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Atiyah 第四章习题解答

Solutions to the exercises of Introduction to Commutative Algebra, Atiyah, Chapter 4.

Primary Decomposition

Exercise 1. If an ideal has a primary decomposition, then has only finitely many irreducible components.

Proof. If has a primary decomposition, then there are finitely many minimal prime ideals containing , so there are finitely many minimal prime ideals in . Thus has finitely many irreducible components.

Exercise 2. If , then has no embedded prime ideals.

Proof. If is a minimal primary decomposition of , and is -primary, then

By 1st uniqueness theorem, is also a minimal primary decomposition of , so each must be minimal along .

Exercise 3. If is absolutely flat, every primary ideal is maximal.

Proof. If is primary, then every zero-divisor of is nilpotent. But is absolute flat, so is absolute flat, and every non-unit in it is a zero-divisor, hence nilpotent. This means has a unique maximal ideal (the nilradical of it), so it’s a local ring. By its absolute flatness, it’s a field. So is maximal.

Exercise 4. In the polynomial ring , the ideal is maximal and the ideal is -primary, but is not a power of .

Proof. is field, so is maximal. has no zero-divisor other than nilpotent, so is primary. But , so is not a power of .

Exercise 5. In the polynomial ring where is a field and are independent indeterminates, let ; and are prime, and is maximal. Let . Show that is a reduced primary decomposition of . Which components are isolated and which are embedded?

Proof. are prime cause are integral. is maxiaml cause is a field. We have Obviously , thus is a reduced primary decomposition of . Here , so is isolated, and are embedded.

Exercise 6. Let be an infinite compact Hasusdorff space, the ring of real-valued continuous function on (Chapter 1, Exercise 26). Is the zero ideal decomposable in this ring?

Proof. For all , there is a prime ideal ; and these are all prime ideals. So there are infinite many minimal prime ideals containing zero, hence zero ideal is not decomposable.

Exercise 7. Let be a ring and let denote the ring of polynomials in one indeterminate over . For each ideal of , let denote the set of all polynomials in with coefficients in .

1.

is the extension of to .

2.

If is a prime ideal in , then is a prime ideal in .

3.

If is a -primary ideal in , then is a -primary ideal in .

4.

If is a minimal primary decomposition in , then is a minimal primary decomposition in .

5.

If is a minimal prime ideal of , then is a minimal prime ideal of .

Proof.

1.

Trivial.

2.

If is a prime ideal in , then is integral. Hence is integral, and is prime in .

3.

If is primary in , then all zero divisors in are nilpotents; so if is a zero divisor, there exists such that ; thus every coefficients of are zero divisors, hence nilpotents, and therefore is nilpotent (Exercise 1.2). This means is primary in . Also, in the nilradical is , so by Exercise 1.2 again, the nilradical of is . So is -primary.

4.

Obviously , and are primary by iii). Hence it’s a primary decomposition of . Obviously it’s a minimal primary decomposition of .

5.

Obvious by iii), iv).

Exercise 8. Let be a field. Show that in the polynomial ring the ideals () are prime and all their powers are primary.

Proof. In the ideal is maximal (the quotient ring is ), so it’s prime and all its powers are primary. Hence is prime and their powers are primary, by Exercise 7.

Exercise 9. In a ring , let denote the set of prime ideals which satisfy the following condition: there exists such that is minimal in the set of prime ideals containing . Show that is a zero divisor for some .

Proof. If and , then there exists , such that is minimal in prime ideals containing ; so is a minimal prime ideal in . By Exercise 3.9 (minimal prime ideals only contains zero divisors), there exists such that ; that is, . So is a zero divisor.

Conversely, if is zero divisor, then is contained in some , so contained in some minimal prime ideal containing .

Let be a multiplicatively closed subsetof , and identify with its image in . Show that

Proof. is minimal prime ideal containing , if and only if is minimal prime ideal containing . So .

If the zero ideal has a primary decomposition, show that is the set of associated prime ideals of .

Proof. If where is -primary, then every is for some ; hence .

Conversely, is a minimal primary decomposition of , and is -primary. So if is a minimal prime ideal containing , then for some . This means .

Exercise 10. For any prime ideal in a ring , let denote the kernel of the homomorphism . Prove that

1.

.

2.

is a minimal prime ideal of .

3.

If , then .

4.

, where is defined in Exercise 9.

Proof.

1.

. Since is prime, and , there must be , so .

2.

the image of in is the nilradical of there is no prime ideal contained in in is minimal in .

3.

If , then is localization of at , so the map can be factor into and , and , i.e. .

4.

If , then there exists such that ; that is, . If , then there is no minimal prime ideal containing , so , and .

Exercise 11. If is a minimal prime ideal of a ring , show that is the smallest -primary ideal.

Proof. Since is a minimal prime ideal, has unique prime ideal ; so in every zero divisor is nilpotent, and so is . Hence is primary. By Exercise 10 ii), is -primary.

