Throughout this chapter, rings are commutative.

You need to be familiar with prime ideals and radical ideals. In particular, the radical of H, written rad(H), is the intersection of the prime ideals containing H, or, equivalently, the set of elements x such that some power of x lies in H.

Primary ideals are a generalization of prime ideals; hence prime ideals are primary. In a laskerian ring, named after Emanuel Lasker, every proper ideal is the finite intersection of primary ideals, and in some cases the product of primary ideals. Let's start by looking at primary rings, then primary ideals will follow from there.

A ring R is primary if each zero divisor is nilpotent.

A subring of a primary ring is primary.

An integral domain, with no zero divisors, is primary.

Let R have one prime ideal P. Let xy = 0. Suppose the powers of x do not reach 0; then drive 0 up to a prime ideal Q missing x. With one prime ideal, Q = P. Thus P does not contain x. Now x is a zero divisor, and is not a unit, hence x is contained in a maximal ideal, that is prime, that cannot equal P. This is impossible, hence x is nilpotent, and R is primary.

By the above, the ring of integers mod pk, with one prime ideal generated by p, is primary.

A proper ideal H is primary if, whenever it contains xy and not x, it contains a power of y. This is symmetric; if H does not contain y it contains a power of x, and if H contains neither x nor y it contains a power of x and a power of y.

Every prime ideal is primary. If P contains xy and not x it contains y.

Let H be a power of a prime ideal in a pid. Say H = {pk}. If xy lies in H, then xy collectively contains k or more instances of p. If x is not in H then pk does not divide x, and y takes up the slack. In other words, p generates y. This means pk divides yk, and yk lies in H. This makes H a primary ideal.

You could probably see this coming a mile away, but here it is. A quotient ring is primary iff its kernel is a primary ideal. Assume R/H is a primary ring, and x is an element not in H, and xy lies in H. If y is in H then we're done, so assume y is not in H, whence x and y are zero divisors in R/H. This means the image of y, raised to the nth power in R/H, equals 0. Thus yn lies in H, and H is a primary ideal. Conversely, if H is primary then select x and y outside of H, with xy in H, representing zero divisors in R/H. Since yn lies in H, the image of y is nilpotent in R/H.

Is there a correspondence under contraction, as there was with prime ideals? Let f be a ring homomorphism from R into S, and let H be a primary ideal of S. Thus S/H = T, a primary ring. Let G be the preimage of H in R. Elements of R/G map forward, through f, to elements of S/H, which builds a subring of T, which is a primary ring. Therefore G is primary, and the contraction of a primary ideal is primary.

A similar proof shows the contraction of a prime ideal is prime. It works because a subring of an integral domain is an integral domain, making G a prime ideal.

The radical of H, written rad(H), is the intersection of prime ideals containing H. Thus the radical of a prime ideal P is P, as that is the intersection of all prime ideals containing P.

As it turns out, the radical of a primary ideal is prime. Let xy lie in rad(H), while x does not lie in rad(H). This means (xy)n lies in H, but no power of x lies in H. Seen another way, xnyn lies in H, but xn does not, hence some power of yn lies in H, a power of y lies in H, and y lies in rad(H). Therefore rad(H) is a prime ideal.

Since rad(H) is an intersection of prime ideals, every prime ideal containing H contains rad(H). With H still primary, rad(H) is itself prime, therefore rad(H) is the smallest prime ideal containing H. There may be many primes containing H; rad(H) is the smallest and is in all the others.

A primary ring has a smallest prime ideal contained in all the other prime ideals, namely rad(0).

Although H has its own prime radical, a prime ideal may be the radical of several different primary ideals. For example, the primary ideals Pk in Z all have the same prime radical P.

Assume P = rad(H), where H is primary. If xy is in H and x is not in H then yn is in H, and y is in P. We normally say xy in P means at least one factor is in P, but now, if xy is in H at least one factor is in P. Think of H as a concentrator for P.

Conversely, assume P = rad(H), and let H be a concentrator for P. If x is not in H then y is in P is in rad(H), yn lies in H, and H is primary. This makes P a prime ideal.

 x P y H xy

When H is primary and P = rad(H) we say H is P primary. Here is a simple theorem about M primary ideals, where M is maximal.

Assume rad(H) = M for M maximal in R. Remember that x is in rad(H) iff some power of x lies in H. In the quotient ring R/H, the image of M is a nil ideal. Let x be an element in R-M, thus x represents an element in R/H that is not in the image of M. By maximality, x and M span 1. Write cx+y = 1, where c is in R and y is in M. In the quotient ring, cx = 1-y, where y is nilpotent. Yet 1-y is a unit, hence x is a unit. All elements in the quotient ring are nilpotent or units, all zero divisors are nilpotent, R/H is a primary ring, and H is a primary ideal.

