# MATH 221 NOTES BRENT HO. Date: January 3, 2009.

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1 MATH 22 NOTES BRENT HO Date: January 3,

2 Table of Contents. Localizations Zariski Topology Local Properties Artinian Rings Nakayama, Locally Free Referred to homework solutions

3 LOCALIZATIONS. Definition.. Let A be a commutative ring. S A is a multiplicative subset if it does not contain 0, it is multiplicatively closed [and it does not contain 0] Examples:. f A non-nilpotent, S = {, f, f 2,... } 2. A is a domain, S = A {0} Remark.. p A is prime if A p is a multiplicative subset. Definition.2. Given a commutative ring A and a multiplicative subset S A, we define the ring A S by defining an equivalence relation on A S by (a, s ) (a 2, s 2 ) if there exists s S such that a s 2 s = a 2 s s. We define multiplation component-wise, denote the equivalence class of (a, s) by a, and define addition by a + a = as +a s. We define a ring s s s ss homomorphism φ univ : A A S by φ univ (a) = a Proposition-Construction.. There exists a bijection between ring homomorphisms f : A B [B is commutative] such that the image of all elements of S are invertible, and ring homomorphisms g : A S B such that g φ univ = f. Proof. Given a map f, we define a map g by g( a) = s f(a)[f(s)]. It is easily checked that it satisfies ring homomorphism axioms. Given a map g, we define f by f(a) = g( a ). We note that f(s) = g( s), so that f(s) has inverse g( ) by homorphism axioms. It is similarly s verifiable that this map satisfies axioms and it clearly commutes. A f B φ univ A S Remark.2. This is the universal property of A S g Proposition.. For a non-nilpotent element f and S = {, f, f 2,... }, A f := A S = A[t]/( tf). Proof. We define a map g : A[t]/( tf) A f by f(at i ) = a. This map vanishes on f i ( tf)a[t] since g( tf) = f = 0. It satisfies ring homomorphisms because f g(at i + bt j ) = a f i + b f j = af j + bt i f i+j = g((af j + bf j )t i+j ) = g(at i + bt j ) I have put in brackets that things that are not in my notes but that I think that Dennis left out. 2

4 since tf = ( tf) + tf =. The check that it satisfies ring multiplication is trivial. This map is surjective because a is the image of at f i. It is injective because for a polynomial p(t) i with coefficients a i, n n a n g(p(t)) = g( a i t i i ) = f = i=0 a if n i = 0 i f n if n i=0 a if n i i=0 = 0, since f is non-nilpotent. But as tf =, n i=0 a if n i n n = g( a i f n i (tf) i ) = g( a i f n t i ) = g(f n p(t)) i=0 i=0 so as f n 0 for all n, this implies that p(t) = 0. Lemma.. Let A be a finitely generated algebra over k, and f A a non-nilpotent element. Then there is a function φ : A k corresponding to some maximal ideal m such that φ(f) 0. Proof. We consider the function A A f, and take any maximal ideal p in A f. Then the composition A A f A f /p = k has nonzero image of f, since f 0 implies that the image of f, f maps to 0, which means that maps to zero, so that this is a zero map. Thus taking the preimage of this maximal ideal, we have the desired map by this composition. Example: (A) := A A {0}, the field of fractions of A. Proposition.2. (k[x, x 2 ]/x x 2 ) x k[x ] x Proof. We define a map f : k[x ] x (k[x, x 2 ]/x x 2 ) x by p p. It is clearly injective, q q and is surjective because x 2 = x 2x x = 0, so that all polynomials with x 2 terms are equivalent to polynomials without those terms. Definition.3. Given an A-module M and a multiplicative subset S, we construct the module M S := A S M. A more explicit description of M S can be obtained by identifying a A m with the fraction am. Thus we denote M s S by the set of fractions m s s S with addition and action of A S inherited in this fashion. Lemma.2. M S = 0 if and only if m M, s S such that sm = 0 Proof. trivial. Example: A = k[t], M = k[t]/t. Then M t = 0 Lemma.3. Let f be a non-nilpotent element. Then M f lim given by multiplication by f. Proof. See 8 on pset 3 2 Corollary.. A = k[t], A t = lim t i k[t] = k[t, t ] Proposition.3. A S is flat. (i.e. localization is an exact functor) 2 See solution at end 3 i=0 for m M and M, with the maps M M

