Advanced Algebra II. Mar. 2, 2007 (Fri.) 1. commutative ring theory In this chapter, rings are assume to be commutative with identity.

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1 Advanced Algebra II Mar. 2, 2007 (Fri.) 1. commutative ring theory In this chapter, rings are assume to be commutative with identity basic definitions. We recall some basic definitions in the section. Definition An element a 0 R is said to be a zero divisor if there is an element b 0 R such that ab = 0. A ring R 0 and has no zero divisor is called an integral domain. Proposition A finite integral domain is a field. Definition A subring I R is an ideal if rx R for all r R, x I. It will be denoted I R. Proposition A subset I is an ideal if and only if for all a, b I, r R, a + b I and ra I. Example For a R, we have (a) := ar, the principal ideal generated by a. Let N R be the subset of all nilpotent elements, i.e. N := {a R a n = 0, for some n > 0}. Then N is an ideal, called the nilradical. Definition An element a R is said to be a unit if ab = 1 for some b R. Note that a is a unit if and only if (a) = R. Definition An ideal p in R is prime if p R and if ab R then either a R or b R. An ideal m in R is maximal if m R and if there is no ideal I R such that m I. It s easy to check that Proposition p R is prime if and only if R/p is an integral domain. m R is maximal if and only if R/m is a field. Also Proposition P R if and only if for any ideal I, J R, IJ p implies that either I p or J p. 1

2 2 Proof. Suppose that P is prime, IJ p and I p. We need to show that J p. To see this, pick any a I p, then ab p for all b J. since p is prime, we must have b p for all b J. Hence J p. Conversely, if p satisfies the above property, we need to show that it is prime. For any a, b R such that ab p, (ab) = (a)(b) p. Thus either (a) p or (b) p. It follows that a p or b p. By direct application of Zorn s Lemma, it s easy to that in a nonzero ring, there exists a maximal ideal. We leave the detail to the readers. A ring with exactly one maximal ideal is called a local ring. We have the following equivalent conditions: 1. (R, m) is local. 2. The subset of non-units is an ideal. 3. If m R is a maximal ideal and every element of 1 + m is a unit. Proposition The nilradical is the intersection of all prime ideals. Proof. Let N be the intersection of all prime ideals. If x N, i.e. x n = 0, then x n = 0 p for all prime p. Hence x p for all prime p, x N. On the other hand, if x N. Let Σ := {I R I {x, x 2,...} = }. Σ is non-empty for 0 Σ. By Zorn e Lemma, there is a maximal element p in Σ. We claim that p is prime. Then x p, hence x N. To see the claim, suppose that ab p but a, b p. We have p p + (a), p + (b). By maximality, we have x n p + (a), and x m p + (b). So x m+n (p + (a))(p + (b)) = p + (ab) = p, a contradiction. Exercise Let I R be an ideal. Let I := {x R x n I}. Then I = p:prime,i p p. Definition we define the Jacobson radical of R, denoted J(R), to be the intersection of all maximal ideals. The Jacobson radical is clearly an ideal. It has the property that: Proposition x J if and only if 1 xy is a unit for all y R. To see this, suppose that x J and 1 xy is not a unit. Then 1 xy m for some maximal ideal m. Since x J m, we have 1 m, a contradiction. Conversely, if x J, then x m for some maximal ideal m. Then m + (x) = R. So we have 1 = xy + u for some u m, and hence 1 xy = u is not a unit. Theorem (Chinese remainder theorem). Let I 1,..., I n be ideals of R such that I i + I j = R for all i j. Given elements x 1,..., x n R, there exists x R such that x x i ( mod I i ) for all i. The proof is left to the readers as an exercise.

3 Corollary Let I 1,..., I n be ideals of R such that I i + I j = R for all i j. Let f : R n i=1 R/I i. Then f is surjective with kernel I i factorization. Definition A non-zero element a R is said to be divide b R, denoted a b if there is c R such that ac = b. Elements a, b R are said to be associate, denoted a b, if a b, b a. The following properties are immediate. 1. a b if and only if (a) (b). 2. a b if and only if (a) = (b). 3. a is a unit if and only if (a) = R. 3 Definition A non-zero non-unit c R is said to be irreducible if c = ab then either a or b is unit. A non-zero non-unit p R is said to be prime if p ab then p a or p b. Then we have the following: Proposition Let p, c are non-zero non-unit elements in R. 1. p is primes if and only if (p) is prime. 2. c is irreducible if and only if (c) is maximal among all proper principal ideals. 3. Prime element is irreducible. 4. If R is a PID, then p is prime if and only if p is irreducible. 5. If a = bu with u a unit, then a b. 6. If R an integral domain, then a b implies a = bu for some unit u. Definition An integral domain is called a unique factorization domain, UFD for short, if every non-zero non-unit element can be factored into products of irreducible elements. And the factorization is unique up to units. Definition A ring R is said to be an Euclidean ring if there is a function ϕ : R {0} N such that: 1. if a, b 0 R and ab 0, then ϕ(a) ϕ(ab). 2. if a, b 0 R, then there exist q, r R such that a = qb + r with either r = 0 or r 0 and ϕ(r) < ϕ(b). Lemma Let R be a PID, then its ideals satisfies Ascending chain condition, i.e. for an ascending chain of ideals I 1 I 2... there is n such that I n = I n+1 =....

4 4 Hint. Considering I := I i. It s an ideal hence I = I n for some n. Theorem Every Euclidean domain, ED for short, is a PID. Every PID is a UFD Localization. Before we going to the study of dimension theory, we need to recall some basic notion of localization. Definition A subset S R is said to be a multiplicative set if (1) 1 S, (2) if a, b S, then ab S. Given a multiplicative set, then one can construct a localized ring S 1 R which I suppose the readers have known this. In order to be self-contained, I recall the construction: In R S, we define an equivalent relation that (r, s) (r, s ) if (rs r s)t = 0 for some t S. Let r denote the equivalent class of s (r, s). One can define addition and multiplication naturally. The set of all equivalent classes, denoted S 1 R, is thus a ring. There is a natural ring homomorphism ı : R S 1 R by ı(r) = r. 1 Remark (1) If 0 S, then S 1 R = 0. We thus assume that 0 S. (2) If R is a domain, then ı is injective. And in fact, S 1 R F naturally, where F is the quotient field of R. (3) Let J S 1 R. We will use J R to denote the ideal ı 1 (J). (If R is a domain, then J R = ı 1 (J) by identifying R as a subring of S 1 R). I would like to recall the most important example and explain their geometrical meaning, which, I think, justify the notion of localization. Example Let f 0 k[x 1,..., x n ] and let S = {1, f, f 2...}. The localization S 1 k[x 1,..., x n ] is usually denoted k[x 1,..., x n ] f. This ring can be regarded as regular functions on the open set U f := A n k V(f). One notices that U f is of course the maximal open subset that the ring k[x 1,..., x n ] f gives well-defined functions. Example Let x = (a 1,..., a n ) A n k and m x = (x 1 a 1,..., x n a k ) be its maximal ideal. Take S = k[x 1,..., x n ] m x, then the localization is denoted k[x 1,..., x n ] mx. It is the ring of regular functions near x. Recall that for a R-module M, one can also define S 1 M which is naturally an S 1 R-module.