Since M = rad(H), H is M primary.

As a corollary, all powers of M are primary ideals, and specifically, they are M primary. We proved this for a pid, but it holds in any ring, for any maximal ideal.

Stepping away from M, a primary ideal need not be a prime power, and a prime power need not be primary.

Let R = Z[x], and let 2 and x generate P, while 4 and x generate Q. Since R/P = Z/2, P is maximal. Since R/Q = Z/4, Q is primary. The image of P in R/Q gives the nil radical, so P = rad(Q), and Q is P primary. Yet P2, generated by 4, 2x, and x2, misses x, and lies properly inside Q, hence Q is not a power of P.

Let R = K[x,y,z] mod z2-xy. A canonical polynomial in R has at most z to the first in each term. Multiply two such polynomials together and replace each z2 with xy to obtain the product. Let x and z generate P. The quotient R/P is K[y], an integral domain, so P is prime. The product xy lies in P2, being equal to z2. However, x is not in P2, and no power of y lies in P2, hence P2 is not primary.

Consider a finite set of P primary ideals H1 H2 H3 etc, and let J be their intersection. Since P contains J, rad(J) = P or something smaller. If some other prime ideal Q contains J it contains the product of the primary ideals, and it contains at least one of them, say H1. P is the smallest prime ideal containing H1, and Q cannot be any smaller. Therefore rad(J) = P.

Let J contain xy, but not x. Thus there is some primary ideal Hi that contains xy, but not x. Since Hi contains a power of y, P contains y. J is a concentrator for P, and J is P primary. The finite intersection of P primary ideals is P primary.

An ideal J has a primary decomposition if it is the intersection of a finite number of primary ideals.

If no ideal in the list contains the intersection of the others, so that each ideal contributes to the intersection, and if all ideals have distinct prime radicals, the decomposition is reduced. Every primary decomposition implies a reduced primary decomposition as follows.

Let P be the radical ideal of one of the primary ideals in the list. Let G be the intersection of all the P primary ideals in the list. As per the previous section, G is P primary. Replace all the P primary ideals with G. Do this for each prime radical, and each P is represented only once.

Finally, if one of the primary ideals contains the intersection of the others, remove it, since it does not change J. Repeat this step until you have a reduced decomposition.

 P1 P2 P3 H1 H2 H3 J

A primary ideal H has a reduced decomposition, namely H.

The associate of an ideal J, written σ(J), is the set of prime ideals P1 P2 P3 etc, such that each Pi is the radical of a conductor ideal [x:J] for some x in R. What does this mean? The conductor ideal is the set of elements that drive x into J. This forms an ideal, which has a radical ideal above it. Thus rad([x:J]) is the set of all y such that xyn lies in J. This radical need not be prime, but if it is, it becomes an associate of J.

A related function, τ(J), brings in every minimal prime ideal over each conductor [x:J]. As long as [x:J] is a proper ideal, minimal primes exist. If rad([x:J]) is prime, then it is the smallest prime over [x:J]. For this x and this J, σ and τ agree, they both include the smallest prime over [x:J]. In general, σ(J) is a subset of τ(J). This because τ might exhibit an x where rad([x:J]) is not itself prime, whence there are many minimal prime ideals over [x:J], which are included in τ, but not in σ.

 P1 P2 P3 [x1:J] [x2:J] [x3:J] J

Here are a couple of definitions: isolated and embedded. If P is minimal with respect to σ(J) it is isolated, else it is embedded.

If x lies in J, [x:J] is the whole ring, which cannot have a prime above it. It is enough to consider x ∉J.

If J is itself a primary ideal, then σ(J) is the prime P over J. Select any y in P, whence yn lies in J. Thus xyn ∈ J, and P belongs to the radical of the conductor ideal. Conversely, if xyn is in J, with x outside of J, then some power of yn is in J, and y is in P. The associate, for any x outside of J, is P, hence σ(J) = τ(J) = P.

There is a connection between associates and intersection. Let J be the intersection of a finite set of primary ideals Hi, having radicals Pi. Fix an x in R and look at the conductor ideals acting through x. Note that cx lies in J iff cx lies in each Hi, hence the conductor [x:J] is the intersection of the conductors [x:Hi]. Take radicals, and remember that radical and intersection commute. Put this all together and rad([x:J]) is the intersection over rad([x:Hi]).

If x is outside of Hi, rad([x:Hi]) = Pi, else it is R. Place x inside, or outside, of the various ideals Hi. This leads to an ideal that is the intersection of some, or all, of our prime ideals Pi. If this intersection is prime it becomes a member of σ(J), otherwise it is the seed for various primes in τ(J), each minimal over rad([x:J]), and minimal over [x:J].