5 Proof. Suppose that T : M M 2, and consider the induced map A S M A S M 2, i.e. the map L : M S M 2S. Then L( m) = T (m) is 0 if s such that s T (m) = 0, by s s definition. But s T (m) = T (s m) = 0 = s m = 0 = m = 0. s Proposition.4. A Noetherian implies that A S is Noetherian. Proof. Take an ideal J A S. It s preimage under the map A A S is an ideal I, so is finitely generated by f i. We claim that f i generate J. This is true because for any element b J, b = a, then by idealness s a = a J, and we have a s s i A such that a = a i f i. Then a = a i f i, so that b = a = a i, as desired. s s fi 4

6 ZARISKI TOPOLOGY 2. Definition 2.. Let A be a commutative ring. We define Spec(A) to be the set of prime ideals of A, with elements ṗ corresponding to primes p. We define the Zariski topology by defining closed sets: V Spec(A) is closed if I A such that V = V (I) := {ṗ p I}. We check that it satisfies topology axioms:. = V (A) is closed. 2. Spec(A) = V (0) is closed. 3. For two closed sets V = V (I ) and V 2 = V (I 2 ), the set V V 2 = V (I I 2 ) is closed. (See following lemma) 4. For closed sets V (I i ), V (I i ) = V ( I i ) is closed. (See following lemma) i i Lemma 2.. V (I ) V (I 2 ) = V (I I 2 ) Proof. Let ṗ (V (I ) V (I 2 )). Then p I or I 2. Thus p I I 2. Suppose p I I 2, and suppose p I i for i =, 2. Then take a I, a 2 I 2 such that a i / I i. Then a a 2 (I I 2 ) = a a 2 p, which contradicts the primeness of p. Lemma 2.2. i V (I i ) = V ( i I i ) Proof. ṗ i V (I i ) p V (I i ) i p I i i p I i Remark 2.. If I I 2, then V (I ) V (I 2 ) Proposition 2.. V (I) = V (rad(i)); radi := {a n Zs.t.a n I} Proof. If p contains rad(i), then as I rad(i), p contains I. If p contains I, then by primeness, p contains rad(i). Corollary 2.. V (I ) = V (I 2 ) if rad(i ) = rad(i 2 ). Let f A. We define U f := {ṗ f / p}. U f = Spec(A) V (f), so is open. Remark 2.2. U f is empty f p for all p f is nilpotent. Proposition 2.2. U f form a basis of the topology. Proof. Take a point ṗ Spec(A), and suppose that ṗ U, an open set. Then consider the set Spec(A) U. By definition, it is closed, so let Spec(A) U = V (I). ṗ / V (I) means that p I, so we find f I such that f / p. Then p U f. Further, U f U because U f (Spec(A) U) = U f V (I) = Lemma 2.3. U f U f2 = U f f 2 Proof. By primeness of elements on either side. Lemma 2.4. Let f,,f n be elements of A. Then the following are equivalent:. f i generate the unit ideal, i.e. a i such that a i f i =. 2. i U fi = Spec(A) 5

7 Proof. U fi = Spec(A) V (f i ) = V ( I i ) =. In general, if V (I) =, then I = A, because if not, I m for some maximal m by Zorn s lemma, which contradicts. Thus Ii = A, as desired. Corollary 2.2. Spec(A) is compact. Proof. Take an open covering, reduce it to basic opens, use the above lemma to express as a finit sum of i j for i j I j. Then we reduce to the basic opens for which these i j are non-zero, which is a finite number by definition, and use the above lemma again to obtain that this is our finite subcover. Remark 2.3. S Spec(A). S = V ( ṗ S p) Remark 2.4. ṗ is closed if and only if p is maximal. Remark 2.5. Suppose A is a domain. Then η := 0 is the generic point of Spec(A). Remark 2.6. Spec(A) is not necessarily T (i.e. for y x, a neighborhood of x that does not contain y). For example, let A be a domain and y = η Example: A = k[t] with k algebraically closed. Then Spec(A) is isomorphic to η k, and if S k is an infinite subset, then S = k since k[t] is a PID. Key results from HW. Given a map φ : A B of rings, we have an associated map Φ : Spec(B) Spec(A) given by p φ (p). This map is continuous, and V (φ (J)) = Φ(V (J)). 2. If A is Noetherian, we have an irreducible decomposition Spec(A) = n i=v (p i ) where p i are all the minimal primes. 6