If is -primary, for any , there exists such that . Since , so , and . This means is smallest -primary ideal.

Let be the intersection of the ideals as runs through the minimal prime ideals of . Show that is contained in the nilradical of .

Proof. If , then . So intersection of all minimal prime ideal. So intersection of all prime ideal (the nilradical of ).

Support that the zero ideal is decomposable. Prove that if and only if every prime ideal of is isolated.

Proof. By Exercise 9, is all minimal prime containing for some . but ; and is exactly the set of associated prime ideals of .

So for every , there exists a minimal ideal such that for any , is contained in some minimal prime ideal only contains minimal prime ideals every prime ideals of is isolated.

Exercise 12. Let be a ring, a multiplicatively closed subset of . For any ideal , let denote the contraction of in . The ideal is called the saturation of with respect to . Prove that

1.

.

2.

.

3.

meets .

4.

.

If has a primary decomposition, prove that the set of ideals (where runs through all multiplicatively closed subsets of ) is finite.

Proof.

1.

, so .

2.

.

3.

if and only if , so it’s equivalent to meets .

4.

By definition, such that . So clearly .

If , then . By Propsition 4.8, , so . So the set of ideals has at most elements.

Exercise 13. Let be a ring and a prime ideal of . The th symbolic power of is defined to be the ideal (in the notation of Exercise 12) where . Show that

1.

is a -primary ideal;

2.

if has a primary decomposition, then is its -primary component;

3.

if has a primary decomposition, then is its -primary component;

4.

is -primary.

Misprint in the book: is -primary; but by i) it’s always -primary.

Proof.

1.

is the maximal ideal in , hence is -primary. So its contraction is -primary.

2.

If has a primary decomposition , where is -primary; then ; so there exists some such that ; and all other . Take , we know is its -primary decomposition. (Theorem 4.11)

3.

Similar to ii), since , it’s sufficient to prove . For any , by definition there exists such that . Hence , so Take we have .

4.

If , then by i) it’s -primary. Conversely if is -primary, then is a primary decomposition of ; so by ii) .

Exercise 14. Let be a decomposable ideal in a ring and let be a maximal element of the set of ideals , where and . Show that is a prime ideal belong to .

Proof. First we show must be prime. If , then ; so there must be ; i.e. if , then . So is prime.

If , where is -primary, then ; so for some . But must be -primary (if it’s not ), hence for some , i.e. is a prime ideal belong to .

Exercise 15. Let be a decomposable ideal in a ring , let be an isolated set of prime ideals belonging to , and let be the intersection of the corresponding primary components. Let be an element of such that, for each prime ideal belonging to , we have , and let be the set of all powers of . Show that for all large .

Proof. If , where is -primary, and . Since , so . Thus .

If , then by , we know there exists such that . Let , then for all ,

Exercise 16. If is a ring in which every ideal has a primary decomposition, show that every ring of fractions has the same property.

Proof. For any ideal , if , then , where each is either or primary ideal. So has a primary decomposition.

Exercise 17. Let be a ring with the following property.

(L1) For every ideal in and every prime ideal , there exists such that , where .

Then every ideal in is an intersection of (possibly infinitely many) primary ideals.

Proof. Let be an ideal in , let be a minimal prime ideal containing . By Exercise 11 (apply to ), is a -primary ideal. By (L1) property, there exists such that . Now , and if , then by , ; so . This means .

Let be a maximal element of the set of ideals such that: and . Then . Repeat the construction starting with , and so on. After steps we have .

If for some , then is a finite intersection of primary ideals. Otherwise define then . Similarly, for any ordinal , construct from by the construction above (such that , and ), and for any limit ordinal define . Then for any ordinal , we have And . By transfinite induction, it’s easy to show (if ). So for some ordinal larger than , there must be , and

Exercise 18. Consider the following condition on a ring :

(L2) Given an ideal and a descending chain of multiplicatively closed subsets of , there exists an integer such that .

Prove that the following are equivalent:

1.

Every ideal in has a primary decomposition.

2.

satisfies (L1) and (L2).

Proof. i)ii): For every ideal in and every prime ideal , if has a primary decomposition , where is -primary, then Let , then ; so there exists an element , i.e. for any . By Exercise 15), for all large . So (L1) holds.

For every ideal , since has a primary decomposition, so by Exercise 12), the set of all is finite (where runs through all multiplicatively closed subsets of ). So any descending chain must be stationary. Thus (L2) holds.

ii)i): construct as in Exercise 17), so that , where is -primary. Let . Since , and for any , so for any , hence , so meets , therefore . This means . If doesn’t have a primary decomposition, then the construction won’t terminate after a finite number of steps, hence , contradicts (L2). So every ideal in has a primary decomposition.

Exercise 19. Let be a ring and a prime ideal of . Show that every -primary ideal contains , the kernel of the canonical homomorphism .

Proof. For any -primary ideal , any , there exists such that . By definition , so .