5 We have: Mar. 9, 2007 (Fri.) Proposition (1) If I R, then S 1 I S 1 R. Moreover, every ideal J S 1 R is of the form S 1 I for some I R. (2) For J S 1 R, then S 1 (J R) = J. (3) S 1 I = S 1 R if and only if I S. (4) There is a one-to-one correspondence between {p Spec(R) p S = } and {q Spec(S 1 R)}. (5) In particular, the prime ideals of the local ring R p are in oneto-one correspondence with the prime ideals of R contained in p. Proof. (1) If x, y S 1 I, that is, x, y I, then x + y = xt+ys S 1 I. s t s t st And if r s S 1 R and x t S 1 I, then r x = rx s t st S 1 I. Hence S 1 I is an ideal. Moreover, let J S 1 R. Let I := ı 1 (J) R. We claim that J = S 1 I. To see this, for x I, x J. Hence x = x 1 J for 1 s 1 s all s S. It follows that S 1 I J. Conversely, if x J, then s x = x s J and x = ı(x). So x 1 s 1 1 s S 1 I. (2) If x J R, then x = y for some y R, i.e. (x sy)t s s 1 0 = 0, for some t 0 S. Then look at y t S 1 (J R). It s clear that y = x 1 J. t s t Conversely, if x J, then x J R. And hence x s 1 s S 1 (J S). (3) If x I S, then 1 = x 1 x S 1 I. Conversely, if 1 = x 1 s S 1 I, then (x s)t = 0. Therefore, xt = st S I. (4) For q Spec(S 1 R). It s clear that q R = ı 1 q Spec(R). (q R) S =, otherwise q = S 1 (q R) = S 1 R which is impossible. Conversely, let p Spec(R) and p S =. We would like to show that S 1 p is a prime ideal. First of all, if 1 1 S 1 p, then 1 = x for some x p. It follows that (x s)t = 0 for some 1 s t S. Thus st = xt p which is a contradiction. Therefore, S 1 p S 1 R. Moreover, if x y S 1 p. Then xy = x for some x p s t st s and s S. Then (xys stx )t = 0 for some t S. Hence xys t p. It follows that xy p since s t p. Thus either x or y in p, and either x or y in s t S 1 p. It remains to show that the correspondence is a one-to-one correspondence. We have seen that ı 1 : Spec(S 1 R) {p Spec(R) p S = } is surjective. By (2), it follows that this is injective. 5

6 6 (5) Let S = R p, then S q = if and only q p. But for I R, then S 1 I R I only. Indeed, if x I R. Then x = xs x S 1 I R. Conversely, for x S 1 I R, then x = y for some 1 t y I. Thus (y xt)s = 0 for some s, t S. We can not get y I in general. However, this is the case if I is prime and S I =. Thus we have Proposition If p R is a prime ideal and S p =. Then S 1 p R = p. Proof. This is an immediate consequence of (1) of the Proposition. Example Let p Spec(R), then R p is a local ring with the unique maximal ideal pr p. To see that, if there is a maximal ideal m. By the correspondence, m = qr p for some q p. Thus m pr p and thus must be equal. A ring with a unique maximal ideal is called a local ring. Thus R p is a local ring. Example More explicitly, we can consider the following example. Let R = k[x, y, x] and p = (x, y). Then there is a chain of prime ideals: 0 (x) (x, y) (x, y, z) R. ( ) If we look at R/p = k[z], we see a chain of primes ideals 0 (z) k[z], which corresponds to (x, y) (x, y, z) R in ( ). This has the following geometric intepreation: by looking at the ring R/p, we understand the polynomial functions on the set defined by x = 0, y = 0, i.e. the z-axis. On the other hand, if we look at R p, we see a chain of primes ideals 0 (x)r p (x, y)r p which corresponds to 0 (x) (x, y) in ( ). Geometrically, it can be think as local functions near a generic point in z-axis. We will come to more precise description of polynomial functions and generic point later. Proposition The operation S 1 on ideals commutes with formation of finite sums, product, intersection and radicals. Proof. These can be checked directly modules. It s essential to study modules in ring theory. One might find that modules not only generalize the notion od ideals but also clarify many things.

7 Definition Let R be a ring. An abelian group M is said to be an R-module if there is a map µ : R M M such that for all a, b R, x, y M, we have: Example a(x + y) = ax + ay (a + b)x = ax + bx a(bx) = (ab)x 1x = x Let R be a ring and I R be an ideal. Then I, R/I are R-modules naturally. Example An abelian group G has a natural Z-module structure by µ(m, g) := mg for all m Z and g G. Note that let M, N be R-modules. By a R-module homomorphism, we mean a group homomorphism ϕ : M N such that ϕ(rx) = rϕ(x). That is, it s an R-linear map. Exercise Given an R-module homomorphism f : M N, then ker(f), im(f), coker(f) are R-modules in a natural way. Let M be an R-module. Given x M, then Rx is a submodule of M. Even Ix is a submodule for any ideal I R. More generally, IN is a submodule of M if N < M and I R. On the other hand, given x M, we may consider the annihilator of x, Ann(x) := {r R rx = 0}. It s clear to be an ideal. Also for any submodule N < M. We can define Ann(N) similarly as {r R rn = 0}. A remark is that for a R-module M, the module structure map R M M also induces a natural map R/Ann(M) M M. Hence M can also be viewed as R/Ann(M)-module. Note that for given x M, the natural map f : R Rx is a R- linear map. And ker(f) = Ann(x). So in fact, we have an R-module isomorphism R/ ker(f) = Rx by the isomorphism theorem. Definition A element x M is said to be torsion if Ann(x) 0. A module is torsion if every non-zero element is torsion. A module is torsion-free if every non-zero element is not torsion. Exercise Given two modules M, N, note that the set of all R-module homomorphisms, denoted Hom(M, N), is naturally an R-module. Another important feature is that 7

