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 p^{k}, 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 = {p^{k}}. If xy lies in H, then xy collectively contains k or more instances of p. If x is not in H then p^{k} does not divide x, and y takes up the slack. In other words, p generates y. This means p^{k} divides y^{k}, and y^{k} 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 n^{th} power in R/H, equals 0. Thus y^{n} 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 y^{n} 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, x^{n}y^{n} lies in H, but x^{n} does not, hence some power of y^{n} 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 P^{k} 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 y^{n} 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), y^{n} lies in H, and H is primary. This makes P a prime ideal. 

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 RM, 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 = 1y, where y is nilpotent. Yet 1y 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 P^{2}, generated by 4, 2x, and x^{2}, misses x, and lies properly inside Q, hence Q is not a power of P.
Let R = K[x,y,z] mod z^{2}xy. A canonical polynomial in R has at most z to the first in each term. Multiply two such polynomials together and replace each z^{2} 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 P^{2}, being equal to z^{2}. However, x is not in P^{2}, and no power of y lies in P^{2}, hence P^{2} is not primary.
Consider a finite set of P primary ideals H_{1} H_{2} H_{3} 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 H_{1}. P is the smallest prime ideal containing H_{1}, and Q cannot be any smaller. Therefore rad(J) = P.
Let J contain xy, but not x. Thus there is some primary ideal H_{i} that contains xy, but not x. Since H_{i} 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.
P_{1}  P_{2}  P_{3} 
H_{1}  H_{2}  H_{3} 
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 P_{1} P_{2} P_{3} etc, such that each P_{i} 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 xy^{n} 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.
P_{1}  P_{2}  P_{3}  
[x_{1}:J]  [x_{2}:J]  [x_{3}: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 y^{n} lies in J. Thus xy^{n} ∈ J, and P belongs to the radical of the conductor ideal. Conversely, if xy^{n} is in J, with x outside of J, then some power of y^{n} 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 H_{i}, having radicals P_{i}. 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 H_{i}, hence the conductor [x:J] is the intersection of the conductors [x:H_{i}]. Take radicals, and remember that radical and intersection commute. Put this all together and rad([x:J]) is the intersection over rad([x:H_{i}]).
If x is outside of H_{i}, rad([x:H_{i}]) = P_{i}, else it is R. Place x inside, or outside, of the various ideals H_{i}. This leads to an ideal that is the intersection of some, or all, of our prime ideals P_{i}. 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 H_{1}, H_{2}, and H_{3}, all primary ideals, having prime radicals P_{1}, P_{2}, and P_{3}, and if there is some x in every combination of these primary ideals, (i.e. a complete venn diagram), then σ(J) includes P_{1}, P_{2}, P_{3}, and their pairwise and threeway intersections, if those turn out to be prime ideals.
If J has a reduced primary decomposition H_{1} H_{2} H_{3} etc, the associates of J are the corresponding prime ideals P_{1} P_{2} P_{3} etc. As shown in the previous section, σ(J) is found by intersecting the various prime ideals. Since the decomposition is reduced, H_{1} does not contain the intersection of the remaining primary ideals. It is possible to place x outside of H_{1}, and inside the other primary ideals. This brings P_{1} into σ(J). Similarly, each P_{i} 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 P_{3}. P_{3} contains [x:H_{3}], hence it also contains [x:J]. Since Q is minimal over [x:J], Q = P_{3}. τ simply reverts back to the primes P_{i}.
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 P_{i} over H_{i}.
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 P_{i} consistent with σ(J). Radical and intersection commute, so rad(J), which is J, is also the intersection of rad(H_{i}), or the intersection of P_{i}. 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 P_{i}, which lie over the primary ideals H_{i}, 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 P_{i}, contains H_{i}, contains J. Since Q is minimal, it must equal P_{i}. The minimal primes containing J are isolated primes in σ(J).
Conversely, an isolated prime P_{i} 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 H_{1} H_{2} H_{3} etc iff J[x] has primary decomposition H_{1}[x] H_{2}[x] H_{3}[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 = RP. This gives the fraction ring wherein the elements outside of P appear as denominators. It is called localization, and the fraction ring is denoted R_{P}.
A ring homomorphism maps R into R_{P} 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 sat_{P}(H).
The prime ideals in P correspond 1 for 1 with all the prime ideals in R_{P}. To illustrate, let P contain the prime ideal Q. Map forward, and P_{P} is the maximal ideal in the local ring R_{P}, Q_{P} is the prime ideal corresponding to Q, Q is the preimage of Q_{P}, Q is the saturation of Q, and Q = sat_{P}(Q).
