has two different minimal primary decompositions M = (y) ∩ (x, y2) = (y) ∩ (x + yy2). The minimal prime is (y) and the embedded prime is (xy).

#### Non-associated prime between two associated primes

In ${\displaystyle k[x,y,z],}$  the ideal ${\displaystyle I=\langle x^{2},xy,xz\rangle }$  has the (non-unique) primary decomposition

${\displaystyle I=\langle x^{2},xy,xz\rangle =\langle x\rangle \cap \langle x^{2},y^{2},z^{2},xy,xz,yz\rangle .}$

The associated prime ideals are ${\displaystyle \langle x\rangle \subset \langle x,y,z\rangle ,}$  and ${\displaystyle \langle x,y\rangle }$  is a non associated prime ideal such that

${\displaystyle \langle x\rangle \subset \langle x,y\rangle \subset \langle x,y,z\rangle .}$

#### A complicated example

Unless for very simple examples, a primary decomposition may be hard to compute and may have a very complicated output. The following example has been designed for providing such a complicated output, and, nevertheless, being accessible to hand-written computation.

Let

{\displaystyle {\begin{aligned}P&=a_{0}x^{m}+a_{1}x^{m-1}y+\cdots +a_{m}y^{m}\\Q&=b_{0}x^{n}+b_{1}x^{n-1}y+\cdots +b_{n}y^{n}\end{aligned}}}

be two homogeneous polynomials in x, y, whose coefficients ${\displaystyle a_{1},\ldots ,a_{m},b_{0},\ldots ,b_{n}}$  are polynomials in other indeterminates ${\displaystyle z_{1},\ldots ,z_{h}}$  over a field k. That is, P and Q belong to ${\displaystyle R=k[x,y,z_{1},\ldots ,z_{h}],}$  and it is in this ring that a primary decomposition of the ideal ${\displaystyle I=\langle P,Q\rangle }$  is searched. For computing the primary decomposition, we suppose first that 1 is a greatest common divisor of P and Q.

This condition implies that I has no primary component of height one. As I is generated by two elements, this implies that it is a complete intersection (more precisely, it defines an algebraic set, which is a complete intersection), and thus all primary components have height two. Therefore, the associated primes of I are exactly the primes ideals of height two that contain I.

It follows that ${\displaystyle \langle x,y\rangle }$  is an associated prime of I.

Let ${\displaystyle D\in k[z_{1},\ldots ,z_{h}]}$  be the homogeneous resultant in x, y of P and Q. As the greatest common divisor of P and Q is a constant, the resultant D is not zero, and resultant theory implies that I contains all products of D by a monomial in x, y of degree m + n – 1. As ${\displaystyle D\not \in \langle x,y\rangle ,}$  all these monomials belong to the primary component contained in ${\displaystyle \langle x,y\rangle .}$  This primary component contains P and Q, and the behavior of primary decompositions under localization shows that this primary component is

${\displaystyle \{t|\exists e,D^{e}t\in I\}.}$

In short, we have a primary component, with the very simple associated prime ${\displaystyle \langle x,y\rangle ,}$  such all its generating sets involve all indeterminates.

The other primary component contains D. One may prove that if P and Q are sufficiently generic (for example if the coefficients of P and Q are distinct indeterminates), then there is only another primary component, which is a prime ideal, and is generated by P, Q and D.

### Geometric interpretation

In algebraic geometry, an affine algebraic set V(I) is defined as the set of the common zeros of an ideal I of a polynomial ring ${\displaystyle R=k[x_{1},\ldots ,x_{n}].}$

An irredundant primary decomposition

${\displaystyle I=Q_{1}\cap \cdots \cap Q_{r}}$

of I defines a decomposition of V(I) into a union of algebraic sets V(Qi), which are irreducible, as not being the union of two smaller algebraic sets.

If ${\displaystyle P_{i}}$  is the associated prime of ${\displaystyle Q_{i}}$ , then ${\displaystyle V(P_{i})=V(Q_{i}),}$  and Lasker–Noether theorem shows that V(I) has a unique irredundant decomposition into irreducible algebraic varieties

${\displaystyle V(I)=\bigcup V(P_{i}),}$

where the union is restricted to minimal associated primes. These minimal associated primes are the primary components of the radical of I. For this reason, the primary decomposition of the radical of I is sometimes called the prime decomposition of I.