8 8 Proposition The operation S 1 is exact. That is, if M f M is an exact sequence of R-module, then is exact as S 1 R-module. g M S 1 M S 1 f S 1 M S 1 g S 1 M Proof. By the construction, it follows directly that S 1 g S 1 f = 0. It suffices to check that ker(s 1 g) im(s 1 f). If x s ker(s 1 g), then g(x) = 0 S 1 M. That is tg(x) = 0 for some t S. We have s tg(x) = g(tx) = 0. And then tx = f(y) for some y M. Therefore, x s = xt st = f(y) st = S 1 f( y st ). Corollary The operation S 1 commutes with passing to quotients by ideals. That is, let I R be an ideal and S the image of S in R := R/I. Then S 1 R/S 1 I = S 1 R. Proof. By considering 0 I R R 0 as an exact sequence of R-modules. We have S 1 R/S 1 I = S 1 R as S 1 R-modules. We claim that there is a natural bijection from S 1 R to S 1 R by x which s x s is compatible with all structures. One can also try to prove this directly. Basically, construct a surjective ring homomorphism S 1 R S 1 R and shows that the kernel is S 1 I. We leave it as an exercise. We now introduce a very important and useful Lemma, Nakayama s Lemma. Theorem Let M be a finitely generated R-module. Let a J(R) be an ideal contained in the Jacobson radical. Then am = M implies M = 0. Perhaps the most useful case is when (R, m) is a local ring. Then J(R) = m is nothing but the unique maximal ideal. The assertion is that if mm = M, then M = 0. Proof. Suppose M 0. Let x 1,..., x n be a minimal generating set of M. We shall prove by induction on n. Note that for an ideal I R, elements in IM can be written as a 1 x a n x n for a i I. Since am = M, we have x 1 = n i=1 a ix i for a i a. Hence we have (1 a 1 )x 1 = n i=2 a ix i. Notice that (1 a 1 ) is a unit, otherwise it s in some maximal ideal which leads to a contradiction. So we have either x 1 = 0 or x 1 is generated by x 2,..., x n. Either one is a contradiction.

9 By applying the Lemma to M/N, and note that a(m/n) = (am + N)/N, we have the following: Corollary Keep the natation as above, if N < M is a submodule such that M = am + N, then M = N. Corollary Let R be a local ring and M be a finitely generated R-module. If x 1,..., x n generates M/mM as a vector space, then x 1,..., x n generates M. Proof. Let N be the submodule generated by x 1,..., x n. Then N + mm = M. We close this section by considering finitely generated modules over PID. We have two important and interesting examples. Example Let G be a finitely generated abelian group. Then it s clearly a Z-module while Z is PID. Moreover, a finite group is clearly a torsion module. Example Let V be a n-dimensional vector space over k. Let A be a n n matrix over k (or a linear transform from V to V ). Then V can be viewed as a k[t]-module via k[t] V V, with f(t)v f(a)v. Note that by Caylay-Hamilton Theorem, f(a) = 0 for f(x) being the characteristic polynomial. In fact, Ann(V ) = (p(x)), where p(x) is the minimal polynomial of A. Therefore, V is a torsion module. Now let M be a finitely generated torsion module over a PID R. One sees that Ann(M) 0. Since R is PID, we have Ann(M) = (p a i i ). For each p i, we consider M(p i ) := {x M p n i x = 0, for some n}. One can prove that M = M(p i ). In fact, for each p i, there exist n 1 n 2... n ji such that M(p i ) = j i k=1 R/(pn k i ). These p n k i are called elementary divisors. Apply this discussion to the example of linear transformation. Then Ann(V ) = (p(x)). We assume that p(x) = (x λ i ) a i splits into linear factors. Then V (λ i ) is nothing but the generalized eigenspace of λ i. And the decomposition V = V (λ i ) is the decomposition into generalized eigenspaces. The further decomposition into cyclic modules corresponds to further decomposition of eigensapces into in invariant subspaces. The restriction of linear transformation into these invariant subspaces gives a Jordan block. More explicitly, suppose that we have v V which corresponds to a generator of R/(x λ i ) m. Then we have independent vectors : 9

10 10 v = v 0, (A λ i )v =: v 1,..., (A λ i ) m 1 v = v m 1. Using this set as part of basis, then we see This gives the Jordan block. Av 0 = λ i v 0 + v 1,. Av m 2 = λ i v m 2 + v m 1, Av m 1 = λ i v m 1

11 11 Mar. 16, 2007 (Fri.) 1.5. tensor product. In this section, we are going to construct tensor product of modules coming from the multilinear algebra consideration. Then we describe its universal property. Lastly, we regard tensor product as a functor and compare its properties with the functor Hom. Let R be a ring and M 1, M 2 be R-modules. We consider a category whose object are (f, N), where N is R-module and f : M 1 M 2 N is a R-bilinear map. A morphism is defined naturally. Definition Keep the notation as above, the universal repelling object is called the tensor product of M 1, M 2, denoted M 1 R M 2. The existence of tensor product can be constructed as following: Let F be the free R-module generated by the set M 1 M 2. Let K be the submodule of F generated by R-bilinear relations, that is (a 1 + a 2, b) (a 1, b) (a 2, b), (a, b 1 + b 2 ) (a, b 1 ) (a, b 2 ), (a, rb) r(a, b), (ra, b) r(a, b). Then we have an induced bilinear map ϕ : M 1 M 2 F/K. We claim that (ϕ, F/K) is the universal object. To see this, note that for any R-bilinear map f : M 1 M 2 N, one easily produce a map h : F N by h (a, b) f(a, b). Since f is bilinear, one sees that h (x) = 0 if x K. Thus we have an induced map h : F/K N. Example Z 2 Z Z 3 = 0. Z 2 Z Z 2 = Z2. Proposition Let M 1, M 2, M 3 be R-modules. Then there exists a unique isomorphism (M 1 M 2 ) M 3 M 1 (M 2 M 3 ) such that (x y) z x (y z). Proposition Let M 1, M 2 be R-modules. There there exists a unique isomorphism M 1 M 2 M 2 M 1 such that x y y x. Proposition Let M R, M R be right R-modules and RN, R N be left R-modules. And let f : M M, g : N N be module homomorphisms. Then there is a unique group homomorphism f g : M R N M R N. Proof. Consider a middle linear map (f, g) : M N M R N by (a, b) f(a) g(b). By the universal property, we are done. There are some more properties:

12 12 Proposition ( n i=1m i ) N = n i=1(m 1 M). In fact, this also holds if the index set in infinite. Also we have Proposition M R R = M Proof. There is a natural map j : N R R N by j(x) = x 1. It s clear that this is an R-homomorphism. We then construct f : R N N by f(r, x) = rx. It s clear that this is middle linear and thus induces a group homomorphism f : R R N N by f(r x) = rx. It s also easy to see that this is a module homomorphism. Therefore, it suffices to check that fj = 1 N (which is clear) and j f = 1 R R N. This mainly due to ri x i = (1 r i x i ) = 1 r i x i. Combining these two, we have Proposition If F is free over R with basis {v i } i I. Then every element of M R F can be written as i I x i v i, with x i M and all but finitely many x i = 0. Moreover, Proposition If M, N are free over R with basis {v i }, {w i } respectively. Then M R N is free with basis {v i w j }. We now consider the base change. That is, if f : R S is a ring homomorphism. Then there are connection between S-modules and R-modules. First, if N is a S-module, then N can be viewed as an R-module by R N N such that (r, x) f(r)x. This operation is called restriction of scalars. For example, a vector space V over Q can be viewed as a Z-module. On the other hand, suppose now that we have M a R-module. S can be viewed as R-module. So we have M S := S R M, which is naturally a S-module. This operation is called base change. For example, let M = Z[x] be a Z-module and S = Q, then M S = Z[x] Z Q = Q[x]. Exercise Let S be a multiplicative set in R. Then we have ı : R S 1 R. Let M be an R-module, then S 1 M = S 1 R R M. Exercise

13 Show that S 1 (M R N) = S 1 M s 1 RS 1 N. In particular, we have (M N) p = Mp Rp N p. Proposition Let 0 M 1 M 2 M 3 0 be an exact sequence of R-modules. And N is an R-module. Then M 1 N M 2 N M 3 N 0 is exact. That is, tensor product ir right exact. Proof. For y N 3, y = g(z) for some z N 2, thus for x M, x y = (1 g)(x z). Hence im(1 g) generate M R N 3. It follows that 1 g is surjective. (1 g)(1 f)(x w) = x gf(w) = x 0 = 0. Therefore, im(1 f) ker(1 g). There is thus an induced map α : M R N 2 /im(1 f) M R N 3. It suffices to show that α is an isomorphism. To this end, we intend to construct the inverse map. Consider x y M R N 3, there is z N 2 such that g(z) = y. We define β 0 : M N 3 M R N 2 /im(1 f) by β 0 (x, y) = x z. We first check that this is well-defined. If z, z N 2 such that g(z) = g(z ) = y, then z z ker g = imf. Thus there is w N 1 such that z z = f(w). One verifies that x z = x (z + f(w)) = x z + x f(w) = x z + (1 f)(x w) = x z. It s routine to check that β 0 is middle linear, hence it induces β : M R N 3 M R N 2 /im(1 f). One can check that αβ(x y) = αx z = x g(z) = x y., βα(x z) = β(x g(z)) = x z. Another way to see it is via the relation with Hom functor. Lemma The sequence M 1 M 2 M 3 0 is exact if and only if 0 Hom(M 3, N) Hom(M 2, N) Hom(M 1, N) is exact for all N. Lemma There is a canonical isomorphism Hom(M N, P ) = Hom(M, Hom(N, P )). Proof of Prop Since M 1 M 2 M 3 0 is exact, we have for all P, 0 Hom(M 3, Hom(N, P )) Hom(M 2, Hom(N, P )) Hom(M 1, Hom(N, P )), is exact. Thus 0 Hom(M 3 N, P ) Hom(M 2 N, P ) Hom(M 1 N, P ), is exact for all P. And hence M 1 N M 2 N M 3 N 0 is exact. 13

14 14 This says that the functor R N is right exact. Similarly, one can see that the functor N R is also right exact. Definition A module is said to be flat if the functor R M is exact. For example, S 1 R is a flat R-module. We have the following easier criterion for flatness. Proposition The following are equivalent: 1. N is flat. 2. If M 1 M 2 is injective, then M 1 N M 2 N is injective. In fact, we can have Theorem M is flat if and only if M is exact with respect to 0 a R R/a 0 for all ideal a. We remark that (R/a) M = M/aM. To see this, note that there is a surjective map R M = M (R/a) M. The kernel of this map if given by the image of ϕ : a M R M = M. The image consists of {ϕ( r i x i ) r i a, x i M} = { r i x i } = am. Proof. We introduce the notion of N-flat if M is exact for any N such that 0 N N. Note that the condition can be rephrased as M is R-flat because every submodule of R is exactly an ideal of R. Step 1. If M is N-flat, then M is N-flat. Step 2. If M is N-flat, then for every submodule S and quotient Q of N, M is S-flat and Q-flat. Let S be a submodule of S. We have that S M N M is injective. This map factors through S M, hence S M S M is also injective and M is S-flat. Let Q be a submodule of Q and N be its preimage in N. We have 0 S N Q 0 0 S N Q 0 Tensoring with M, we get 0 K S M N M Q M 0 = 0 S M N M Q M

15 By Snake Lemma, we have K = 0. Thus M is Q-flat. Step 3. Since every module is quotient of free modules, i.e there is a surjection i I R N. So M is R-flat implies that M is i I R-flat by Step 1. And then by Step 2, M is N-flat. Thus M is flat and we are done. 15