Select an ideal H, and the elements driven into H by RP are also driven into H by RQ. Therefore, sat_{Q}(H) contains sat_{P}(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 sat_{P}(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 sat_{Q}(0) in R/H. Conversely, let the image of x lie in sat_{Q}(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 sat_{P}(H).
In summary, the image of sat_{P}(H), under the quotient map R/H, equals sat_{Q}(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 sat_{P}(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 sat_{P}(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 sat_{P}(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 sat_{P}(0) contains x if ux = 0. Taking the radical of this ideal, x could be an n^{th} power. In other words, x is in the radical of sat_{P}(0) if ux^{n} = 0 for some n and some u outside of P. Turn this around, and x is not in rad(sat_{P}(0)) if the powers of x, times the elements of RP, never reach 0. Fold x into RP 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(sat_{P}(H)).
If P is a minimal prime, G = sat_{P}(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 GH. 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 sat_{P}(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 = sat_{P}(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 H_{i}, below primes P_{i}.
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 H_{i}. If u_{i}x lies in H_{i}, then set v equal to the product over u_{i}, and vx lies in J. In other words, saturation and intersection commute. Find the saturation of each H_{i} with respect to T, and take their intersection.
If H_{i} is P_{i} primary, ask whether T intersects P_{i}. If they intersect in u, a power of u winds up in H_{i}, T intersects H_{i}, and the saturation is all of R.
If T and P_{i} do not intersect, ratchet T up to the complement of P_{i}, which can only make the saturation larger. Since T contains 1, the saturation is at least H_{i}. P_{i} is minimal over H_{i}, and the saturation through the complement of P_{i} gives the smallest primary ideal between H_{i} and P_{i}. This has to be H_{i}. Therefore the saturation of H_{i} with respect to T yields H_{i}.
Everything depends on whether T intersects P_{i}. 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 2^{n} 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 f^{0} = 1. T intersects P_{i} iff P_{i} 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), sat_{P}(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 x^{2} 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 yax and x^{2} generate H, for some fixed nonzero value a in K.
Since x^{2} 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 x^{2}, which is in P and H. Subtract this away, and the rest must be divisible by x and yax. These are both irreducible in the ufd, thus f is divisible by xyax^{2}. 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 x^{2} 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 x^{2} is in H. Multiply yax by y+ax and add in a^{2}x^{2} to find y^{2} 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: P_{1} P_{2} P_{3} etc and Q_{1} Q_{2} Q_{3} etc. Assume the first set is closed under containment. Any primes in σ(J) that are contained in P_{3}, for instance, are included in the first set.
We want an f that is in each Q_{i}, and not in any P_{i}. Suppose there is no such f. Let W be the intersection of the prime ideals Q_{i}. If any part of W misses all the primes in P_{i}, we have found our element f. Otherwise W is contained in the finite union of prime ideals P_{i}, and W lies in one of them. Let W lie in P_{3}. Thus the intersection of the primes Q_{i} lies in P_{3}. This means some prime, say Q_{7}, lies in P_{3}. This contradicts P_{i} 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 H_{i}, for each H_{i} beneath P_{i}. The primary ideals below P_{i} participate in the intersection; the primary ideals below Q_{i} 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 H_{i} under P_{i}.
Can we select a particular power of f that drives J′ into J?
Let x lie in each H_{i} below P_{i}. In other words, x belongs to J′. Consider an outlying prime Q_{k}. Since f is in Q_{k}, which is the radical of its H_{k}, some power of f winds up in H_{k}. When this is multiplied by x, the result stays in H_{k}.
Choose n so that f^{n} lies in every primary ideal below Q_{k}. Now xf^{n} lies in all these primary ideals.
Next consider some P_{i} in our preferred set. Since x is in J′, x is already in the corresponding primary ideal H_{i} beneath P_{i}, and xf^{n} is in same. Put this all together and xf^{n} is in J.
Notice that n does not depend on x. Every x in J′ is driven into J by f^{n}. Since all of T drives J′ into J, f^{n} won't drive anything else into J. Therefore J′ is the conductor ideal [f^{n}: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 H_{1} be P_{1} primary and let H_{2} be P_{2} primary. Since P_{1} and P_{2} are maximal, write x_{1} + x_{2} = 1, where x_{1} is in P_{1} and x_{2} is in P_{2}. (You can do this whenever P_{1} and P_{2} are coprime ideals, but that is assured if they are maximal.) Choose n so that x_{1}^{n} is in H_{1} and x_{2}^{n} is in H_{2}, and raise the equation to the 2n.