If J is the intersection of H1, H2, and H3, all primary ideals, having prime radicals P1, P2, and P3, and if there is some x in every combination of these primary ideals, (i.e. a complete venn diagram), then σ(J) includes P1, P2, P3, and their pairwise and three-way intersections, if those turn out to be prime ideals.

If J has a reduced primary decomposition H1 H2 H3 etc, the associates of J are the corresponding prime ideals P1 P2 P3 etc. As shown in the previous section, σ(J) is found by intersecting the various prime ideals. Since the decomposition is reduced, H1 does not contain the intersection of the remaining primary ideals. It is possible to place x outside of H1, and inside the other primary ideals. This brings P1 into σ(J). Similarly, each Pi belongs to σ(J).

If x is anywhere else (other than in J), the associate for x is the intersection of two or more of these prime ideals. This intersection is proper, with a proper radical ideal, below some maximal prime ideal, so place it in a minimal prime ideal Q. In other words, Q belongs to τ(J). Q contains a finite intersection of primes, and a finite product of primes, hence Q contains one of them, say P3. P3 contains [x:H3], hence it also contains [x:J]. Since Q is minimal over [x:J], Q = P3. τ simply reverts back to the primes Pi.

This is also the case if Q in fact equals (not floating above) rad([x:J]), i.e. Q is a member of σ(J). Q is once again minimal over [x:J], and the above reasoning applies. Thus σ also reverts back to Pi over Hi.

If J has a reduced primary decomposition, σ(J) is the prime radicals over the primary ideals in the reduced decomposition of J, and τ(J) is same.

You might think there are many ways to represent J as the intersection of primary ideals, but there is a catch; each reduced decomposition has to exhibit a specific set of prime radicals, i.e. the primes in σ(J). This collection of primes depends only on J. Every reduced decomposition must conform to this constraint.

Assume J = rad(J), and J is the intersection of finitely many prime ideals. Since prime ideals are primary, this is a primary decomposition for J. Discard any duplicate primes, or primes that contain other primes. If a prime contains the intersection of the others it contains their product, and contains one of them, a situation I have already dealt with. Therefore the decomposition is reduced, and the primes become σ(J).

Conversely, assume J = rad(J), and J has a primary decomposition. Cut this down to a reduced decomposition, with prime ideals Pi consistent with σ(J). Radical and intersection commute, so rad(J), which is J, is also the intersection of rad(Hi), or the intersection of Pi. Thus J is the intersection of finitely many prime ideals.

A radical ideal is the intersection of finitely many prime ideals iff it has a reduced decomposition, whereupon the prime ideals are precisely σ(J). In this case all the primes are isolated, and none are embedded.

Assume J has a primary decomposition, and Q is a minimal prime containing J. Since radical and intersection commute, rad(J) is the intersection over the primes Pi, which lie over the primary ideals Hi, which intersect in J. This radical lies in Q, hence the intersection of prime ideals lies inside a prime ideal. The intersection includes the product, hence Q contains some Pi, contains Hi, contains J. Since Q is minimal, it must equal Pi. The minimal primes containing J are isolated primes in σ(J).

Conversely, an isolated prime Pi is minimal over J, else a smaller prime Q would have to be a smaller prime in σ(J).

As a corollary, there are finitely many minimal primes over J, specifically, the isolated primes in σ(J).

The ring R is primary iff the polynomial ring R[x] is primary. One direction is implied by R a subring of R[x], so let z be a zero divisor in R[x]. The coefficients are zero divisors, hence they are nilpotent. The terms are nilpotent, and their sum, z, is nilpotent, hence R[x] is primary.

By looking at lead coefficients, R[x] is an integral domain iff R is an integral domain.

Let H be an ideal in R. This implies an ideal H[x] in R[x], the polynomials with coefficients in H. This in turn implies a quotient ring R[x]/H[x], which is the same as (R/H)[x], the polynomials with coefficients in R/H.

H is prime iff R/H is an integral domain, iff (R/H)[x] is an integral domain, iff R[x]/H[x] is an integral domain, iff H[x] is prime.

A similar proof shows H is primary iff H[x] is primary.

Since radical and extension commute, H is P primary iff H[x] is P[x] primary.

Finally, extension and intersection commute, thus J has primary decomposition H1 H2 H3 etc iff J[x] has primary decomposition H1[x] H2[x] H3[x] etc. Furthermore, the decomposition of J is reduced iff the corresponding decomposition of J[x] is reduced.

