Primary Ideals, Laskerian, l1, and l2

Laskerian

The ring R is laskerian (biography) if every proper ideal H in R has a primary decomposition.

l1

The ring R is l1 if, given any proper ideal H and any prime P, SPH = [H:f] for some f not in P. The saturation of H through R-P is the same as the conductor ideal that drives f into H.

If H is not contained in P then SPH = R. All of R is driven in to H by something outside of P. At the same time, let f be anything in H-P, and f drives all of R into H. This satisfies the l1 constraint. With this case handled, let's assume P contains H.

Laskerian Implies l1

Start with a prime Q over an ideal J, and assume a primary decomposition for J. Thus J is the intersection of primary ideals Hi, below primes Pi.

Remember that l1 makes an assertion about the saturation of J with respect to a multiplicatively closed set T = R-Q. These saturations have been completely characterized. The saturation only depends on the primes in σ(J) that miss T.

Partition the primes in σ(J) into two sets, those that are completely inside Q and those that are not. Ratchet T up to the complement of the union of the primes in Q. since T misses the same primes it missed before, the saturation of J has not changed. This saturation equals the intersection of the primary ideals below the primes in Q.

This saturation, or intersection, is a specific conductor ideal [J:f], for some f outside the primes that are inside Q, and inside the primes that do not lie in Q. So we're pretty much done, unless f lies in Q.

Let W be the intersection of those primes in σ(J) that don't lie in Q. If W lies entirely in Q then one of the primes that we used to build W lies in Q, even though none of them is contained in Q. This is a contradiction, so W contains an f not in Q.

If all the primes in σ(J) lie in Q then their intersection is simply J. Set f = 1, which is outside of Q, and [J:1] = J.

This completes the proof; laskerian implies l1.

l1 and Noetherian Implies Laskerian

Given a proper ideal J, use l1 to write J as a (possibly infinite) intersection of primary ideals. This is done step by step, starting with an initial primary ideal containing J, then building another, and another, and another. If this goes on forever, the resulting infinite cascade of primary ideals implies yet another primary ideal over J. This sets the stage for transfinite induction, mapping ordinals to primary ideals. Since R is a set with a fixed cardinality, we can't keep extracting different subsets forever. The process must terminate, making J the intersection of a (possibly infinite, possibly uncountable) set of primary ideals.

Ok, it's time to create the first primary ideal over J. Let P0 be a minimal prime ideal containing J. (Think of J as J0.) Let H0 equal SP0J. This is a primary ideal below P0, and containing J. If J = H0 we can stop, having found a primary decomposition for J. So let's assume H0 properly contains J. There are more primary ideals to be found.

Use the fact that R is l1 to select x1 outside of P0, such that x1 drives H0 into J0. Let Y be the ideal generated by J and x1. Suppose Y contains something in H0-J. Subtract any generators of Y that come from J, and cx1 lives in H0, but not in J. In particular, c lives in the saturation of H0 with respect to the complement of P0. This saturation reproduces the primary ideal H0, hence c lies in H0. This forces cx1 into J, a contradiction. Therefore Y intersects H0 in precisely J.

With Y as a base, select larger and larger ideals that intersect H0 in J. If you find an infinite ascending chain of ideals, their union still intersects H0 in J. Use zorn's lemma to find a maximal ideal containing J and x1, that intersects H0 in J. Call this ideal J1.

Since J1 does not contain all of H0, it is a proper ideal. Also, H0∩J1 = J. This isn't a primary decomposition, but we're on our way.

Let P1 be a minimal prime over J1. Since P1 contains x1 it is not contained in P0; we have a new prime.

Let H1 be the saturation of J1 relative to R-P1. This becomes the smallest primary ideal between J1 and P1. If H1 = J1 then H1∩H0 = J, and we're done. Assume this is not the case, and move on.

Use l1 to select x2 outside of P1, such that x2 drives H1 into J1. Let Y be the ideal generated by J1 and x2. Reason as above to show that Y intersects H1 in J1.

Combine the earlier equation J = H0∩J1 with J1 = H1∩Y to get J = H0∩H1∩Y. Push Y up to a maximal ideal J2 satisfying H0∩H1∩J2 = J. This is a little closer to a decomposition; we have two primary ideals. (The reason for taking the maximal ideal at each step will become clear when we investigate l2.)

The sequence J0 J1 J2 etc is an ascending chain of ideals.