(x_{1} + x_{2})^{2n} = 1
Expand the left hand side using the binomial theorem. Something from H_{1} plus something from H_{2} equals 1. Thus H_{1} and H_{2} 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 l_{1} if, given any proper ideal J and any prime Q, sat_{Q}(J) = [f:J] for some f not in Q. The saturation of J through RQ is the same as the conductor ideal that drives f into J.
If J is not contained in Q then sat_{Q}(J) = R. At the same time, let f be anything in JQ, and f drives all of R into J. This satisfies the l_{1} constraint.
Next assume Q contains J. Assume there is a primary decomposition for J. Thus J is the intersection of primary ideals H_{i} below primes P_{i}.
Remember that l_{1} makes an assertion about the saturation of J with respect to a multiplicatively closed set T = RQ. 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 [f^{n}: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 RQ. Set f = 1, which is outside of Q, and [1:J] = J.
That completes the proof: if R is laskerian it is also l_{1}.
Given a proper ideal J, use l_{1} 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 P_{0} be a minimal prime ideal containing J.
Let H_{0} equal sat_{P0}(J).
This is the saturation of J through RP_{0},
and with P_{0} minimal over J, H_{0} is a P_{0} primary ideal containing J.
If J = H_{0} we can stop, having found a primary decomposition for J. So assume H_{0} properly contains J. There are more primary ideals to be found. Use the fact that R is l_{1} to select x_{1} outside of P_{0}, such that x_{1} drives H_{0} into J_{0}. Let Y be the ideal generated by J and x_{1}. Y is then elements of J plus multiples of x_{1}. Suppose Y contains something in H_{0}J. This means cx_{1} lives in H_{0}, but not in J. This puts c in the saturation of H_{0} through RP_{0}. But H_{0} is already saturated; it is the saturation of J. A second saturation doesn't change anything. Thus c is in H_{0}. By definition, x_{1} sends H_{0} into J. This forces cx_{1} into J, which is a contradiction. Therefore Y intersects H_{0} in precisely J. 

With Y as a base, select larger and larger ideals that intersect H_{0} in J. If there is an infinite ascending chain of ideals, their union still intersects H_{0} in J. Use zorn's lemma to find a maximal ideal containing J and x_{1}, that intersects H_{0} in J. Call this ideal J_{1}.
Since J_{1} does not contain all of H_{0}, it is a proper ideal. Also, H_{0}∩J_{1} = J. This may not be a primary decomposition, but we're on our way.
Let P_{1} be a minimal prime over J_{1}. Since P_{1} contains x_{1}, not in P_{0}, P_{1} is a new prime. (P_{1} could contain P_{0}, if P_{1} is embedded.)
Let H_{1} be the saturation of J_{1} through RP_{1}. This becomes the smallest primary ideal between J_{1} and P_{1}. If H_{1} = J_{1} then H_{1}∩H_{0} = J, and we're done. Assume this is not the case, and move on.
H_{1} spills over into H_{0} above J. It must, because J_{1} is a maximal ideal intersecting H_{0} in J. This spillover seems concerning at first, but it will all work out.
Use l_{1} to select x_{2} outside of P_{1}, such that x_{2} drives H_{1} into J_{1}. Let Y be the ideal generated by J_{1} and x_{2}. Reason as above to show that Y intersects H_{1} in J_{1}.
Combine the earlier equation J = H_{0}∩J_{1} with J_{1} = H_{1}∩Y to get J = H_{0}∩H_{1}∩Y. Push Y up to a maximal ideal J_{2} satisfying H_{0}∩H_{1}∩J_{2} = J. This is a little closer to a decomposition, two primary ideals and J_{2}.
The sequence J_{0} J_{1} J_{2} etc is an ascending chain of ideals.
Let P_{2} be a minimal prime over J_{2}. Remember that P_{2} contains x_{2} and P_{1} does not. Thus P_{1} cannot contain P_{2}. If P_{0} contains P_{2} then P_{0} contains J_{2} contains J_{1} contains x_{1}, which is a contradiction. At the n^{th} step, P_{n} is not contained in any of the earlier primes. Each prime is new.
Let H_{2} be the saturation of J_{2} through RP_{2}. If H_{2} = J_{2} then J = H_{0}∩H_{1}∩H_{2}, and we're done. Assume this is not the case, and move on.
