The following theory was originally developed by world chess champion Emmanuel Lasker in his doctoral thesis under David Hilbert and then greatly simplified (and generalised to noetherian rings) by Emmy Noether.
An ideal is called primary ideal if and only if the following holds:
Clearly, every prime ideal is primary.
We have the following characterisations:
Theorem 19.5 (characterisations of primary ideals):
Let . The following are equivalent:
- is primary.
- If , then either or or .
- Every zerodivisor of is nilpotent.
1. 2.: Let be primary. Assume and neither nor . Since , for a suitable . Since and , for a suitable .
2. 3.: Let be a zerodivisor of , that is, for a certain such that . Hence , that is, for a suitable .
3. 1.: Let . Then either or or is a zerodivisor within , which is why for a suitable .
1. 3.: Let be primary, and let be a zerodivisor within . Then for a and hence for a suitable .
3. 2.: Let . Assume neither nor . Then both and are zerodivisors in , and hence are nilpotent, which is why for suitable and hence .
2. 1.: Let . Assume not and not . Then in particular , that is, for suitable .
If is any primary ideal, then is prime.
Let . Then for a suitable . Hence either and thus or for a suitable and hence .
Existence in the Noetherian case
Following the exposition of Zariski, Samuel and Cohen, we deduce the classical Noetherian existence theorem from two lemmas and a definition.
An ideal is called irreducible if and only if it can not be written as the intersection of finitely many proper superideals.
In a Noetherian ring, every irreducible ideal is primary.
Assume there exists an irreducible ideal which is not primary. Since is not primary, there exist such that , but neither nor for any . We form the ascending chain of ideals
this chain is ascending because . Since we are in a Noetherian ring, this chain eventually stabilizes at some ; that is, for we have . We now claim that
Indeed, is obvious, and for we note that if , then
for suitable and , which is why , hence , since thus , , hence and . Therefore .
Furthermore, by the choice of and both and are proper superideals, contradicting the irreducibility of .
In a Noetherian ring, every ideal can be written as the finite intersection of irreducible ideals.
Assume otherwise. Consider the set of all ideals that are not the finite intersection of irreducible ideals. If we are given an ascending chain within that set
this chain has an upper bound, since it stabilizes as we are in a Noetherian ring. We may hence choose a maximal element among all ideals that are not the finite intersection of irreducible ideals. itself is thus not irreducible. Hence, it can be written as the intersection of strict superideals; that is
for appropriate . Since is maximal, each is a finite intersection of irreducible ideals, and hence so is , which contradicts the choice of .
In a Noetherian ring, every ideal can be written as the finite intersection of primary ideals.
Combine lemmas 19.8 and 19.9.
Let be an ideal in a ring, and let
be a primary decomposition of . This decomposition is called minimal if and only if
- there does not exist with , and
- for all , (that is, the radicals of the prime ideals are pairwise distinct).
In fact, once we have a primary composition for a given ideal, we can find a minimal primary decomposition of that ideal. But before we prove that, we need a general fact about radicals first.
Let be ideals. Then
One could phrase this lemma as "radical interchanges with finite intersections".
Let . For each , choose such that . Set
Then , hence .
Note that for infinite intersections, the lemma need not (!!!) be true.
Let be an ideal in a ring that has a primary decomposition. Then also has a minimal primary decomposition.
First of all, we may exclude all primary ideals for which
the intersection won't change if we do that, for intersecting with a superset changes nothing in general.
Then assume we are given a decomposition
and for a fixed prime ideal set
due to theorem 19.6,
We claim that is primary, and . For the first claim, note that by the previous lemma
For the second claim, let . If there is nothing to prove. Otherwise let . Then there exists such that , and hence for a suitable . Thus , and hence for all and suitable . Pick
Then . Hence, is primary.
In general, we don't have uniqueness for primary decompositions, but still, any two primary decompositions of the same ideal in a ring look somewhat similar. The classical first and second uniqueness theorems uncover some of these similarities.
Theorem 19.14 (first uniqueness theorem):
Let be an ideal within a ring , and assume we are given a minimal primary decomposition
Then the prime (theorem 19.6) ideals are exactly the prime ideals among the ideals and hence are independent of the choice of the particular decomposition. That is, the ideals are uniquely determined by .