The components of a primary decomposition (as well as of the algebraic set decomposition) corresponding to minimal primes are said isolated, and the others are said embedded.

For the decomposition of algebraic varieties, only the minimal primes are interesting, but in intersection theory, and, more generally in scheme theory, the complete primary decomposition has a geometric meaning.

## Primary decomposition from associated primes

Nowadays, it is common to do primary decomposition of ideals and modules within the theory of associated primes. Influential Bourbaki's textbook Algèbre commutative, in particular, takes this approach.

Let R be a ring and M a module over it. By definition, an associated prime is a prime ideal appearing in the set ${\displaystyle \{\operatorname {Ann} (m)|0\neq m\in M\}}$  = the set of annihilators of nonzero elements of M. Equivalently, a prime ideal ${\displaystyle {\mathfrak {p}}}$  is an associated prime of M if there is an injection of an R-module ${\displaystyle R/{\mathfrak {p}}\hookrightarrow M}$ .

A maximal element of the set of annihilators of nonzero elements of M can be shown to be a prime ideal and thus, when R is a Noetherian ring, M is nonzero if and only if there exists an associated prime of M.

The set of associated primes of M is denoted by ${\displaystyle \operatorname {Ass} _{R}(M)}$  or ${\displaystyle \operatorname {Ass} (M)}$ . Directly from the definition,

• If ${\displaystyle M=\bigoplus _{i}M_{i}}$ , then ${\displaystyle \operatorname {Ass} (M)=\bigcup _{i}\operatorname {Ass} (M_{i})}$ .
• For an exact sequence ${\displaystyle 0\to N\to M\to L\to 0}$ , ${\displaystyle \operatorname {Ass} (N)\subset \operatorname {Ass} (M)\subset \operatorname {Ass} (N)\cup \operatorname {Ass} (L)}$ .[3]
• If R is a Noetherian ring, then ${\displaystyle \operatorname {Ass} (M)\subset \operatorname {Supp} (M)}$  where ${\displaystyle \operatorname {Supp} }$  refers to support.[4] Also, the set of minimal elements of ${\displaystyle \operatorname {Ass} (M)}$  is the same as the set of minimal elements of ${\displaystyle \operatorname {Supp} (M)}$ .[4]

If M is a finitely generated module over R, then there is a finite ascending sequence of submodules

${\displaystyle 0=M_{0}\subsetneq M_{1}\subsetneq \cdots \subsetneq M_{n-1}\subsetneq M_{n}=M\,}$

such that each quotient Mi/Mi−1 is isomorphic to ${\displaystyle R/{\mathfrak {p}}_{i}}$  for some prime ideals ${\displaystyle {\mathfrak {p}}_{i}}$ , each of which is necessarily in the support of M.[5] Moreover every associated prime of M occurs among the set of primes ${\displaystyle {\mathfrak {p}}_{i}}$ ; i.e.,

${\displaystyle \operatorname {Ass} (M)\subset \{{\mathfrak {p}}_{0},\dots ,{\mathfrak {p}}_{n}\}\subset \operatorname {Supp} (M)}$ .[6]

(In general, these inclusions are not the equalities.) In particular, ${\displaystyle \operatorname {Ass} (M)}$  is a finite set when M is finitely generated.

Let ${\displaystyle M}$  be a finitely generated module over a Noetherian ring R and N a submodule of M. Given ${\displaystyle \operatorname {Ass} (M/N)=\{{\mathfrak {p}}_{1},\dots ,{\mathfrak {p}}_{n}\}}$ , the set of associated primes of ${\displaystyle M/N}$ , there exist submodules ${\displaystyle Q_{i}\subset M}$  such that ${\displaystyle \operatorname {Ass} (M/Q_{i})=\{{\mathfrak {p}}_{i}\}}$  and

${\displaystyle N=\bigcap _{i=1}^{n}Q_{i}.}$ [7][8]

A submodule N of M is called ${\displaystyle {\mathfrak {p}}}$ -primary if ${\displaystyle \operatorname {Ass} (M/N)=\{{\mathfrak {p}}\}}$ . A submodule of the R-module R is ${\displaystyle {\mathfrak {p}}}$ -primary as a submodule if and only if it is a ${\displaystyle {\mathfrak {p}}}$ -primary ideal; thus, when ${\displaystyle M=R}$ , the above decomposition is precisely a primary decomposition of an ideal.