16 16 Mar. 23, 2007 (Fri.) We can consider the following local properties Proposition The following are equivalent: 1. M = M p = 0 for all prime ideal p 3..M m = 0 for all maximal ideal m. One can think of this as a function is zero if its value at each point is zero. Proof. It suffices to show that 3 1. Suppose that M 0, let x 0 M. Then Ann(x) R. Thus Ann(x) m for some m. Now x = 0 M 1 m, which means that sx = 0 for some s (R m) Ann(x). This is a contradiction. Proposition Let ϕ : M N be an R-homomorphism. The following are equivalent: 1. ϕ : M N is injective. 2. ϕ p : M p N p is injective for all prime ideal p 3. ϕ m : M m N m is injective for all maximal ideal m. Proof. Since R p is exact, we have is trivial. It suffices to show that 3 1. Let K := ker(ϕ), then K m = ker(ϕ m ) = 0 since R m is exact. Thus we have K = 0 by Proposition Proposition The following are equivalent: 1. M is flat. 2. M p is a flat R p -module for all prime ideal p 3..M m is a flat R m -module for all maximal ideal m. Proof. For 1 2, if suffices to check that M p is R p -flat. Let b R p be an ideal, then b = ar p. Now ar p M p = (a M) p = (a M) R p M R p, is injective because M is flat and R p is flat. To see 3 1, for any a R, we consider am M. Localize it, we have (a R M) m = ar m Rm M m M m. This is injective by our assumption. By Proposition , we see that am M is injective chain conditions. In this section, we are going to survey some basic properties of Noetherian and Artinian modules. Let (Σ, ) be a partial ordered set, then the following two conditions are equivalent

17 (1) Every increasing sequence of x 1 x 2... in Σ is stationary, i.e. there exist n such that x n = x n+1 =... (2) Every non-empty subset of Σ has a maximal element. Definition Let M be an R-module. Let Σ be the set of submodules of M. We say that M is a Noetherian (resp. Artinian) R-module if the P.O. set (Σ, ) (resp. (Σ, )) satisfies the above condition. Remark In the case of (Σ, ) we say condition (1) is a.c.c. (ascending chain condition) and condition (2) is maximal condition. While in the case of (Σ, ) we say condition (1) is d.c.c. (descending chain condition) and condition (2) is minimal condition. Proposition Let 0 M M M 0 be an exact sequence of R-modules. Then M is Noetherian (resp. Artinian) if and only if both M, M are Noetherian (resp. Artinian). Proof. We leave it to the readers as an exercise. Corollary If M i is Noetherian (resp. Artinian), then so is n i=1m i. Proposition M is a Noetherian R-module if and only if every submodule of M is finitely generated. Proof. Let N < M be a submodule, let Σ be the set of finitely generated submodules of N. By the maximal condition, there is an maximal element N 0 Σ. We claim that N 0 = N then we are done. To see the claim, let s suppose on the contrary that N 0 N. Pick any x N N 0, then N 0 N 0 + Rx < N. And clearly, N 0 + Rx is finitely generated. This contradict to the maximality of N 0. Conversely, given an ascending chain M 1 < M 2 <... of submodules of M. Let N = M i. It s clear that N is a submodule of M. Thus N is finitely generated, say N = Rx Rx r. For each x i, x i M ji for some j i. Let n = max i=1,...,r {j i }. Then it s easy to see that M n = M n+1 =... and hence we are done. Definition A ring R is said to be Noetherian (resp. Artinian) if R is a Noetherian (resp. Artinian) R-module. Or equivalently, the ideals of R satisfies ascending (resp. descending) chain condition. Example A field if both Noetherian and Artinian. 2. The ring Z is Noetherian but not Artinian. 3. More generally, a PID is always Noetherian. To see this, suppose that we have a 1 a 2... an ascending chain of ideals. Let a := a i. Then a is an ideal, hence a = (x) for some x. Now x a n for some n, thus we have a a n. Example

18 18 Consider R = k[x 1, x 2,...] the polynomial ring of infinitely many indeterminate. There is an ascending chain of ideal (x 1 ) (x 1, x 2 ) (x 1, x 2, x 3 )... So R is not Noetherian. Let K be its quotient field, then clearly K is Noetherian. Thus a subring of a Noetherian ring is not necessarily Noetherian. However, Noetherian and Artinian properties are preserved by taking quotient. Proposition If R is Noetherian or Artinian, then so is R/a for any a R. Proof. Use the correspondence of ideals. readers. We leave the detail to the Indeed, by using the correspondence of ideals one can also show that if R is Noetherian (resp. Artinian) and then so is S 1 R. Proposition Let R be a Noetherian (resp. Artinian) ring, and M a finitely generated R-module. Then M is Noetherian (resp. Artinian). Proof. By Proposition and One important result is the following: Theorem (Hilbert s basis theorem). If R is Noetherian, then so is R[x]. Proof. Let b R[x] be an ideal. We need to show that it s finitely generated. Let a be the set of leading coefficients of b, it s easy to see that it s an ideal. Let a 1,..., a n be a set of generators. Then there are f i = a i x r i +... b. Let r = max{r i }. And let b = (f 1,..., f n ). For any f = ax r +... b of degree r m, a = c i a i, for some c i R. Thus f c i x r r i f i b has degree < r. Inductively, we get a polynomial g with degree < r and f = g + h with h b. Lastly, consider M = R + Rx Rx r 1 a finitely generated R-module. Then b M is a finitely generated R-module. So b = (b M) + b is clearly a finitely generated R[x]-module. An immediate consequence is Corollary If R is Noetherian, then so is R[x 1,..., x n ]. Also any finitely generated R-algebra is Noetherian. By almost the same argument, one can show that R[[x]] is Noetherian if R is Noetherian. We now turn into the consideration of Artinian rings.

19 Proposition Let R be an Artinian ring, then every prime ideal is maximal. Proof. Let p be a prime ideal, then B := R/p is an Artinian integral domain. For any x 0 B, we consider (x) (x 2 )... Since B is Artinian, we have (x n ) = (x n+1 ) for some n. In particular, x n = x n+1 y. Hence xy = 1, that is x is a unit. Thus B is a field. Proposition Let R be an Artinian ring, then R has only finitely many maximal ideals. Proof. Let Σ be the set of finite intersection of maximal ideals. There is a minimal element, say µ := m 1... m n. For any maximal ideal m, one sees that m µ = µ. Thus m µ m 1...m n. It follows that m m i, and hence m = m i for some i. Therefore, there are n maximal ideals. We now turn into the consideration of Artinian rings. Proposition Let R be an Artinian ring, then every prime ideal is maximal. Proof. Let p be a prime ideal, then B := R/p is an Artinian integral domain. For any x 0 B, we consider (x) (x 2 )... Since B is Artinian, we have (x n ) = (x n+1 ) for some n. In particular, x n = x n+1 y. Hence xy = 1, that is x is a unit. Thus B is a field. Proposition Let R be an Artinian ring, then R has only finitely many maximal ideals. Proof. Let Σ be the set of finite intersection of maximal ideals. There is a minimal element, say µ := m 1... m n. For any maximal ideal m, one sees that m µ = µ. Thus m µ m 1...m n. It follows that m m i, and hence m = m i for some i. Therefore, there are n maximal ideals. We close this section by comparing Artinian rings and Noetherian rings. Theorem R is Artinian if and only R is Noetherian and dim R = 0. Recall that dim R = sup{n p 0 p 1... p n SpecR} Proof. Step 1. If R is Artinian, then every prime is maximal. So dim R = 0. 19