We begin by deducing an equation. According to theorem 19.2 and lemma 19.12,
Now we fix and distinguish a few cases.
- If , then obviously .
- If (where again ), then if we must have since no power of is contained within .
- If , but , we have , since
In conclusion, we find
Assume first that is prime. Then the prime avoidance lemma implies that is contained within one of the , , and since , .
Let now for be given. Since the given primary decomposition is minimal, we find such that , but . In this case, by the above equation.
This theorem motivates and enables the following definition:
Let be any ideal that has a minimal primary decomposition
Then the ideals are called the prime ideals belonging to .
We now prove two lemmas, each of which will below yield a proof of the second uniqueness theorem (see below).
Let be an ideal which has a primary decomposition
and let again for all . If we define
then is an ideal of and .
Let . There exists such that without , and a similar with an analogous property in regard to . Hence , but not since is prime. Also, . Hence, we have an ideal.
Let . There exists such that
In particular, . Since no power of is in , .
Let be multiplicatively closed, and let
be the canonical morphism. Let be a decomposable ideal, that is
for primary , and number the such that the first have empty intersection with , and the others nonempty intersection. Then
by theorem 9.?. If now , lemma 9.? yields . Hence,
Application of on both sides yields
since holds for general maps, and means , where and ; thus , that is . This means that
Hence , and since no power of is in ( is multiplicatively closed and ), .
Let be an ideal which admits a primary decomposition, and let be a set of prime ideals of that all belong to . is called isolated if and only if for every prime ideal , if is a prime ideal belonging to such that , then as well.
Theorem 19.19 (second uniqueness theorem):
Let be an ideal that has a minimal primary decomposition. If is a subset of the set of the prime ideals belonging to which is isolated, then
is independent of the particular minimal primary decomposition from which the are coming.
Note that applied to reduced sets consisting of only one prime ideal, this means that if all prime subideals of a prime ideal belonging to also belong to , then the corresponding is predetermined.
Proof 1 (using lemma 19.16):
We first reduce the theorem down to the case where is the set of all prime subideals belonging to of a prime ideal that belongs to . Let be any reduced system. For each maximal element of that set (w.r.t. inclusion) define to be the set of all ideals in contained in . Since is finite,
this need not be a disjoint union (note that these are not maximal ideals!). Hence
Hence, let be an ideal belonging to and let be an isolated system of subideals of . Let be all the primary ideals belonging to not in . For those ideals, we have , and hence we find . For each take large enough so that . Then
which is why
From this follows that
where is the element in the primary decomposition of to which is associated, since clearly for each element of the left hand side, and thus , but also . But on the other hand, implies . Hence for any such lemma 19.16 implies
which in turn implies
Proof 2 (using lemma 19.17):
Let be an isolated system of prime ideals belonging to . Pick
which is multiplicatively closed since it's the intersection of multiplicatively closed subsets. The primary ideals of the decomposition of which correspond to the are precisely those having empty intersection with , since any other primary ideal in the decomposition of must contain an element outside all , since otherwise its radical would be one of them by isolatedness. Hence, lemma 19.17 gives
and we have independence of the particular decomposition.
Characterisation of prime ideals belonging to an ideal
The following are useful further theorems on primary decomposition.
First of all, we give a proposition on general prime ideals.
Let be a (commutative) ring, and let be a prime ideal. If contains
- either the intersection or the product
of certain arbitrary ideals, then it contains one of the completely.
Since the product is contained in the intersection, it suffices to prove the theorem under the assumption that .
Indeed, assume none of the is contained in . Choose for . Since is prime, . But it's in the product, contradiction.
This proposition has far-reaching consequences for primary decomposition, given in Corollary 19.22. But first, we need a lemma.
Let be a primary ideal, and assume is prime such that . Then .
If , then .
Let be an ideal admitting a prime decomposition
If is any prime ideal that contains , then it also contains a prime ideal belonging to . Further, the prime ideals belonging to are exactly those that are minimal with respect to the partial order induced by inclusion on .
The first assertion follows from proposition 19.20 and lemma 19.21. The second assertion follows since any prime ideal belonging to contains .