Taking ${\displaystyle N=0}$ , the above decomposition says the set of associated primes of a finitely generated module M is the same as ${\displaystyle \{\operatorname {Ass} (M/Q_{i})|i\}}$  when ${\displaystyle 0=\cap _{1}^{n}Q_{i}}$  (without finite generation, there can be infinitely many associated primes.)

## Properties of associated primes

Let ${\displaystyle R}$  be a Noetherian ring. Then

• The set of zero-divisors on R is the same as the union of the associated primes of R (this is because the set of zerodivisors of R is the union of the set of annihilators of nonzero elements, the maximal elements of which are associated primes).[9]
• For the same reason, the union of the associated primes of an R-module M is exactly the set of zero-divisors on M, that is, an element r such that the endomorphism ${\displaystyle m\mapsto rm,M\to M}$  is not injective.[10]
• Given a subset ${\displaystyle \Phi \subset \operatorname {Ass} (M)}$ , M an R-module , there exists a submodule ${\displaystyle N\subset M}$  such that ${\displaystyle \operatorname {Ass} (N)=\operatorname {Ass} (M)-\Phi }$  and ${\displaystyle \operatorname {Ass} (M/N)=\Phi }$ .[11]
• Let ${\displaystyle S\subset R}$  be a multiplicative subset, ${\displaystyle M}$  an ${\displaystyle R}$ -module and ${\displaystyle \Phi }$  the set of all prime ideals of ${\displaystyle R}$  not intersecting ${\displaystyle S}$ . Then
${\displaystyle {\mathfrak {p}}\mapsto S^{-1}{\mathfrak {p}},\,\operatorname {Ass} _{R}(M)\cap \Phi \to \operatorname {Ass} _{S^{-1}R}(S^{-1}M)}$
is a bijection.[12] Also, ${\displaystyle \operatorname {Ass} _{R}(M)\cap \Phi =\operatorname {Ass} _{R}(S^{-1}M)}$ .[13]
• Any prime ideal minimal with respect to containing an ideal J is in ${\displaystyle \mathrm {Ass} _{R}(R/J).}$  These primes are precisely the isolated primes.
• A module M over R has finite length if and only if M is finitely generated and ${\displaystyle \mathrm {Ass} (M)}$  consists of maximal ideals.[14]
• Let ${\displaystyle A\to B}$  be a ring homomorphism between Noetherian rings and F a B-module that is flat over A. Then, for each A-module E,
${\displaystyle \operatorname {Ass} _{B}(E\otimes _{A}F)=\bigcup _{{\mathfrak {p}}\in \operatorname {Ass} (E)}\operatorname {Ass} _{B}(F/{\mathfrak {p}}F)}$ .[15]

## Non-Noetherian case

The next theorem gives necessary and sufficient conditions for a ring to have primary decompositions for its ideals.

Theorem — Let R be a commutative ring. Then the following are equivalent.

1. Every ideal in R has a primary decomposition.
2. R has the following properties:
• (L1) For every proper ideal I and a prime ideal P, there exists an x in R - P such that (I : x) is the pre-image of I RP under the localization map RRP.
• (L2) For every ideal I, the set of all pre-images of I S−1R under the localization map RS−1R, S running over all multiplicatively closed subsets of R, is finite.

The proof is given at Chapter 4 of Atiyah–MacDonald as a series of exercises.[16]

There is the following uniqueness theorem for an ideal having a primary decomposition.

Theorem — Let R be a commutative ring and I an ideal. Suppose I has a minimal primary decomposition ${\displaystyle I=\cap _{1}^{r}Q_{i}}$  (note: "minimal" implies ${\displaystyle {\sqrt {Q_{i}}}}$  are distinct.) Then

1. The set ${\displaystyle E=\left\{{\sqrt {Q_{i}}}|1\leq i\leq r\right\}}$  is the set of all prime ideals in the set ${\displaystyle \left\{{\sqrt {(I:x)}}|x\in R\right\}}$ .
2. The set of minimal elements of E is the same as the set of minimal prime ideals over I. Moreover, the primary ideal corresponding to a minimal prime P is the pre-image of I RP and thus is uniquely determined by I.