Let J have a primary decomposition, and let a minimal prime ideal IN R[x] contain J[x]. Contract this to a prime Q in R containing J. Q[x] is contained in our minimal prime ideal, and is itself prime, and contains J[x], so the minimal prime ideal must be Q[x]. If there is a prime between Q and J in R, it would lead to a prime between Q[x] and J[x]. Thus Q is a minimal prime over J, and is isolated in σ(J). By correspondence, Q[x] is isolated in σ(J[x]).

Of course there are many other ideals within the polynomial ring; this is merely a characterization of those ideals that are extensions of ideals in R.

A prime ideal P in R is in the set pan(R) if it is a minimal prime ideal containing the annihilator [x:0] for some x. The notation pan(R) is shorthand for primes over annihilators.

Take a moment to verify that pan(R) = τ(0). This notation is a convenience, not a new concept. Just as nil(R) = rad(0) in R, pan(R) = τ(0) in R.

As with any ideal, τ(0) = σ(0) if 0 has a primary decomposition.

Setting x to 0 is pointless, since [0:0] becomes R, and there are no prime ideals containing R. In contrast, set x to a unit, whence [x:0] = 0. Thus pan(R) includes the minimal primes containing 0, which are the minimal primes of R.

If there is a prime ideal P containing [x:0] for some x, P descends to a minimal prime ideal containing [x:0], thus giving a member of pan(R).

Let y be a zero divisor, hence y is in [x:0] for some x. Since x is not 0, [x:0] is proper, and contained in a maximal ideal. Pull this down to a minimal prime ideal containing [x:0]. If y is a zero divisor, pan(R) includes at least one prime that contains y.

Conversely, let y belong to P in pan(R). Let P be minimal over [x:0]. Mod out by the annihilator and P maps to a minimal prime ideal in the quotient ring. If y maps to 0 then y is in the annihilator, and y becomes a zero divisor. If y maps to a nonzero element in the quotient ring, it is a zero divisor. Pull this back to R and yz lies in the annihilator, whence yzx = 0. This makes y a zero divisor.

In summary, each zero divisor is contained in a prime in pan(R), and each prime in pan(R) consists entirely of zero divisors. The primes in pan(R) comprise all the zero divisors of R. The isolated primes are the minimal prime ideals of R. Embedded primes may exist over these minimal primes, but they still consist of zero divisors. This reaffirms the earlier theorem, wherein the zero divisors are a union of prime ideals, including the minimal prime ideals.

Let S be a multiplicatively closed set in R. Thus R/S is the ring of fractions with elements of S in the denominator.

Map prime ideals and annihilators over to R/S. If P intersects S its image becomes all of R/S, so assume P misses S. With this constraint in mind, prime ideals in R and in R/S correspond.

If one ideal contains another in R, the same is true in R/S. Thus the minimal prime containing an annihilator implies a corresponding minimal prime in R/S containing the image of the annihilator. We only need ask whether the image of an annihilator is in fact an annihilator.

If y is in [x:0] then all the fractions of y kill x/1 in R/S. The image of [x:0] lies in the annihilator of x/1, relative to R/S.

Conversely, if y/w times x/1 = 0, then uyx = 0 for some u in S. Thus uy annihilates x, and lies in P. Since P misses S, u cannot lie in P, hence P contains y. y might not be in [x:0], but it lies in P, a minimal prime over [x:0], and when we move to R/S, the image of P contains the fractions of y, which are part of the annihilator of x in the fraction ring. Prime ideals containing annihilators map forward to prime ideals containing annihilators.

How about the other direction? A fraction kills x/1 iff it kills x/w, so it is enough to consider annihilators of x/1 in R/S. If x/1 = 0 then ux = 0 for some u in S. This causes [x:0], and any prime over [x:0], to intersect with S. We are only considering primes disjoint from S. So x/1 is not 0, and x/1 produces a proper annihilator in R/S, with a prime P/S above it, which pulls back to P over [x:0] in R. All annihilators in R/S have been considered.

Put this all together and pan(R/S) equals pan(R)/S, excluding any primes that intersect S.

The previous section pushed pan(R) into R/S. As a special case, let P be a prime ideal of R and let S = R-P. This gives the fraction ring wherein the elements outside of P appear as denominators. It is called localization, and the fraction ring is denoted RP.

A ring homomorphism maps R into RP via R/1. The kernel of this map includes a/b iff a/b = 0/1, iff ua = 0 for some u in S. Thus ua lies in P, and u is not in P, so a lies in P. The kernel consists of zero divisors a in P, with some u in S, such that ua = 0. In this characterization of the kernel, we could let a range over all of R, for as mentioned above, a has to lie in P. This is precisely the saturation of 0.