8 LOCAL PROPERTIES 3. Definition 3.. We say that a property Q is local if given an A-module M, [ f A s.t. ṗ U f and Q holds for M f as an A f -module p] = [Q holds for M]. Equivalently, a property is local if for some f i that generate the unit ideal, the property holding for M fi implies that the property holds for M. Lemma 3.. The property of being 0 is local. Proof. Suppose M is a module s.t. p, f s.t. p U f and M f = 0. Then for all p, we take the set U fp that satisfies this property. This is an open cover of Spec(A), so as Spec(A) is quasi-compact, we have a finite subcover U fi, i =,..., n of Spec(A) such that M fi = 0. However, M fi = 0 = n i s.t. f n i i m = 0 for all m M. By Lemma 2.4, these f i generate A, so that there is some combination a i f i =. But now we have m = m = n max{ni} = ( a i f i ) n max{ni} m = 0. Lemma 3.2. The property of being surjective/injective/bijective is local. Proof. Suppose we have a map M N. Surjective. Let L be the cokernel, so that we have the exact sequence M N L 0. Applying the previous lemma, if for all p, there is some f such that ṗ U f and M f N f is surjective, then L f is zero for these f, so L is 0. Injective. Let L be the kernel, so that we have the exact sequence 0 L M N. Then again, if the M f N f are injective, then L f is zero, so L is zero. Lemma 3.3. The property of being finitely generated is local. Proof. Suppose that for some f i that generate the unit ideal, M fi is finitely generated over A fi. Let these generators be m j i. We can assume that these generators lie in the image of the map M M fi by clearing denominators (replace m ms with ). In other words, for each i, s we have a surjection A fi M fi. We claim that the preimages of all of these m j i generate j M over A. This is equivalent to the assertion that the map i which is true because it is locally surjective. Definition 3.2. Given a prime p, we denote A p := A A p and M p := M Ap. Definition 3.3. A ring A is called local if it has only one maximal ideal. j A M is surjective, Lemma 3.4. m A. Then A is local with unique maximal ideal m if and only if every element in A m is invertible. Proof. Suppose that A is local with unique maximal ideal m. Suppose that f A m is not invertible. Then (f) is an ideal not equal to A, so by Zorn s lemma, is in a maximal ideal m. But since this m has f, m m, which contradicts localness. Let m be a maximal ideal, and m m. Then m m by maximality, so a m such that a / m. Then (a) m, but as a is invertible, (a) = A - contradiction. Example: Let A = C[x], and p = (x). Then A p = C[x] C[x] (x) is not finitely generated (do infinite prime trick on denominators) 7

9 Lemma 3.5. A p is local. Proof. Consider the prime p p A p. Every element in A p p p is invertible by construction. Result from homework 5.: The property of being flat is local Definition 3.4. Given an A-module M, we define supp(m) = {ṗ M p 0} Spec(A) to be the support of M Proposition 3.. Let M be a finitely generated A-module. Then supp(m) = V (I), where I =Ann(M) := {a A am = 0 m m}. Proof. We prove the following lemma: M S = 0 if and only if S Ann(M). Take m i generators of A over M. Then for all i, s i S such that s i m i = 0. Then s i kills all m M, so s i Ann(M), but as S is a multiplicative subset, it is closed under multiplication, so s i S. Take s S Ann(M). Then m s M S, m s = sm ss = 0. In light of this lemma, we see that M p = 0 if and only if (A p) Ann(M), which occurs if and only if p Ann(M), i.e. p / V (Ann(M)). Remark 3.. The above local property stuff is for the definition above, not for the general definition of a p-local property, which is that if a property holds for all localizations at primes p, then it holds in general. These defintions are not equivalent, as we shall see. For an example of a module that is finitely generated at localizations M p over all primes p, but is not itself finitely generated, see problem 5.4(c) Extra Stuff not in notes: Definition 3.5. A property Q is said to be p-local if [Q holds for M] if and only if [Q holds for M p for all primes p]. From pg 40-4 in Atiyah, we have that the property of being 0 is p-local. In fact, it is m-local (i.e. for maximals ideals), as is the property of being injective/surjective/bijective and flatness. Lemma 3.6. (Serre) Let A be a ring, f,..., f n elements that generate the unit ideal. For an A-module M, we define a map M φ i M fi. The map φ is injective. 2. The image of M inside i the images of m i and m j coincide in M fi f j. by compiling the maps M M fi. We claim that M fi are elements (m i ) such that when mapped further onto M fi f j, i,j M M fi Proof. M fj 8 M fi f j

