• Primary Ideal
• The Radical of a Primary Ideal
• The Finite Intersection of P Primary Ideals
• Primary Decomposition
• Associate Primes
• First Uniqueness
• Self Radical
• Minimal Primes over J
• Polynomial Ring
• pan(R)
• pan(R) in the Fraction Ring
• Saturation
• The Intersection of the Saturations of 0
• Minimal Primes and the Radical of the Saturation
• The Smallest P Primary Ideal
• Finitely Many Saturations
• Second Uniqueness
• An Embedded Prime
• Saturation Through a Single Element
• Finite Intersection Equals Finite Product
• Laskerian is l₁
• l₁ and Noetherian Implies Laskerian
• Laskerian is l₂
• l₁ and l₂ Implies Laskerian
• Noetherian Implies Laskerian
• Primary Modules

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 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 P^k in Z all have the same prime radical P.

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 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.

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 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 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.

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 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 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 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 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 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 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 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 R-P are also driven into H by R-Q. 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 uxⁿ = 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 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(sat_P(H)).

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 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_ix 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 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 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 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 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 P₃. Thus the intersection of the primes Q_i lies in P₃. This means some prime, say Q₇, lies in P₃. 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 fⁿ lies in every primary ideal below Q_k. Now xfⁿ 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 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, sat_Q(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 sat_Q(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 H_i below primes P_i.

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.

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 n^th step, P_n 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 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 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 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 G₀. 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 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 P_n, as P_n does not contain x_n+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 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 l₂.

Let R be both l₁ and l₂. Given J, use l₁ 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 l₂, 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 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 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 b₁ 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 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 H_i. Conversely, if some yⁿ takes M into each H_i, 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 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), 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 H_i, take a moment to show [b:J] is the intersection of [b:H_i]. 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: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.