20 20 Step 2. Moreover, R has only finitely many maximal ideals say m 1,..., m n. Then N = m i. We claim that N r = 0 for some r. To see this, notice that we have a descending chain N N 2... We claim that N r = 0 for some r. Since R is Artinian, we have N r = N r+1 =..., call it a. We would like to show that a = 0. If not, pick an minimal ideal b such that ab 0. There is x 0 b such that xa 0. So by minimality, (x) = b. Also, (xa)a = xa 0 and (x)a (x). So (x) fa = (x) by minimality as well. Thus there is y a so that x = xy. Then we have x = xy = xy 2 =... Since y N is nilpotent, we have x = 0. Step 3. We show that R is a Noetherian R-module. Consider ( m i ) r N r = 0, then we have a finite filtration 0... m i...m 1 R. And each quotient is Artinian R-module, hence also a Artinian R/m i -module. It follows that each quotient is also Noetherian R/m i -module and hence R-module, too. So R is also Noetherian. Step 4. If R is Noetherian, we have a primary decomposition of 0 = n i=1q i. Taking radical, we have N = n i=1p i. Since dim R = 0, so every prime is maximal. We see that p i are the only primes ideals. Step 5. Moreover, since N is finitely generated, we see that N r = 0 for some r. Then we proceed as above. We are done.

21 1.7. integral extension. In this section, we are going to explore more about the notion of integral extension. The goal is to show the going up and going down theorems. Let A, B be rings. We say B is an extension over A if A B. An element x B is said to be integral over A if it satisfies a monic polynomial in A[x]. B is integral over A if every element of B is integral over A. The following Properties are more or less parallel to the theory of algebraic extensions. Proofs are similar. The reader should find no difficulty working them out. Proposition Let A B be an extension. The followings are equivalent: (1) x B is integral over A. (2) A[x] is a finitely generated A-module. (3) A[x] is contained is a subring C B such that C is a finitely generated A-module. Corollary Let A B be an extension. If x i B is integral over A for i = 1,.., n. Then A[x 1,..., x n ] is a finitely generated A-module. Corollary Let A B C be extensions. If C is integral over B and B is integral over C, then C is integral over A. Corollary Let A B be an extension. The integral closure of A in B, which is the set of elements in B integral over A, is a ring (subring of B). Let A B be an extension. A is said to be integrally closed in B if the integral closure of A is A itself. Corollary Let A B be an extension. And let C be the integral closure of A in B. Then C is integrally closed in B. Example Consider Z Q Q( 1). The integral closure of Z in Q( 1) is Z[ 1]. Exercise Determine the integral closure of Z in Q( d) for an integer d. Proposition Let A B be an integral extension. (1) If b B and a := b A, then B/b is integral over A/a. (2) Let S be a multiplicative set of A (hence of B), then S 1 B is integral over S 1 A. Proof. (1) For b B/b, one notices that b B and b n + a n 1 b n a 0 = 0 for some a i A. It s clear that b n +a n 1 bn ā 0 = 0 B := B/b. And hence b is integral over Ā := A/a. 21

22 22 (2) For b s S 1 B. One notice that b n + a n 1 b n a 0 = 0 for some a i A. Hence ( b s )n + a n 1 s (b s )n a 0 s n = 0. And we are done. Lemma Let A B be an integral extension and B is an domain. Then A is a field if and only if B is a field. Proof. Suppose that A is a field. For any b 0 B, b n + a n 1 b n a 0 = 0 for some a i A. We may assume that a 0 0 A because B is a domain. Then b(b n 1 + a n 1 b n a 1 ) = a 0. Therefore, b is invertible in B and so B is a field. Conversely, let B be a field. For a 0 A, a 1 B. Thus (a 1 ) n + a n 1 (a 1 ) n a 0 = 0. In particular, a 1 = (a n a 0 a n 1 ) A. Let A B be an integral extension, and q SpecB, p SpecA. We say that q is lying over p if q A = p. We are going to study the relation between prime ideals of integral extension. Proposition Keep the notation as above with q is lying over p. Then q is maximal if and only if p is maximal. Proof. B/q is again integral over A/p. Since B/q is a field if and only if A/p is a field, we are done. An consequence is the following corollary which assert the uniqueness in a chain of prime ideal : Corollary Keep the notation as above. If q 1 q 2 are prime ideals lying over p. Then q 1 = q 2. Proof. Let S = A p. (Note that q i S = for i = 1, 2.) Then we have B p integral over A p. Moreover, q 1 B p q 2 B p. Note that q i B p A p = (q i A)A p = pa p is maximal for i = 1, 2. Hence both q 1 B p q 2 B p are maximal ideal lying over pa p. We then have q 1 B p = q 2 B p. By the correspondence of prime ideals, we have q 1 q 2. It s also desirable to have existence of prime ideal lying over a specific one. Proposition Let A B be an integral extension. Let p Spec(A). Then there exist a q Spec(B) lying over A.

