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Solutions to the exercises of Introduction to Commutative Algebra, Atiyah, Chapter 1.

## Rings and Ideals

Exercise 1. Sum of a nilpotent and a unit is a unit.

Proof. If then , so is a unit. If is a unit and is a nilpotent, then is a nilpotent, hence is a unit.

Exercise 2. Let . Prove:

 1 is a unit in is a unit in , and are nilpotent. 2 is nilpotent are nilpotent. 3 is a zero-divisior there exists such that . 4 is said primitive if . Prove that is primitive and are both primitive.

Proof.

 1 : Suppose such that , then hence are both units. We Prove for all by induction on r.If this is true for all , consider the -th coefficient of , we have , so . Multiply to both side, then by induction hypothesis we get .In partial, , hence is nilpotent, and so is . By Ex.1.1, is a unit, so repeat the proof above we have are nilpotent.: Repeat Ex.1.1 for times. 2 If , then consider -th coefficient of we have . Then is also nilpotent, hence are all nilpotent.: is sums of nilpotents, also a nilpotent. 3 : Choose of least degree such that , then , so is of at most degree . But , by the choice of we have . Then we prove by induction on . For it’s proved above, and if it is true for , then we have , so similarly . So we have .: Trivial. 4 : if all coefficients of are in an ideal , then obviously all coefficients of are also in , so is not primitive. Left and both primitive. Suppose is not primitive, that is, . Let be a prime ideal contains , and let be the least number that (cause are primitive, these number exist), then , contradiction.

Exercise 3. Generate results in Ex.1.2 to a ring with several indeterminates.

Proof. Just consider as a polynomial ring over . Repeat the proof above, we have: if , then

 1 is a unit the constant term of is a unit, all other coefficients are nilpotent. 2 is nilpotent all coefficients of are nilpotent. 3 is a zero-divisior there exists such that 4 is primitive are both primitive.

Exercise 4. In the ring , the Jacobson radical is equal to nilradical.

Proof. Let belong to Jacobson radical, then for all , is a unit. In particular, is a unit, hence any coefficient of is nilpotent, and is nilpotent, i.e. belongs to nilradical of .

Exercise 5. Let be the ring former power series over , and let . Show that

 1 is a unit of is a unit of . 2 If is nilpotent, then is nilpotent for all . Is the converse true? (See Chapter 7,Exercise 2.) 3 belongs to Jacabson radical of belongs to Jacabson radical of . 4 The contraction of a maximal ideal of is a maximal ideal of , and is generated by and . 5 Every prime ideal of is the contraction of a prime ideal of .

Proof.

 1 : Trivial.: Let where , then , so is a unit. 2 Induction on . if is nilpotent, then so is (Cause ). So there exists such that , hence . 3 Suppose belongs to Jacabson radical of , then for all , is a unit. In particular for all , is a unit, so by i) is a unit, so belongs to Jacabson radical of ; and vice versa. 4 If , then is a ideal of , which contains all series of constant term , so , hence and . 5 Let be a prime ideal of , then is the contraction of . So it is sufficient to prove is a prime ideal of .If , and , i.e. , then either or belongs to . So either or belongs to , hence is a prime ideal of .

Exercise 6. Let be a ring such that every ideal not contained in the nilradical contains a nonzero idempotent (that is, an element such that ). Prove that the nilradical and Jacobson radical of are equal.

Proof. If the Jacobson radical is not contained in the nilradical, then there exists a nonzero idempotent belongs to the Jacobson radical, so is a unit. but , hence , so , contradiction.

Exercise 7. Let be a ring in which every element satifies for some . Show that every prime ideal of is maximal.

Proof. Let be a prime ideal, then we have is a intergal domain, and for all there exists such that , hence . If , then , so is a unit. Hence is a field, and is maximal.

Exercise 8. Let be a ring . Show that the set of prime ideals of has minimal elements with repect to inclusion.

Proof. Forall chains of prime ideals , if we let , then is an ideal. Suppose , then for all we have , hence either or belongs to , so at least one of them belongs to infinite , hence belongs to all , i.e. either or belongs to . So all chains of prime ideals of has a lower bound, so by Zorn’s Lemma there exists a minimal elements along them.

Exercise 9. Let be an ideal in a ring . Show that is an intersection of prime ideals.

Proof. : Let be the intersection of all prime ideals containing . If , then in , belongs to all prime ideals, so for some , i.e. . so is a intersection of prime ideals.