In general, the saturation of an ideal H, denoted sat(H), pushes H forward into the fraction ring, then pulls it back, and consists of all a in R where some u in S satisfies ua ∈ H. If H lies in P, and ua lies in H, then a also lies in P, as mentioned above. It doesn't matter if we consider all a in R or a in P, the saturation of H is the same. Also, H in P means the image of H is in the image of P, and the saturation of H is in the saturation of P. But the saturation of P is P by prime ideal correspondence, hence the saturation of every ideal inside P remains nicely inside P. Furthermore, the saturation of H inside any prime ideal Q missing S remains nicely inside Q.

Since we may be localizing about different primes, I will denote this saturation as satP(H).

The prime ideals in P correspond 1 for 1 with all the prime ideals in RP. To illustrate, let P contain the prime ideal Q. Map forward, and PP is the maximal ideal in the local ring RP, QP is the prime ideal corresponding to Q, Q is the preimage of QP, Q is the saturation of Q, and Q = satP(Q).

Select an ideal H, and the elements driven into H by R-P are also driven into H by R-Q. Therefore, satQ(H) contains satP(H). Saturate H through a smaller prime, and you could get a larger ideal.

With H contained in P, consider the quotient ring R/H. Map P forward to its corresponding prime ideal Q downstairs. Let x belong to satP(H), hence yx lies in H for some y outside of P. In the quotient ring, yx = 0, and y lies outside of Q. This is the definition of satQ(0) in R/H. Conversely, let the image of x lie in satQ(0) in R/H, as directed by an element outside of Q, which has a preimage y outside of P. Now yx lies in H, and x is in satP(H).

In summary, the image of satP(H), under the quotient map R/H, equals satQ(0) in the quotient ring. As a result, theorems about sat(0) often generalize to sat(H) by pulling back through a quotient ring.

Consider all the primes in pan(R). For each prime P in pan(R), evaluate satP(0), then take the intersection. The result is 0.

Suppose x is a nonzero element in the intersection. Place the annihilator [x:0] in a maximal ideal, then pull this down to a minimal prime ideal P in pan(R). Since x is in satP(0), ux = 0 for some u outside of P. That puts u in the annihilator of x, which lives in P. This is a contradiction, hence x cannot exist, and the intersection is empty.

The radical of satP(0) = P iff P is minimal.

One direction is easy. The saturation lies within P, and its radical is in P, and if P is minimal then the radical is indeed P.

Remember that satP(0) contains x if ux = 0. Taking the radical of this ideal, x could be an nth power. In other words, x is in the radical of satP(0) if uxn = 0 for some n and some u outside of P. Turn this around, and x is not in rad(satP(0)) if the powers of x, times the elements of R-P, never reach 0. Fold x into R-P to create a larger multiplicatively closed set. The complement of this larger closed set includes a smaller prime.

Given an ideal H in P, apply the above to the quotient ring R/H. Pull the result back to R, and P is a minimal prime containing H iff P = rad(satP(H)).

If P is a minimal prime, G = satP(0) is the smallest P primary ideal. As per the previous section, rad(G) = P, so we only need show G is primary.

Start with xy in G, with x outside of G. Thus uxy = 0. If ux = 0 then x is in G, so ux is nonzero, and ux and y are zero divisors. If y lies outside of P, uy drives x into 0, and x is in G, which is a contradiction. Thus y is in P. This puts y in rad(G), hence some power of y is in G, and G is primary.

Now G is suppose to be the smallest P primary ideal. Suppose there is some other ideal H, not containing G, that is P primary. Let x lie in G-H. Now ux = 0, and some power of u lies in H. This places u in rad(H), inside P, yet u lies outside of P. Therefore G is the smallest P primary ideal.

Conversely, if G, the saturation of 0, is known to be P primary, then P = rad(G), and P is minimal, as per the previous section, and all of the above applies.

In summary, with G set to satP(0), P is minimal iff P = rad(G), iff G is P primary - and in that case G is the smallest P primary ideal.

Next let P be a prime ideal containing a smaller ideal J, and let G = satP(J). Pass to the quotient ring R/J and apply the above. Pull the results back to R, and P is minimal over J iff P = rad(G), iff G is P primary - and in that case G becomes the smallest P primary ideal containing J.

Let J be a proper ideal and assume a primary decomposition for J. Thus J is the intersection of primary ideals Hi, below primes Pi.

If T is a multiplicatively closed set containing 1, what can we say about the saturation of J with respect to T?

Look at J as an intersection. If ux lies in J then ux lies in each Hi. If uix lies in Hi, then set v equal to the product over ui, and vx lies in J. In other words, saturation and intersection commute. Find the saturation of each Hi with respect to T, and take their intersection.

If Hi is Pi primary, ask whether T intersects Pi. If they intersect in u, a power of u winds up in Hi, T intersects Hi, and the saturation is all of R.