10 . We localize the function at each f i, and obtain that since (M fi ) fi = M fi, then the function φ i : M fi (M fj ) fi is injective (look at the i th coordinate, where we have a bijection), so that since injectiveness is a local property, φ is injective. 2. We denote the set of these kinds of elements M. It is clear that Im(M) M by the commutative diagram above. Case : Suppose that f is invertible. Then A f = A, so M f = M. For an element (m,..., m n ) in M, we see that m is an element of M. By the definition of M, the image of m in M f f j = (M f ) fj = M fj is the same as the image of m j in M fj. Thus the element (m i ) is the image of m M. Thus M Im(M). Case 2: We see that for an element g, M g = the set of elements (m i ) M fi g such that the image in i,j M fi f j g coincide. Thus we take g = f i for each i, apply case, and now obtain that M f i (Im(M)) fi for all i. Thus as injectiveness is a local property, we have M Im(M), as desired. 9

11 ARTINIAN RINGS 4. Definition 4.. A commutative ring A is Artinian if every descending chain of ideals I I 2... stabilizes. Examples. Z is not Artinian Z 2Z 4Z... does not stabilize. 2. C[t] is not Artinian (t) (t 2 )... does not stabilize. 3. C[t]/t 0000 is Artinian as it is a finite dimensional vector space over a field 4. For V a finite dimensional vector space, the subalgebra of End(V ) consisting of elements T i such that T i T j = T j T i with multiplication consisting of composition is Artinian because it is a subspace of End(V ), which is finite dimensional over a field. Proposition 4.. Let A be Artinian. Then every prime ideal is maximal. Proof. Consider A/p, for a prime ideal p. We note that A/p is also Artinian, since chains of ideals can be lifted to chains of ideals in A. Further, since p is prime, A/p is a domain. We take an element a A/p, and consider the descending chain (a) (a 2 ).... It stabilizes, so that we have (a n ) = (a n+ ) for some n. In particular, we have that b such that a n = a n+ b, or a n ( ab) = 0. Since A/p is a domain, this implies that either a n or ab is zero. But domainness again implies that a n = 0 = a = 0, so we can ignore this case. Thus b is an inverse for a, so that A/p is a field, i.e. p is maximal. Corollary 4.. If A is also Noetherian, then as all points are closed, 6 on problem set 4 implies that from our decomposition into irreducibles, there are a finite number of maximal ideals. Thus A has a finite number of prime ideals. Lemma 4.. Let Spec(A) = V V 2. Then there is a unique decomposition A = A A 2 such that V i Spec(A i ). Proof. See 6 on pset 5. Lemma 4.2. Let A be a local Artinian ring. Then n such that m n = 0, for m its maximal ideal. Proof. Since the intersection of maximal/prime ideals is the set of nilpotent elements, we have that every element of m is nilpotent. By quasi-compactness of Spec(A), we take a finite number of elements f i that generate the unit ideal. Then by nilpotence, n i such that f n i i = 0. Thus m max{ni} = 0. Remark 4.. From Homework 5.7, every Artinian ring is Noetherian Random Lemma Lemma 4.3. Suppose that B is rational (i.e. not nilpotent). Then for φ, φ 2 : A B, we look at the associated maps Φ i : Spec(B) Spec(A). If Φ = Φ 2, then φ = φ 2. Proof. We want to show that φ (f) = φ 2 (f) for all f A. For every maximal ideal ṁ Specm(B), consider the map ψ m : B B/m. From the assumption, we have ψ m (φ (f)) = ψ m (φ 2 (f)). Thus by the Nullstellensatz, φ (f) φ 2 (f) is nilpotent. Thus by assumption of rationality, it is 0. 0