Suppose that satisfies the following condition: for every prime ideal , the intersection of all -primary ideals of is equal to . (Noetherian rings satisfy this condition: see Chapter 10.) Let be distinct prime ideals, none of which is a minimal prime ideal of . Then there exists an ideal in whose associated prime ideals are .

Proof. Induction on . The case is trivial (let ).

Suppose , and is maximal in . By inductive hypothesis, there exists an ideal whose associated ideals are . If , choose a minimal ideal contained in , then . Take radicals we have , hence for some . Since is minimal, , contradiction (cause none of is minimal). So , hence for some -primary ideal .

Let (write , where is -primary), then it is sufficient to show: for any , Let , cause for any , meets . So . So , hence .

Primary decomposition of modules

Practically the whole of this chapter can be transposed to the context of modules over a ring . The following exercises indicate how this is done.

Exercise 20. Let be a fixed -module, a submodule of . The radical of in is defined to be Show that . In particular, is an ideal.

State and prove the formulas for analogous to (1.13).

Proof. And clearly:

1.

.

2.

.

3.

.

4.

.

5.

if then .

Exercise 21. An element defines an endomorphism of , namely . The element is said to be a zero-divisor (resp. nilpotent) in if is not injective (resp. is nilpotent). A submodule of is primary in if and every zero-divisor in is nilpotent.

Show that if is primary in , then is a primary ideal and hence is a prime ideal . We say that is -primary (in ).

Proof. Since , we have . If and , then , so is a zero-divisor in . Since is primary in , is also a nilpotent in , i.e. for some . Thus ; so is a primary ideal in .

Prove the analogues of (4.3) and (4.4).

Proof. Analogues of (4.3): If are -primary, then is -primary.

First we have . If is a zero divisor in , then there exists such that . Since , there exists such that , but , thus . Hence is -primary.

Analogues of (4.4): If is -primary submodule of , then

1.

if then ;

2.

if then is a -primary ideal.

3.

if then .

i) and iii): by definition.

ii): if , then . So . Taking radicals, we have . If but , then , hence , and . So is -primary.

Exercise 22. A primary decomposition of in is a representation of as an intersection of primary submodules of ; it is a minimal primary decomposition if the ideals are all distinct and if none of the components can be omitted from the intersection, that is if .

Prove the analogue of (4.5), that the prime ideals depend only on (and ). They are called the prime ideals belonging to in . Show that they are also the prime ideals belonging to in .

Proof. We will prove that the prime ideals are exactly all prime ideals which occur in the set of ideals ; so that they only depend on (and ).

Suppose is a minimal primary decomposition of , where is -primary. For any , we have If it’s a prime ideal , then for some . But (if not ), hence . Conversely, for each , there exists , . In this case we have .

Now if the image of in is , then , so particularly . Hence the prime ideals belonging to in are precisely the prime ideals belonging to in . (Equivalently, if is a minimal primary decomposition of in , then is also a minimal primary decomposition of in ; and clearly .)

Exercise 23. State and prove the analogues of (4.6) - (4.11) inclusive. (There is no loss of generality in taking .)

(4.6)

Let be a decomposable submodule of . Then any prime ideal contains a minimal prime ideal belonging to in , and thus the minimal prime ideals of in are precisely the minimal elements in the set of all prime ideals containing .

Proof. If , then for some . Hence contains a minimal prime ideal of in .

(4.7)

Let be a decomposable submodule of , let be a minimal primary decomposition, and let . Then where . In particular, if , then the set of zero-divisors in of is the union of the prime ideals belonging to .

Proof. Without loss of generality, we can assume . Clearly . For each , Hence for some , and thus . Conversely, for each , there exists , so .

(4.8)

Let be a multiplicatively closed subset of , and let be a -primary submodule.

1.

if , then .

2.

if , then is a -primary submodule of (as a -module); and its preimage (contraction) under the canonical map is . Hence primary -submodules of correspond to primary -submodules of .

Proof. If , then there exists satisfies , so there is such that . Hence for each , .

If . First we show . Suppose , then also , so there is such that . In particular, for each , , hence exists that . So . Since , we have . Therefore and . The other direction follows immediately from definition.

Suppose but . Then there exist such that , but . hence and so . Thus , and is a -primary submodule.

Finally, if , then there exists such that . But , hence . Therefore the preimage of in is .

(4.9)

Let be a multiplicatively closed subset of and let be a decomposable submodule of . Let be a minimal primary decomposition of . Let and suppose the numbered so that meets but not . Then

Proof. Clearly by (4.8).

(4.10)

Let be a decomposable submodule of , let be a minimal primary decomposition of , and let be an isolated set of prime ideals of in . Then is independent of the decomposition.

Proof. Take in (4.9), then , hence independent of the decomposition.

In particular,

(4.11)

: The isolated primary components ( in (4.10), the corresponding prime ideal is minimal) are uniquely determined by (and ).