If T and Pi do not intersect, ratchet T up to the complement of Pi, which can only make the saturation larger. Since T contains 1, the saturation is at least Hi. Pi is minimal over Hi, and the saturation through the complement of Pi gives the smallest primary ideal between Hi and Pi. This has to be Hi. Therefore the saturation of Hi with respect to T yields Hi.

Everything depends on whether T intersects Pi. The saturation of J becomes the intersection of the saturations through some of the primes in σ(J), specifically, those primes that miss T. The saturation is the intersection of the corresponding primary ideals. If T misses all the primes, then the saturation is J.

There are finitely many cases, finitely many ways to miss some of the primes in σ(J). If σ(J) has n primes, there are at most 2n different saturations of J.

Sometimes it is helpful to change T in some way. As long as T intersects the same prime ideals in σ(J), the saturation has not changed.

T could be the powers of an element f, including f0 = 1. T intersects Pi iff Pi contains f. The saturation of J by T depends only on which prime ideals contain f. This technique will be applied later.

Suppose J has two different reduced primary decompositions, such that two different primary ideals, G and H, lie under the same isolated prime P. Let T be the complement of P. Since T intersects every other prime in σ(J), satP(J) = G, and H. This is a contradiction.

J determines the primes (first uniqueness), and the primary ideals under the isolated primes (second uniqueness). There is no third uniqueness however; the primary ideals beneath the embedded primes are not uniquely determined by J. An example is given below.

Let K be a field and let R be the polynomial ring K[x,y]. This is a ufd and a finitely generated K algebra, a fairly nice ring.

Let x2 and xy generate J, and build a primary decomposition for J as follows.

Let x generate P, an ideal that is both prime and primary.

Let y-ax and x2 generate H, for some fixed nonzero value a in K.

Since x2 and xy are in P ∩ H, J is contained in P ∩ H.

If f is a polynomial in both P and H, part of it could be generated by x2, which is in P and H. Subtract this away, and the rest must be divisible by x and y-ax. These are both irreducible in the ufd, thus f is divisible by xy-ax2. This is also in J. Therefore J is the intersection of P and H.

P does not contain H, nor does H contain P.

Look at the quotient ring R/H. If a term contains x2 it vanishes, and y can be replaced with ax wherever it appears. We are left with first degree polynomials in x. The nil radical consists of multiples of x, and if you mod out by this nil radical you get K, a field. Thus the nil radical is a maximal ideal. This is a primary ring, hence H is a primary ideal.

Let Q = rad(H). Clearly x2 is in H. Multiply y-ax by y+ax and add in a2x2 to find y2 in H. Thus Q is generated by x and y, and Q is a maximal ideal. H is Q primary.

P is properly contained in Q. Q is embedded and P is isolated. Second uniqueness fixes the primary ideal beneath the isolated prime P. Let's follow along as an exercise. We need to saturate J with the complement of P. Which elements of P are driven into J by elements outside of P? Multiply x by y and find xy in J. That's everything; the saturation of P is P. Sure enough, P is the primary ideal beneath P.

Second uniqueness does not constrain the primary ideal beneath Q. In fact each nonzero value of a in K leads to a different primary ideal, and none of these ideals contains the other.

Let J have a primary decomposition, and partition σ(J) into two sets: P1 P2 P3 etc and Q1 Q2 Q3 etc. Assume the first set is closed under containment. Any primes in σ(J) that are contained in P3, for instance, are included in the first set.

We want an f that is in each Qi, and not in any Pi. Suppose there is no such f. Let W be the intersection of the prime ideals Qi. If any part of W misses all the primes in Pi, we have found our element f. Otherwise W is contained in the finite union of prime ideals Pi, and W lies in one of them. Let W lie in P3. Thus the intersection of the primes Qi lies in P3. This means some prime, say Q7, lies in P3. This contradicts Pi closed under containment. Therefore there is an f that meets our criteria.

Let T consist of the powers of f, and saturate J through T. The saturation is determined by those primes that do or do not contain f. In this case the saturation of J through f is the intersection of Hi, for each Hi beneath Pi. The primary ideals below Pi participate in the intersection; the primary ideals below Qi do not. This is an intersection of some, but not all, of the primary ideals that combine to create J. The result is a larger ideal J′ containing J. By construction, J′ has a primary decomposition, namely Hi under Pi.

Can we select a particular power of f that drives J′ into J?

Let x lie in each Hi below Pi. In other words, x belongs to J′. Consider an outlying prime Qk. Since f is in Qk, which is the radical of its Hk, some power of f winds up in Hk. When this is multiplied by x, the result stays in Hk.