23 Proof. Let S = A p. We consider A p B p. (This is injective since S 1 is exact.) Take a maximal ideal m of B p. We claim that q := m B is a prime ideal lying over p. To see this, q A = (m B) A = (m A p ) A = pa p A = p. This is because m lying over a maximal ideal of A p and the only maximal ideal of A p is pa p. Theorem (Going-up theorem). Let B be an integral extension over A. Let p 1 p 2 Spec(A) and q 1 Spec(B) lying over p 1. Then there is q 2 Spec(B) containing q 1 lying over p 2. Proof. Let Ā := A/p 1 and B := B/q 1. Then B is integral over Ā. There is a prime ideal q 2 lying over p 2. Lift to B, then we are done. As we have seen, let B be an integral extension over A. Then every chain of distinct prime ideals of B restricts to a chain of distinct prime ideals of A and conversely, every chain of distinct prime ideals of A extends to a chain of distinct prime ideals of B. It follows that dima = dimb. 23

24 24 Apr. 13, 2007 (Fri.) Before we going to dimension theory. We would like to investigate integral extension a little bit more which will be useful in Dedekind domain and DVRs. Integral extension is very similar to algebraic extension. Definition A domain A is said to be integrally closed if it is integrally closed in its quotient field. For the rest of this section, we are going to assume that B is a domain integral over A and A is integrally closed (i.e. in its quotient field K). We remark that this setting is closely related to algebraic field extensions. Lemma Let A B be an extension and C is the integral closure of A in B. Let a A be an ideal. We say that b B is integral over a if b satisfies a polynomial with coefficient (except the leading term) in a. Then the integral closure of a in B is ac. Proof. If b B is integral over a, then Since b C, we have b n + a n 1 b n a 0 = 0. b n = a n 1 b n 1... a 0 ac. Thus b ac. Conversely, let b ac. Then b n = r i=1 a ix i, with a i a, x i C. Then A[x 1,..., x r ] is a finite module over A. Since b n A[x 1,..., x r ] aa[x 1,..., x r ]. It follows that the multiplication by b n behave like a matrix on A[x 1,..., x r ] with entries in a. Hence b n satisfies the characteristic polynomial with coefficient in a. It follows that b is integral over a. Remark Keep the notation as above, let f(x) be an integral polynomial of b B and p(x) be its minimal polynomial over K. We remark that there is NO notion of minimal integral polynomial in general because the division algorithm doesn t holds in A. However, if A is UFD, then by Gauss lemma, we have p(x) f(x) not only in K[x] but also in A[x]. Hence the minimal polynomial is integral. Lemma Keep the notation as above, the minimal polynomial of b B is in A[x]. If b B is integral over a A, i.e. the integral polynomial has coefficients in a, then the minimal polynomial of b B is in a[x]. Proof. Assume now b is integral over a with minimal polynomial p(x). Take a splitting field of p(x), say L/K. And let b = b 1,..., b n be conjugates of b. Then they satisfy f(x) as well. Thus b i is integral over

25 a. It s not difficult to see that p(x) = n i=1 (x b i) m i for some m i 0. The coefficient are combination of b i hence integral over a. Apply the above Lemma to the extension A K, we have integral closure of a is a. Hence minimal polynomial is in a[x]. Let A be an integrally closed domain with quotient field K. Given an extension L/K, one can consider B to be the integral closure of A in L. (We may assume that L is algebraic over K, or even to be the splitting field of B.) Especially, in number theory, we usually consider a number field which is a finite extension over Q. And let O be the domain of algebraic integers. The extension Z O justify our setup. Proposition Let A be an integrally closed domain with quotient field K. Given a normal extension L/K, one can consider B to be the integral closure of A in L. Let p SpecA, then prime ideals in B lying over p are conjugate. That is, for q 1, q 2 SpecB lying over p, there is σ Aut K L such that σ(q 1 ) = q 2. Proof. We will only prove this under the assumption that B is finitely generated over A. (Then we may assume that [L : K] is finite). If q 2 σ j (q 1 ) for all j then q 2 σ j (q 1 ) for all j. claim. there is x q 2 such that x σ j (q 1 ) for all j. We leave this claim as an exercise. Let y = σ j Aut K L σ j(x). Then y is invariant under Aut K L. We may assume that y pl is separable over K for some l 0. (If char(k)=0, then l = 0.) Let S be the separable closure of K in L. Then S is Galois over K with Aut K S = Aut K L (cf. Thm. 12 of lecture 1). Hence y pl K. One notice that B K = A. It follows that y pl A. And then y K is integral over A. Therefore, y A. We Clearly, y = x y q x 2. Hence y A q 2 = p q 1. Hence σ j (x) q 1 for some j. This leads to a contradiction. Corollary Keep the notation as above. Assume furthermore that B is finitely generated over A. Then for p SpecA, there are only finitely many prime ideal in B lying over p. And any two of them are conjugate to each other. Proof. Let L be the splitting field of B over K. One sees that L/K is finite. Let C be the integral closure of A in L. Clearly, A B C. We know that prime ideal in C lying over p is conjugate to each other. Since the Galois group is finite. There are only finitely many prime ideals. Restrict to B, then there are only finitely many prime ideals in B lying over p. Theorem (Going down theorem). Keep the notation as above, i.e. B is an domain integral over an integrally closed domain A. If there are p 2 p 1 SpecA and q 1 in B lying over p 1. Then there is q 2 q 1 SpecB lying over p 2 25

26 26 Proof. Let L be the normal closure of B over K and C be the integral closure of A in L. It s clear that C is integral over B. There is a prime ideal r 1 in C lying over q 1. And there is a prime ideal r 2 in C lying over p 2. By Going up theorem, there is a prime ideal r 1 r 2 in C lying over p 1. Therefore, there is σ Aut K L such that σ(r 1 ) = r 1. Then σ 1 (r 2 ) r 1. Let q 2 := σ 1 (r 2 ) B. Then we are done. We close this section by recall Theorem (Noether normalization theorem). Let R be a finitely generated domain over a field k. Let r be the transcendental degree of R over k. Then there exists y 1,..., y r in R, algebraically independent over k, such that R is integral over k[y 1,..., y r ]. Proof. Let F be the field of quotients of R. Then r is defined to be the transcendental degree of F over k. Let first consider the case r = 1 and see what could be the problem. If r = 1, then we can pick y R which is transcendental over k. The hope is to show that x R is integral over k[y]. However, all we know so far is x n + a n 1 x n a 0 = 0 for some a i = q i(y) p i k(y). Eliminate the denominators, we get (y) f n (y)x n f 0 (y) = 0, ( ) which is not enough to show that x is integral over k[y]. A possible way out is to consider z := y x m for some m 0, e.g. m > n. Then the equation is now cx md+i + g(x, z) = 0, with c k, d = max{deg y (f i (y))} and deg x g(x, z) (d 1)m + i < dm. Hence x is integral over k[z]. More generally, if x 1,..., x t is a set of generator of R over k and r = 1. We can pick m 0 so that it normalize x i simultaneously. Finally, by induction on r. We are able to prove the theorem. We leave the induction step to the readers.