: If , then .

Exercise 10. Let be a ring, its nilradical. Show that the following are equivalent:

 1 has exactly one prime ideal. 2 every element of is either a unit ot nilpotent. 3 is a field.

Proof.

 i) ii): If is not a unit, then there exists a prime ideal containing , so is the onlyprime ideal. But then , contradiction. ii) iii): Every elements is a unit, so every elements of is a unit, therefore is a field. iii) i): is a maximal ideal and the intersection of all prime ideals. So the only prime ideal of is itself.

Exercise 11. A ring is Boolean if for all . In a Boolean ring , show that

 1 for all . 2 every prime ideal is maximal, and is a field with two elements. 3 every finitely generated ideal in is principal.

Proof.

 1 , so for all . 2 Suppose . By we have . So , and only contains two elements, hence is a field, and is maximal. 3 For any , let , then , so .

Exercise 12. A local ring contains no idempotent .

Proof. Let be the only maximal ideal of . For any idempotent , if is a unit then . If is not unit, then (the Jacabson radical of ), so is a unit. but is also idempotent, so , therefore .

Exercise 13. field, the set of all irreducible monic polynomials f of one indeterminate with coefficients in . Let be the polynomial ring over generated bt indeterminates , one for each . Let be the ideal of generated by the polynomials for all . Show that .

Let be a maximal ideal of containing , and let . Then is an extension field of in which each has a root. Repeat the construction with in place of , obtaining a field , and so on. Let , Then is a field in which each splits completely into linear factors. Let be the set of all elements of which are algebraic over . Then is an algebraic closure of .

Proof. Let , then can be written in a finite sum of finite products of -s. Choose one form containing least -s, suppose it contains . then but .

Hence , so , i.e. exists such that . Obviously is an intergal domain, so ; but then , contradiction.

If , then , so in , and can be written as . Repeat this progress then in , splits into linear factors.

Exercise 14. In a ring a, let be all ideals in which every element is a zero-divisior. Show that the set has maximal elements, and every maximal element of is a prime ideal. Hence the set of zero-divisiors in A is a union of prime ideals.

Proof. For every chain in , it has a upperbound . So by Zorn’s Lemma has maximal elements.

Suppose is a maximal element of , and . Consider and . If neither belongs to , then there exists such that are both not zero-divisior. But is a zero-divisior, contradiction. So at least one of belongs to , i.e. or .

Exercise 15. Let be a ring and let be the set of all prime ideals of . For eah subset of , let denote the set of all prime ideals of which contain . Prove that

 1 if is the ideal generated by , then . 2 . 3 if is any family of subsets of , then 4 for any ideals .

Proof.

 1 If then also is . If and then by definition , so . The inverse is trivial. 2 Trivial. 3 Trivial. 4 If then or . So . -s are trivial.

Exercise 16. Draw pictures of .

Proof.

 1 In there is countable infinite closed points , and a generic point . 2 Cause is a field, it have only one prime ideal . So is a trivial topology space with only one point. 3 In a prime ideal is or with , it’s similar to but with uncountable infinite points. 4 In a prime ideal is or with , actually it’s isomorphic to 5 In there are three sorts of prime ideals: or ; where is a irreducible polynomial over (or equivalentlly irreducible polynomial over ); where is a monic irreducible polynomial over . For the Zariski topology, there is a closed base of it: for all ; for all irreducible ; for all irreducible over .

Exercise 17. For each , let denote the complement of in . The sets are open. Show that they form a basis of the Zariski topology, and that

 1 . 2 is nilpotent. 3 is a unit. 4 . 5 is quasi-compact (every open covering of has a finite sub-covering) 6 More generating, each is quasi-compact. 7 An open subset of is quasi-compact if and only if it is a finite union of sets .The sets are called basic open sets of .

Proof. For any open set of (that is, all prime ideals which doesn’t contain ), and a prime ideal , we must find a such that and . Just select an element .

 1 is the complement of , i.e.. 2 for all prime ideals is contained in the nilradical, i.e. is nilpotent. 3 for all prime ideals is a unit. 4 If then , so . On the other hand is the nilradical of , hence is intersection of all ideals which contains . So implies . 5 It’s sufficient to show: if is covered by some for , then this open covering has a finite sub-covering. But means the ideal generated by is , so for some (finite many) . so these also cover . 6 is covered by if and only if for some . 7 An open set of is then union of some . If it’s quasi-compact, then it’s a finite union of some . Conversely, finite union of quasi-compact sets is quasi-compact.