Choose n so that fn lies in every primary ideal below Qk. Now xfn lies in all these primary ideals.

Next consider some Pi in our preferred set. Since x is in J′, x is already in the corresponding primary ideal Hi beneath Pi, and xfn is in same. Put this all together and xfn is in J.

Notice that n does not depend on x. Every x in J′ is driven into J by fn. Since all of T drives J′ into J, fn won't drive anything else into J. Therefore J′ is the conductor ideal [fn:J].

As a special case, assume all the primes in σ(J) are in our designated set. There are no outlying primes, and f is not well defined. However, we don't have to saturate J at all, because J is already the intersection of the designated primary ideals. Set f = 1, and J is the saturation of J through f.

If a finite set of primary ideals exhibits distinct maximal prime radicals, the intersection equals the product.

Consider any pair of primary ideals. Let H1 be P1 primary and let H2 be P2 primary. Since P1 and P2 are maximal, write x1 + x2 = 1, where x1 is in P1 and x2 is in P2. (You can do this whenever P1 and P2 are coprime ideals, but that is assured if they are maximal.) Choose n so that x1n is in H1 and x2n is in H2, and raise the equation to the 2n.

(x1 + x2)2n = 1

Expand the left hand side using the binomial theorem. Something from H1 plus something from H2 equals 1. Thus H1 and H2 are also coprime.

Given a finite set of pairwise coprime ideals, their product equals their intersection. The proof is part of the chinese remainder theorem. Therefore a reduced primary decomposition for J, presenting maximal prime ideals, is also a factorization for J.

A ring R is laskerian if every proper ideal J in R has a primary decomposition.

A ring R is l1 if, given any proper ideal J and any prime Q, satQ(J) = [f:J] for some f not in Q. The saturation of J through R-Q is the same as the conductor ideal that drives f into J.

If J is not contained in Q then satQ(J) = R. At the same time, let f be anything in J-Q, and f drives all of R into J. This satisfies the l1 constraint.

Next assume Q contains J. Assume there is 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 [f:J], for some f outside the primes that are inside Q, and inside the primes that do not lie wholly in Q. (This was called [fn:J] in an earlier section, but I'll just call it [f:J] here.) 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.

As a special case, assume all the primes in σ(J) lie in Q. The intersection of the corresponding primary ideals is simply J. This is also the saturation of J through R-Q. Set f = 1, which is outside of Q, and [1:J] = J.

That completes the proof: if R is laskerian it is also l1.

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. Let H0 equal satP0(J). This is the saturation of J through R-P0, and with P0 minimal over J, H0 is a P0 primary ideal containing J.

If J = H0 we can stop, having found a primary decomposition for J. So 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. Y is then elements of J plus multiples of x1. Suppose Y contains something in H0-J. This means cx1 lives in H0, but not in J. This puts c in the saturation of H0 through R-P0. But H0 is already saturated; it is the saturation of J. A second saturation doesn't change anything. Thus c is in H0. By definition, x1 sends H0 into J. This forces cx1 into J, which is a contradiction. Therefore Y intersects H0 in precisely J.

 P0 H0 Y J

With Y as a base, select larger and larger ideals that intersect H0 in J. If there is 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 may not be a primary decomposition, but we're on our way.

Let P1 be a minimal prime over J1. Since P1 contains x1, not in P0, P1 is a new prime. (P1 could contain P0, if P1 is embedded.)

Let H1 be the saturation of J1 through 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.

H1 spills over into H0 above J. It must, because J1 is a maximal ideal intersecting H0 in J. This spillover seems concerning at first, but it will all work out.

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, two primary ideals and J2.

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 through 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.

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. Further assume J has a primary decomposition, thus there are finitely many saturations of J, as shown earlier. The saturations can only decrease, and so they must stabilize beyond some Sm.

Let R be both l1 and l2. Given J, use l1 to build the sequences Jn, Pn, and Hn, as n marches up through the ordinals. 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 i < n. (Remember that n could be a limit ordinal.)

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 this intersection. Remember that Jn, intersect the foregoing primary ideals, equals J. Thus W∩Jn = J. Advancing to the next level, J = W ∩Jn = W∩(Hn∩Jn+1) = (W∩Hn)∩Jn+1 = W∩Jn+1. The point is, we should have used Jn+1 instead of Jn. By construction, Jn is maximal in the chain of ideals that produce an intersection of J, and now Jn+1 is larger and gives the same intersection. This is a contradiction, hence l2 causes the sequence to terminate, and J has a primary decomposition.

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

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 satP(H) is the ideal that is driven into H by elements outside of P. Ideal? Yes; if ua lies in H and vb lies in H then uv(a+b) lies in H, as does uca. Thus the saturation is an ideal.