12 NAKAYAMA, LOCALLY FREE 5. Lemma 5.. (Nakayama) Let A be a local ring with maximal ideal m, M a finitely generated A-module. Consider M A/m = M/mM. Then M/mM = 0 implies that M = 0. A Proof. let m,..., m n generate M over A By assumption, we have M/mM = 0, i.e. mm = M. Therefore, we can represent each b i as an element of the form am, with m M and a m. Representing m with the generators, we obtain b i = n j= aa jb j. As a m, we have aa j m. Thus we have b i = n j= a ijb j for a ij m. We consider the commutative diagram A n M T A n M Where the map T is given by T (e i ) = e i n j= a ije j, and the maps across are given by e i b i. The matrix for T is given by a a 2... a n a2... an a nn Taking the determinant of this matrix, we obtain an element of the form + m, where m m. Thus this element is invertible, so is nonzero, so T is invertible. Therefore T is an isomorphism. However, going down and across in the diagram, we obtain e i b i n j= a ijb j = b i b i = 0, so for this diagram to commute, we put the zero map on the right hand side. However, T being an isomorphism means that this map is an isomorphism, so thus M = 0, as desired. Corollary 5.. Let A be a local ring, M and M 2 finitely generated modules. If φ : M M 2 is a function such that M /mm M 2 /mm 2, then φ is surjective. Proof. Let coker(φ) = K. Then K/mK is the cokernel of the map M /mm M 2 /mm 2. Thus it is zero, so by Nakayama, K is zero. Example: A = Z p, M = pz p. Then Q p is a module for which Nakayama s does not hold (Z p is not local), since Q p /pq p = 0. Lemma 5.2. M free M projective M flat. Proof. If M is free, then M = A. Since A is projective and direct sums of projective modules are projective, then M is projective. Now suppose that M is projective, i.e. is a direct summand of a free module, A = M N. Then as A is flat, we have that for an injection M M 2, M A M 2 A so (M M) (M N) (M 2 M) (M 2 N), so in particular, as each summand maps to each summand, M M M 2 M. Theorem 5.. Let A be a local Noetherian ring and M a finitely generated A-module. Then TFAE. M is free 2. M is projective 3. M is flat

13 Proof. 2 3 by lemma above. 3 Take n such that A n M. Tensoring with A/m, we obtain a map (A/m) n M/mM. Since these are both vector spaces, we can adjust n so that this map is an isomorphism. Then by the corrolary to Nakayama above, for this n, we still have A n M. We thus consider the exact sequence 0 M A n M 0 By Noetherianess, M, as a submodule of a finitely generated module, is finitely generated. Therefore, showing that M /mm = 0 shows that M = 0. However, from number 4 on the tensor problem homework, we obtain that M is flat gives up the exact sequence 0 M A/m A n A/m M A/m 0 Thus since n was chosen so that the latter arrow was an isomorphism, we have M A/m = M /mm = 0. Thus the map A n M is an isomorphism. Definition 5.. Let A be a commutative ring, M an A-module. Then M is locally free if ṗ Spec(A), f s.t. ṗ U f s.t. M f is locally free over A f. Equivalently, for f i that generate the unit ideal, M fi are free over A fi. Lemma 5.3. Suppose that M is finitely generated and locally free. Then for the f i in the definition, M fi A n f i = M fj A n f j, i.e. M can be said to be locally free of rank n (all the localizations at f i are isomorphic to A n f i ). Proof. Take i j, and consider M fi = A n i f i (M fi ) fj = M A A fi A and M fj = A n j f j. Then as A fj = M A A fj A A fi = (M fj ) fi Thus we have A n i f i f j = (A n i f i ) fj = (A n j f j ) fi = A n j f i f j, so that n i = n j, as desired. Theorem 5.2. Let A be Noetherian, M finitely generated. Then the following are equivalent. M is locally free 2. M is projective 3. M is flat Proof. From 2(d) on problem set 5, the property of being finitely generated and projective is local. Localizing, therefore, we obtain that M is locally projective, and further, as M is finitely generated, then M f = M A f is finitely generated over A f (take the same generators). Thus M is projective. From the lemma above, we have 2 3. Thus it remains to show that 3. 3 M flat over A implies, as localizations are flat, that M p is flat over A p. Then the previous lemma, as A p is local, implies that M p is free over A p, so suppose that M p = A n p. We prove the following lemma Let N be a finitely generated module over A such that N p = 0 for some p. Then there is some f / p such that N f = 0. Proof. N p = 0 = p / supp(n). By proposition 3., supp(n) is closed, so it s complement is open, so there is some open set containing p such that this set does not intersect supp(n). Taking this set to be a basic open, we have p U f such that U f supp(n) = U f V (Ann(N)) =, which means that f Ann(M). Thus N f = 0. In light of this lemma, we look at the map A n p M p. We can take the basis elements e i of A n p to map to elements of the form m by ring action of A p, so that we can consider the corresponding map A n M and exact sequence 0 K A n M C 0 2