Exercise 18. For psychological reasons it is sometimes convenient to denote a prime ideal of by a letter such as and when thinking of it as a point of . When thinking of as a prime ideal of , we denote it by (logically, of course, it is the same thing). Show that

 1 The set is closed ( is a “closed point”) in is maximal; 2 ; 3 ; 4 is a -space.

Proof.

 1 is a closed point for some is maximal. 2 . 3 . 4 If are distinct points of , then either or . Suppose the former, then , so is a neighborhood of which seperate .

Exercise 19. A topological space is said to be irreducible if and if every pair of non-empty open sets in intersect, or equivalently if every non-empty open set is dense in . Show that is irreducible if and only if the nilradical of is a prime ideal.

Proof. is irreducible if and only if for every two prime ideal , there is a prime ideal . This means minimal prime ideal of is unique, so it’s contained in every prime ideal. So ; and obviously , so is a prime ideal.

Exercise 20. Let be a topological space.

 1 If is an irreducible subspace of , then the closure of in is irreducible. 2 Every irreducible subspace of is contained in a maximal irreducible subspace. 3 The maximal irreducible subspace of are closed and cover . They are called the irreducible components of . What are the irreducible components of a Hausdorff space? 4 If is a ring and , then the irreducible components of are the closed sets , where is a minimal prime ideal of .

Proof.

 1 Every open subset of is dense in , hence dense in . 2 If are irreducible subspaces, then is irreducible, so by Zorn’s lemma, every irreducible subspace of is contained in a maximal irreducible subspace. 3 If is a maximal irreducible subspace, then is irreducible, so , that means is closed. All sets of one point are irreducible, hence contained in some maximal irreducible subspace. Hence maximal irreducible subspaces cover . If is Hausdorff and is irreducible, then must contains only one point. So irreducible components of a Hausdorff space is all one-point subspace. 4 If is a irreducible component of , then is closed and irreducible, so there is a unique minimal prime ideal in and . The converse is trivial.

Exercise 21. Let be a ring homomorphism. Let and . if , then is a prime ideal of , i.e. a point of . Hence induces a mapping . Show that:

 1 If then , and hence that is continuous. 2 If is an ideal of , then . 3 If is an ideal of , then . 4 If is surjective, then is a homeomorphism of onto the closed subset of . (In particular, ). 5 If is injective, then is dense in . More precisely, is dense in . 6 Let be another ring homomorphism, then . 7 Let be an intergal domain with just one non-zero prime ideal , and let be the field of fractions of . Let . Define by , where is the image in . Show that is bijective but not a homeomorphism.

Proof.

 1 . 2 . 3 if and only if for every , we have . But , so it’s equivalent to ,i.e.. 4 If is surjective, then , so we can identify with . Now is the quotient map, and prime ideals of is one-to-one corresponding with primes ideals of which containing . Also is same as , hence is a homeomorphism. 5 If is injective, then . So is dense in . is dense , i.e. , 6 Trivial. 7 There are two points in , where is a closed point but isn’t. Also there are two points . maps to and to , so it’s bijective. But in two points are both closed, so can’t be a homeomorphism.

Exercise 22. Let be the direct product of rings . Show that is the disjoint union of open (and closed) subspaces , where is canonically homeomorphic with .

Conversely, ler be any ring. Show that the following statements are equivalent:

 1 is disconnected. 2 where neither of the rings is zero ring. 3 contains an idempotent .

Proof. A prime ideal of is in the form , where is a prime of . So is disjoint union of . Also is closed.

Conversely, If is disconnected, then where and both is non-empty open (and closed) set. So for some ideal , and . So select such that , then there exists n s.t. . so are coprime and , and . So i) ii).

If , then is an idempotent . Conversely, if is an idempotent , then , and . So .

Exercise 23. Let be a Boolean ring (Exercise 11), and let .

 1 For each , the set is both open and closd in . 2 Let , Show that for some . 3 The sets are the only subsets of which are both open and closed. 4 is a compact Hausdorff space.

Proof.

 1 Obviously is open. By and we have and . Show is closed. 2 The complement of is . By Exercise 11, there exists such that , So . 3 If is open, that is, is the disjoint union of and some . So and . Let where , then and . This means . 4 If and is two distinct point of , we may assume , then there exists . So . Since is both open and closed, and are seperated by neighborhoods. is compact by Exercise 17.