Since every ideal in a noetherian ring is finitely generated, let b1 through bn generate satP(H). 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 satP(H), into H. Conversely, if fY lies in H then Y is a member of satP(H) by definition.

The element f satisfies the l1 constraint, hence R is l1, and laskerian.

You thought your head was spinning before - hang on to your hat. A module can be primary, just as an ideal can be primary, and most of the theorems in this chapter extend to modules.

Given an R module M and an element y in R, call y a zero divisor if the module endomorphism y*M is not injective. In other words, yb = 0 for some nonzero b in M.

Call y nilpotent if yn kills M. yn could be 0, that's ok. Thus the endomorphism y*M, invoked n times, produces 0. Since the last invocation of y kills some b in M, y is a zero divisor.

A proper submodule H in M is primary if every 0 divisor of M/H is nilpotent in M/H. Here is an equivalent definition in terms of b and y. If y is a zero divisor, so that yb lies in H for some b outside of H, then yn drives M into H.

Let M = R, so that H becomes an ideal, and review the above definition. If y is in H it becomes 0 in R/H. But if y and b are outside of H, with yb in H,then some power of y drives R into H. Since R contains 1, some power of y lies in H. Therefore H is a primary ideal. The definitions are consistent.

Let H be a primary submodule of M, and let Q be the set of elements of R that drive M into H. Note that Q is an ideal, in fact it is a proper ideal, since Q does not contain 1.

Let xy lie in Q, so that xyM lies in H. Assume y does not lie in Q. Thus yb is not in H for some b in M. Since x drives yb into H, x becomes a zero divisor, and is nilpotent. Some power of x drives M into H, and xn lies in Q. Therefore Q is a primary ideal.

If M = R, only H can drive 1 into H, whence Q = H.

Returning to H a submodule of M, let H be a P primary module if P = rad(Q). Thus y is in P iff some power of y drives M into H.

A proper submodule J has a primary decomposition if it is the finite intersection of primary submodules. This decomposition is reduced if none of the primary modules contains the intersection of the others, and all the prime radicals are distinct.

Let H1 H2 H3 etc be a finite set of P primary modules, and let J be their intersection. If yn takes M into J it takes M into each Hi. Conversely, if some yn takes M into each Hi, select the largest n, and yn takes M into J. The radical of J is P.

Is J P primary? Let y be a zero divisor for M/J. If yb lies in J, and b is not in J, b is not in at least one of the primary modules Hi. A power of y drives M into Hi, y lies in P, and some power of y drives M into J. Therefore J is a P primary module.

Replace a set of P primary modules with their intersection, until each prime radical is represented only once. Remove any redundant modules, that contain the intersection of the others. The result is a reduced primary decomposition. A primary decomposition implies a reduced primary decomposition.

If H is a submodule, and b is in M, the conductor ideal [b:H] is the ideal in R that drives b into H. Note that b is in H iff [b:H] = R. (Remember, modules are assumed unitary, so 1*M = M.)

Assume H is primary, b lies outside of H, xy is in [b:H], and x is not. With xyb in H, y is a zero divisor relative to M/H, y is nilpotent (H being primary), yn drives M into H, ynb lies in H, and yn is in [b:H]. This makes [b:H] a primary ideal.

The radical of [b:H] includes y iff ynb lies in H. This brings in rad(H), where powers of elements of R drive all of M into H. Conversely, if ynb lies in H, then yn is a zero divisor, and nilpotent, hence y is in rad(H). For any b outside of H, rad([b:H]) = rad(H).

If J is the intersection of several modules Hi, take a moment to show [b:J] is the intersection of [b:Hi]. If x1b is in H1 and x2b is in H2, then x2x1b is in H1 and x1x2b is in H2. Radical and intersection commute in R, so rad([b:J]) is the intersection of rad([b:Hi]).

Review the proof of first uniqueness; it applies here. If J has a primary decomposition, b ∈ M can be in none, some, or all of the primary modules Hi. When b is in all but one of them, missing Hi, rad([b:J]) produces Pi, the radical of Hi. If b misses multiple primary modules, rad([b:J]) becomes the intersection of several prime ideals. If this is prime it has to equal one of the prime ideals. Therefore the prime radicals over the various conductors of J, denoted σ(J), are the prime radicals over the primary modules in the reduced decomposition of J. This depends only on the structure of M, hence the primes in the reduced decomposition are fixed. First uniqueness extends to modules.

Second uniqueness can also be generalized to modules, and this can be used to characterize laskerian modules, where every proper submodule has a primary decomposition. I'm going to stop here, before this section becomes as long and involved as all the material that has come before. I hope I have provided enough of an outline for you to fill in the details if you are interested.