14 Then since A n p M p, we have K p = C p = 0. Thus by our lemma, K f and C f are zero, so that the map A n f M f is an isomorphism, as desired. Remark 5.. In light of this theorem, we have M locally free if and only if M is flat, and as flatness is a p-local property and localizations are flat, M is flat if and only if M p is flat for all p, which by the theorem before is equivalent to M p free over A p. Thus M is locally free if and only if M p is free over A p for all p. Lemma 5.4. Let B be a domain. N B n such that N F rac(b) = 0 implies that N = 0 Proof. Take n N. Suppose that this is nonzero, i.e. as a submodule of B n, 0 f B such that fn 0. Then N F rac(b) = 0 implies that 0 b B such that bfn = 0 = bf = 0, as B is a domain. But this is impossible because B is again a domain. Theorem 5.3. Let A be a local Noetherian domain and M a finitely generated module. TFAE:. M is free. 2. dim A/m (M/mM) =dim F rac(a) (M F rac(a)). Proof. 2 Suppose that M is free. Then M = A n, so M/mM = M A/m = A n A/m = (A/m) n, so dim A/m (M/mM) = n. Also, M F rac(a) = A n F rac(a) = (F rac(a)) n, so that the RHS is also n. 2 Suppose that dim A/m (M/mM) = n =dim F rac(a) (M F rac(a)). Then we have the surjection (A/m) n M/mM. From the corollary to Nakayama, we have the surjection A n M. Thus we consider the short exact sequence Tensoring this with F rac(a), we obtain 0 M A n M 0 0 M F rac(a) A n F rac(a) = F rac(a) n M F rac(a) = F rac(a) n 0 by assumption, so we M F rac(a) = 0. As A is a domain, and M A n, we can apply the previous lemma to obtain that this implies that M = 0, i.e. A n M, as desired. Theorem 5.4. Let A be a Noetherian domain and M a finitely generated A-module. Then TFAE:. M is locally free of rank n. 2. There exists n such that dim(m p A p A p /p p A p ) = n as a vector space over A p /p p A p for all p. Proof. 2 From the lemma in the theorem above, we have A n f M f n implies that An p M p. Thus tensoring both sides with A p /p p, we get the desired result. 2 By assumption, we have (A p /p p ) n M p /p p M p. By Nakayama, this implies that the function A n p M p is an isomorphism (the cokernel and kernel are 0 by Nakayama). Thus M p is flat over A p, and as flatness is a p-local property, then M is flat, so by the previous lemma, M is locally free, as desired. Definition 5.2. M p /p p M p = M A p /p p A p is called the fiber of M at p A Definition 5.3. M p is called the stalk of M at p. 3

15 SOLUTIONS 6. f f 3.8 To define a map l im (M 0 M M2...) M f (the M i = M, just indexed conveniently by i), we define maps φ i : M i M f by φ 0 (m) = m, φ i(m) = m, and see that the diagram: f i f M f M f M... φ 0 φ φ 2 M f commutes, and that the map is a map of modules. Then by the universal property of the direct limit, there is a unique map φ univ : l im (M 0 f M f M2...) M f s.t. φ i = φ univ f i. To map M f l im (M 0 f M f M2...), we define a map ψ by ψ( m f i ) = (0,.., m,..) which is nonzero in the k position. It is clearly a map of modules, and under this map, images of elements of the set {, f i } are invertible, as f i (ψ( m f i )) = (0,.., f i m,..) (, 0, 0..). Finally, by problem set we have that all elements of the direct limit can be expressed as (0,..m,..), so ψ(φ((0,..m,..))) = ψ( m ) = (0,..m,..) f i and we have the desired isomorphism. φ(ψ( m f i )) = φ((0,..m,..)) = m f i 5. Suppose that for an A-module M, for every p A, f / p s.t. p U f and M f is flat. Then we claim that M is flat. Consider the covering of Spec(A) by these U f for each prime. Then by quasi-compactness f...f n s.t. U f...u fn cover Spec(A), so by homework before, f...f n generate A. Now for any injection between A-modules N and N 2, we consider the exact sequences: Ker N M N 2 M Ker fi N M fi N 2 M fi Where Ker is the kernel of this map. Then by flatness of M fi, Ker fi being 0 is local, we have that Ker is 0, as desired. is 0, and as the property of 5.4(c) Let M be (C[x]/(x a)). It is not finitely generated over C[x], as it is an infinite a C direct sum. Primes of C[x] are ideals generated by x b, where b C. Now consider M (x b) = a C (C[x]/(x a)) x b, so consider (C[x]/(x a)) x b. If b a, x a C[x] ((x b)), so (C[x]/(x a)) x b = 0. If b = a, (x a) (C[x] ((x b)) =, so by the last pset, (C[x]/(x a)) x a 0. Further, we claim that (C[x]/(x a)) x a C[x]/(x a). We define a map from the right side to the left by sending p(x) to p(x). This map is injective because p(x) = g(x) implies that there exists an element s(x) of the multiplicative set C[x] ((x a)) such that s(x)g(x) = s(x)p(x). But since s(x)p(x) = p(x) this implies that g(x) = p(x), so the map is injective. To prove surjectivity, we first note that C[x]/x a C. Now take an element 4