Now, for any commutative ring R, an ideal I and a minimal prime P over I, the pre-image of I RP under the localization map is the smallest P-primary ideal containing I.[17] Thus, in the setting of preceding theorem, the primary ideal Q corresponding to a minimal prime P is also the smallest P-primary ideal containing I and is called the P-primary component of I.

For example, if the power Pn of a prime P has a primary decomposition, then its P-primary component is the n-th symbolic power of P.

This result is the first in an area now known as the additive theory of ideals, which studies the ways of representing an ideal as the intersection of a special class of ideals. The decision on the "special class", e.g., primary ideals, is a problem in itself. In the case of non-commutative rings, the class of tertiary ideals is a useful substitute for the class of primary ideals.

## Notes

1. ^ Primary decomposition requires testing irreducibility of polynomials, which is not always algorithmically possible in nonzero characteristic.
1. ^ Ciliberto, Ciro; Hirzebruch, Friedrich; Miranda, Rick; Teicher, Mina, eds. (2001). Applications of Algebraic Geometry to Coding Theory, Physics and Computation. Dordrecht: Springer Netherlands. ISBN 978-94-010-1011-5.
2. ^ In other words, ${\displaystyle {\sqrt {Q_{i}}}=(I:g_{i})}$  is the ideal quotient.
3. ^ Bourbaki, Ch. IV, § 1, no 1, Proposition 3.
4. ^ a b Bourbaki, Ch. IV, § 1, no 3, Corollaire 1.
5. ^ Bourbaki, Ch. IV, § 1, no 4, Théorème 1.
6. ^ Bourbaki, Ch. IV, § 1, no 4, Théorème 2.
7. ^ Bourbaki, Ch. IV, § 2, no. 2. Theorem 1.
8. ^ Here is the proof of the existence of the decomposition (following Bourbaki). Let M be a finitely generated module over a Noetherian ring R and N a submodule. To show N admits a primary decomposition, by replacing M by ${\displaystyle M/N}$ , it is enough to show that when ${\displaystyle N=0}$ . Now,
${\displaystyle 0=\cap Q_{i}\iff \emptyset =\operatorname {Ass} (\cap Q_{i})=\cap \operatorname {Ass} (Q_{i})}$
where ${\displaystyle Q_{i}}$  are primary submodules of M. In other words, 0 has a primary decomposition if, for each associated prime P of M, there is a primary submodule Q such that ${\displaystyle P\not \in \operatorname {Ass} (Q)}$ . Now, consider the set ${\displaystyle \{N\subseteq M|P\not \in \operatorname {Ass} (N)\}}$  (which is nonempty since zero is in it). The set has a maximal element Q since M is a Noetherian module. If Q is not P-primary, say, ${\displaystyle P'\neq P}$  is associated with ${\displaystyle M/Q}$ , then ${\displaystyle R/P'\simeq Q'/Q}$  for some submodule Q', contradicting the maximality. Thus, Q is primary and the proof is complete. Remark: The same proof shows that if R, M, N are all graded, then ${\displaystyle Q_{i}}$  in the decomposition may be taken to be graded as well.
9. ^ Bourbaki, Ch. IV, § 1, Corollary 3.
10. ^ Bourbaki, Ch. IV, § 1, Corollary 2.
11. ^ Bourbaki, Ch. IV, § 1, Proposition 4.
12. ^ Bourbaki, Ch. IV, § 1, no. 2, Proposition 5.
13. ^ Matsumura 1970, 7.C Lemma
14. ^ Cohn, P. M. (2003), Basic Algebra, Springer, Exercise 10.9.7, p. 391, ISBN 9780857294289.
15. ^ Bourbaki, Ch. IV, § 2. Theorem 2.
16. ^ Atiyah–MacDonald 1969
17. ^ Atiyah–MacDonald 1969, Ch. 4. Exercise 11