27 2. basic algebraic geometry 2.1. affine varieties. The main object in algebraic geometry is algebraic variety. One can consider it as zero locus of a set of polynomial, roughly. To set it up, let s first fix an algebraically closed field k. The affine n-space over k, denoted A n k, is the set of all n-tuples. To study A n, the polynomial ring A := k[x 1,.., x n ] is a convenient tool. They are closely connected via the following operation: (1) Given a set of polynomials T, one can define V(T ), the common zero locus of T. We call such V(T ) an algebraic set. (2) Given a subset Y of affine space, one can define I(Y ) which consists of polynomials vanish along Y. It s immediate that I(Y ) is an ideal. We remark that V(T ) = V(< T >), where < T > denotes the ideal generated by T. These two operation give connection between ideals and algebraic sets. It s not difficult to see that one can define a topology on A n with algebraic sets as closed sets. This topology is called the Zariski topology. Can one also construct a topology on the algebraic side? The answer is yes, with some extra care. To see these, one can verify the following: Proposition Keep the notation as above. We have the following: (1) V(0) = A n. (2) V(A) =. (3) V(I 1 ) V(I 2 ) = V(I 1 I 2 ). (4) V(I α ) = V( I α ). One notices that different ideals might give the same algebraic set, for example, the ideal (x) and (x 3 ) do. Among all ideals defining the same algebraic set, there is a maximal one, the radical ideal. So we have an one-to-one correspondence between radical ideals and algebraic sets. Moreover, we have the following Theorem (Hilbert s Nullstellensatz, weak form). Every maximal ideal of A = k[x 1,..., x n ] is of the form (x 1 a 1,..., x n a n ). Proof. Let m be a maximal ideal of A, we consider R := A/m. R is clearly a finitely generated k-algebra. By Noether normalization theorem, there exists y 1,..., y r R such that R is integral over k[y 1,..., y r ]. Since R is a field, so is k[y 1,..., y r ]. This is possible only when r = 0. So R is integral over k, hence algebraic over k. But k is algebraically closed. So R = k. Theorem (Hilbert s Nullstellensatz). Let k be an algebraically closed field and A = k[x 1,..., x n ] be the polynomial ring. Let a be an 27

28 28 ideal in A, then I(Z(a)) = a. Corollary There is an one-to-one correspondence between algebraic sets and radical ideals. Furthermore, the algebraic set is irreducible (resp. a point) if and only if its ideal is prime (resp. maximal). Definition An algebraic set X is irreducible it if can t be written as union of two algebraic set in a non-trivial way. More precisely, if X = X 1 X 2 with X i being algebraic sets, then either X = X 1 or X = X 2. An affine variety is an irreducible algebraic set in A n. An open subset of an affine variety is a quasi-affine variety. Since A is Noetherian, it s easy to see that every algebraic set can be written as finite union of affine varieties. (This is basically what primary decomposition does). For an algebraic set X, it has the induced Zariski topology. It s easy to see that it has descending chain condition for closed subsets, i.e. for any sequence Y 1 Y 2... of closed subsets, there is an integer r such that Y r = Y r+1 =... A topological space is called Noetherian if ithas d.c.c for closed subsets. With the correspondence in mind, we can define the concept of dimension geometrically and algebraically. Definition For a Noetherian topological space X, the dimension of X, denoted dimx, is defined to be the supremum (=maximum) of the length of chain of closed subvarieties. For an affine variety XinA n, the polynomial functions A restrict to X is nothing but the homomorphism π : A A/I(X). The ring A/I(X) is called the coordinate ring of X, denoted A(X). One can recover the geometry of X from A(X) by considering Spec(A(X)), which consist of prime ideals in A(X). One can give the Zariski topology on Spec(A(X)) which is closely related to the Zariski topology on X. This is actually the construction of affine scheme. And affine variety can be viewed as a nice affine scheme. Exercise The coordinate ring of an affine variety is a domain and a finitely generated k-algebra. Conversely, a domain which is a finitely generated k-algebra is a coordinate ring of an affine variety. One can also similarly define the Krull dimension or simply dimension to be the supremum of length of chain of prime ideals of a ring. It s easy to see that for an algebraic set X, then dimx = dima(x). However, it s not trivial to prove that dima n = n.

29 2.2. Dimension theory. In this section, we are going to explore dimension theory a little bit. Let first recall the definition. Definition Let R be a ring and p Spec(R). We define And for an ideal I R, we define We define ht(p) := sup{n p n... p 0 = p}. ht(i) := inf{ht(p) I p}. dimr := sup{ht(p) p Spec(R)}. Example Let R := k[x 1,..., x n ] with k algebraically closed. We will see later that dimr is n. Let p be a prime ideal, then V(p) defines a variety in A n k. Then ht(p) is nothing but the codimension of V(p) in A n k because there is a one-to-one correspondence between prime ideals and subvarieties. Let I k[x 1,..., x n ] be an ideal, then V(I) is not necessarily irreducible. We usually define the dimension of V(I) to be the dimension of irreducible component of maximal dimension. That is, we are looking for dimv(i) := max{dimy Y is an irreducible component in V(I)}. Hence the codimension of V(I) corresponds to inf{ht(p) I p}. One has the following property immediately by the correspondence we ve been built up. Proposition (1) htp = dimr p. (2) dimr/i + ht(i) dimr. Theorem Let R be a finitely generated domain over a field k. Then tr.d. k R = dimr. (where tr.d. k R := tr.d. k F, where F is the quotient field of R.) Proof. We first claim that dimr tr.d. k R. To see this, it suffices to show that if p q Spec(R), then tr.d. k R/p tr.d. k R/q. Let {β 1,..., β r } be a transcendental basis of R/q. Then it lifts to {α 1,..., α r } which is algebraically independent in R/p. This is because there is a surjective homomorphism ϕ : R/p R/q. If there is an algebraic relation among {α 1,..., α r } then it gives an relation among {β 1,..., β r } via the homomorphism ϕ, which is absurd. Hence we have shown that tr.d. k R/p tr.d. k R/q. Assume now that tr.d. k R/p = tr.d. k R/q. Then {α 1,..., α r } is an basis. Lift to R, we have an algebraically independent set {y 1,..., y r } 29