Exercise 24. Let be a lattice, in which the sup and inf of two elements are denoted by and respectively. is a Boolean lattice (or Boolean algebra) if

 1 has a least element and a greatesr element (denoted by respectively). 2 Each of is distributive over the other. 3 Each has a unique "complement" such that and

Let be a Boolean lattice. Define addition and multiplication in L by the rules Verify that in this way become a Boolean ring, say .

Conversely, starting from a Boolean ring , define an ordering on a as follows: means that . Show that, with respect to this ordering, is a Boolean lattice [The sup and inf are given by and , and the complement by .] In this way we obtain a one-to-one correspondence between Boolean rings and Boolean lattices.

Proof. Verify directly.

Exercise 25. From the last two exercise deduce Stone’s theorem, that every Boolean lattice is isomorphic to the lattice of open-and-closed subsets of some compact Hausdorff topological space.

Proof. If is a Boolean lattice then is a Boolean ring. Let , then is Hausdorff, and there is a one-to-one correspondence between open-and-closed subsets of and elements of : .

We have and . Also ; this shows that lattice of ’s is isomorphic to .

Exercise 26. Let be a ring. The subspace of consisting of the maximal ideals of , with the induced topology, is called the maximal spectrum of and is denotedby . For arbitrary commutative rings it does not have the nice functorial properties of (See Exercise 21), because the inverse image of a maximal ideal under a ring homomorphism need not be maximal.

Let be a compact Hausdorff space and let denote the ring of all real-valued continuous functions on (add and multiply functions by adding and multiplying their values). For each , let be theset of all such that . The ideal is maximal, because it is the kernel of the (surjective) homomorphism which takes to . If denotes , we have therefore defined amapping , namely .

We shall show that is a homeomorphism of onto .

 1 Let be any maximal ideal of , and let be theset of common zeros of the functions in . That is, Suppose that is empty. Then for each there exists such that . Since is continus, there is an open neighborhood of in on which does not vanish. By compactness a finite number of neighborhoods, say , cover . Let Then doest not vanish at any point of , hence is a unit in . But this contracdicts , hence is not empty. Let be a point of . Then , hence because is maximal. Hence is surjective. 2 By Urysohn’s lemma (this is the only non-trivial fact required in the argument) the continuous functions seperate the points of . Hence , and therefore is injective. 3 Let ; let and let Show that . The open sets (resp. ) form a basis of the topology of (resp. ) and therefore is a homeomorphism.Thus can be reconstructed from the ring of functions .
Proof. . But every is some cause of i). So . If and open, then by Urysohn’s lemma there exists a function such that and for all we have , so . This shows form a basis of . Also forms a basis of , so is a homeomorphism.

### Affine algebraic varieties

Exercise 27. Let be an algebraically closed field and let be a set of polynomial equations in variables with coefficients in . The set of all points which satisfy these equations is an affine algebraic variety.

Consider the set of all polynomials with the proerty that for all . the set is an ideal in the polynomial ring, and is called the ideal of the variety . The quotient ring is the ring of polynomial functions on , because two polynomial define the same polynomial function on is and only if vanishes at every point of , that is, if and only if .

Let be the image of in . The are the coordinate functions on : if , then is the th coordinate of . is generated as a -algebra by the coordinate functions, and is called the corrdinate ring (or affine algebra) on .

As in Exercise 26, for each let be the ideal of all such that ; it is a maximal ideal of . Hence, if , we have defined a mapping , namely .

It’s easy to show that is injective: if , we must have for some , and hence is in but not in , so that . What is less obvious (but still true) is that is surjective. This is one form of Hilbert’s Nullstellensatz (See Chapter 7).

Exercise 28. Let be elements of . They determine a polynomial mapping : if , the coordinates of are .

Let be affine algebraic varieties in respectively. A mapping is said to be regular if is the restriction to of a polynomial mapping from .

If is a polynomial function on , then is a polynomial function on . Hence induces a -algebra homomorphism , namely . Show that in this way we obtain a one-to-one correspondence between the regular mappings and the -algebra homomorphism .

Proof. Let denote coordinates on . For any -algebra homeomorphism , let , and find any such that the image of in is .

Let be the mapping induced by , and be the -algebra homomorphism from to . Then we have on . Cause the -algebra is generated by , we have . This shows the way above gives a one-to-one correspondence between regular mappings and the -algebra homomorphism .