Continue this process, building a sequence of J_{n}, P_{n}, and H_{n}. If R is noetherian the increasing sequence J_{n} cannot continue forever. When it stops, J has a primary decomposition. Thus l_{1} 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 G_{0} be the union of all the foregoing J_{n} 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 G_{0}. Find the least n such that J_{n} contains v. Now v lies in J_{n}, and all the primary ideals prior to H_{n}, hence v lies in J. Conversely, if v lies in J it lies in each J_{n}, and in each H_{n}, and in G_{0}. Therefore J is the intersection of G_{0} and all the foregoing primary ideals.
If we can find a primary decomposition for G_{0}, it can be combined with the earlier primary ideals to build a decomposition for J. Apply the process to G_{0}, just as we did for J_{0}. Place a prime ideal Q_{0} over G_{0}, and find a primary ideal in between. Note that Q_{0} cannot fit into P_{n}, as P_{n} does not contain x_{n+1}, which is part of G_{0}. 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 l_{2} 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 S_{n} to drive things into J. However, at some point the saturation bottoms out. It is what it is, for all S_{n} beyond some S_{m}.
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 S_{m}.
Laskerian implies l_{2}.
Let R be both l_{1} and l_{2}. Given J, use l_{1} to build the sequences J_{n}, P_{n}, and H_{n}, as n marches up through the ordinals. Let S_{n} be the complement of the union of all prime ideals up to, but not including, P_{n}. If P_{n} 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, S_{n}, forms a descending chain of multiplicatively closed sets.
Apply l_{2}, and the saturation of J, with respect to S_{n}, is fixed beyond some S_{m}.
Since saturation and intersection commute, sat(J) is the intersection of sat(H_{i}), for i < n, and sat(J_{n}). Since J_{n} and S_{n} intersect in x_{n}, the saturation of J_{n} is simply R, so don't worry about that. The saturation of H_{i} by a set outside of P_{i} is simply H_{i}. Therefore sat(J), with respect to S_{n}, is the intersection of the primary ideals H_{i}, 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 H_{n} does not change the intersection of the primary ideals. Thus H_{n} contains the intersection of the foregoing primary ideals. Let W be this intersection. Remember that J_{n}, intersect the foregoing primary ideals, equals J. Thus W∩J_{n} = J. Advancing to the next level, J = W ∩J_{n} = W∩(H_{n}∩J_{n+1}) = (W∩H_{n})∩J_{n+1} = W∩J_{n+1}. The point is, we should have used J_{n+1} instead of J_{n}. By construction, J_{n} is maximal in the chain of ideals that produce an intersection of J, and now J_{n+1} is larger and gives the same intersection. This is a contradiction, hence l_{2} causes the sequence to terminate, and J has a primary decomposition.
Put this all together and R is laskerian iff R is l_{1} and l_{2}.
Let R be a noetherian ring. In such a ring, l_{1} implies laskerian, hence it is sufficient to prove R is l_{1}.
Let H be any ideal and P any prime ideal containing H. Now sat_{P}(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 b_{1} through b_{n} generate sat_{P}(H). Let c_{i} be an element outside of P that drives b_{i} into H. Let f be the product over c_{i}. Now f drives each b_{i}, and all of sat_{P}(H), into H. Conversely, if fY lies in H then Y is a member of sat_{P}(H) by definition.
The element f satisfies the l_{1} constraint, hence R is l_{1}, 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 y^{n} kills M. y^{n} 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 y^{n} 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 x^{n} 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 H_{1} H_{2} H_{3} etc be a finite set of P primary modules, and let J be their intersection. If y^{n} takes M into J it takes M into each H_{i}. Conversely, if some y^{n} takes M into each H_{i}, select the largest n, and y^{n} 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 H_{i}. A power of y drives M into H_{i}, 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), y^{n} drives M into H, y^{n}b lies in H, and y^{n} is in [b:H]. This makes [b:H] a primary ideal.
The radical of [b:H] includes y iff y^{n}b lies in H. This brings in rad(H), where powers of elements of R drive all of M into H. Conversely, if y^{n}b lies in H, then y^{n} 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 H_{i}, take a moment to show [b:J] is the intersection of [b:H_{i}]. If x_{1}b is in H_{1} and x_{2}b is in H_{2}, then x_{2}x_{1}b is in H_{1} and x_{1}x_{2}b is in H_{2}. Radical and intersection commute in R, so rad([b:J]) is the intersection of rad([b:H_{i}]).
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 H_{i}. When b is in all but one of them, missing H_{i}, rad([b:J]) produces P_{i}, the radical of H_{i}. 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.