16 m m s(x), and let s(x) = n. Then as s(x) = m n, so it is mapped to by the element m n, so the map is also surjective. Also, we claim that C[x]/(x a) C (this is true as it is a -dimensional field extension of C, but we ll show it more explicitly). We do this by mapping p(x) to its remainder when divided by x a. This map is well defined because if p(x) q(x), m(x) (x a) s.t. p(x) = q(x) + m(x), so their images are the same because the remainder of q(x) + m(x) is the remainder of q(x) plus the remainder of m(x), and the remainder of m(x) is zero. Further, this map is injective because the images of p(x) and q(x) are the same if and only if m(x) (x a) that is their difference. It is also surjective because for any b C, b is the image of b. Therefore, for any prime p, M p C is finitely generated.(c is finitely generated by, with the action of C[x] given by multiplication of constants), but as mentioned before, M is not finitely generated. 5.6 Suppose V i = V (I i ). Then Spec(A) = V (I ) V (I 2 ) = V (I I 2 ) implies that elements of I I 2 are nilpotent. Further, V (I ) V (I 2 ) = V (I + I 2 ) = implies that I + I 2 = A. Thus as A, then i, i 2 s.t. i + i 2 =. As both are nilpotent, let raising each to the n kill both. Now consider the element of A, (i + i 2 ) 2n = i 2n + a i 2n i 2...a n i n in 2 + a n+i n i n i 2n 2, where the a i are constants. Then the term a n i n in 2 is zero. Let the terms before equal a, and the terms after equal b. Then a I, b I 2, a b = 0, a + b =. Now let A = A/(a), A 2 = A/(b). Then we claim A A 2 = A. We define a map from A A A 2 by sending an element to its projections. Then if an element c is sent to (0, 0), then c (a) and c (b), which implies that bc = ac = 0 = (a + b)c = c = 0. Thus this map is injective. Finally, given an element of the image, (m, n), we have that (m, n) is the image of an + bm, so the map is surjective. Now, we claim that V (I ) = V (0 A 2 ), where we view 0 A 2 as its preimage in A, (a), i.e. we claim that V (I ) = V (a). If p I, then as a I, p (a) Suppose p (a). Then as Spec(A) = V (I ) V (I 2 ), where this is a disjoin union, p either I or I 2. However, if p I 2, then as b I 2, b p, but then this would imply that a, b p, and as (a) + (b) = A, this would imply that p = A, which is a contradiction. Thus p I. Now, we claim that V ((a)) Spec(A ). This is true because we have the surjection φ : A A = A/(a), so by 2. on the last problem set, Spec(A ) is homeomorphic to V (ker(φ)) = V ((a)), as desired. We do similarly for A 2. TP.4 In class, we showed that tensoring is a right-exact functor. Take an R module N 3, and surject onto from a free module N 2 module, and let N be the kernel of this projection, and inject it into N 2. Then 0 N N 2 N 3 0 is exact. Then we have the following exact commutative 5

17 diagram: N M N M 2 2 N M N 2 M 6 7 N 2 M 2 N 2 M N 3 M N 3 M 2 N 3 M 3 We want to show that is injective. Suppose that a N 3 M maps to zero in, a 0. As 8 is surjective, it has a nonzero preimage b N 2 M. By injectivity of 6, 6(b) = c N 2 M 2 is nonzero, and by commutativity, 9(c) = 0 c ker(9) c Im(4) nonzero d N M 2 s.t. 4(d) = c. But as c Im(6), c ker(7), so 7(c) = 0. Then as 5 is injective, this implies by commutivity that 2(d) = 0 d ker(2) d Im(). But b / ker(8) b / Im(3), so we have a contradiction. 0 6

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