Let P2 be a minimal prime over J2. Remember that P2 contains x2 and P1 does not. Thus P1 cannot contain P2. If P0 contains P2 then P0 contains J2 contains J1 contains x1, which is a contradiction. At the nth step, Pn is not contained in any of the earlier primes. Each prime is new.

Let H2 be the saturation of J2 relative to R-P2. If H2 = J2 then J = H0∩H1∩H2, and we're done. Assume this is not the case, and move on.

Continue this process, building a sequence of Jn, Pn, and Hn.

If R is noetherian the increasing sequence Jn cannot continue forever. When it stops, J has a primary decomposition. Thus l1 and noetherian implies laskerian.

Just for grins, assume R is not noetherian, and run the process through transfinite induction. What do we do in the limit?

For notational convenience, I'm going to select a new letter. Let G0 be the union of all the foregoing Jn ideals. This is larger than any of the ideals in the ascending chain.

Let v lie in each primary ideal in the sequence, and in G0. Find the least n such that Jn contains v. Now v lies in Jn, and all the primary ideals prior to Hn, hence v lies in J. Conversely if v lies in J it lies in each Jn, and in each Hn, and in G0. Therefore J is the intersection of G0 and all the foregoing primary ideals.

If we can find a primary decomposition for G0, it can be combined with the earlier primary ideals to build a decomposition for J. Apply the process to G0, just as we did for J0. Place a prime ideal Q0 over G0, and find a primary ideal in between. Note that Q0 cannot fit into Pn, as Pn does not contain xn+1, which is part of G0. Once again we have a new prime.

Continue this process through successors and limits, until the process terminates, restricted by the cardinality of R. This assures a complete, albeit infinite, primary decomposition for J.

l2

A ring R is l2 if, given any descending chain of multiplicatively closed sets, and an ideal J, the successive saturations of J become constant. Clearly the saturations can only get smaller, as there are fewer and fewer elements in each Sn to drive things into J. However, at some point the saturation bottoms out. It is what it is, for all Sn beyond some Sm.

If J is all of R then every saturation is R, which is constant. That's not very interesting, so assume J is a proper ideal.

Laskerian Implies l2

Since J has a primary decomposition, there are finitely many saturations. Clearly the saturations stabilize beyond some Sm.

l1 and l2 Implies Laskerian

Given J, use l1 to build the sequences Jn, Pn, and Hn, as described above. Let Sn be the complement of the union of all prime ideals up to (but not including) Pn. If Pn is contained in the union of the foregoing prime ideals it is contained in one of them, and that is a contradiction. Each prime brings new elements into the union, and the complement, Sn, forms a descending chain of multiplicatively closed sets.

Apply l2, and the saturation of J, with respect to Sn, is fixed beyond some Sm.

Since saturation and intersection commute, sat(J) is the intersection of sat(Hi), for i < n, and sat(Jn). Since Jn and Sn intersect in xn, the saturation of Jn is simply R, so don't worry about that. The saturation of Hi by a set outside of Pi is simply Hi. Therefore sat(J), with respect to Sn, is the intersection of the primary ideals Hi, as i runs from 0 to n-1.

Assume we have reached the point where the saturation is fixed. Bringing in Hn does not change the intersection of the primary ideals. Thus Hn contains the intersection of the foregoing primary ideals.

Let W be the intersection of H0 through Hn-1. Thus W∩Jn = J. Advancing to the next level, W∩Hn∩Jn+1 = W ∩Jn+1 = J.

The point is, we should have used Jn+1 instead of Jn - since Jn was suppose to be maximal. After all, Jn+1 is larger, and gives the desired intersection. This is a contradiction, hence the sequence must terminate, J has a primary decomposition, and R is laskerian.

Put this all together and R is laskerian iff R is l1 and l2.

Noetherian Implies Laskerian

Let R be a noetherian ring. In such a ring, l1 implies laskerian, hence it is sufficient to prove R is l1.

Let H be any ideal and P any prime ideal containing H. Now SPH is the ideal that is driven into H by elements outside of P.

Since every ideal in a noetherian ring is finitely generated, let b1 through bn generate SPH. Let ci be an element outside of P that drives bi into H. Let f be the product over ci. Now f drives each bi, and all of SPH, into H. Conversely, if fY lies in H then Y is a member of SPH.

Our element f satisfies the l1 constraint, hence R is l1, and laskerian. Since artinian implies noetherian, we can write the following.

artinian → noetherian → laskerian ⇔ l1 and l2