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

Size: px
Start display at page:

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

Transcription

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

### Reid 5.2. Describe the irreducible components of V (J) for J = (y 2 x 4, x 2 2x 3 x 2 y + 2xy + y 2 y) in k[x, y, z]. Here k is algebraically closed.

Reid 5.2. Describe the irreducible components of V (J) for J = (y 2 x 4, x 2 2x 3 x 2 y + 2xy + y 2 y) in k[x, y, z]. Here k is algebraically closed. Answer: Note that the first generator factors as (y

### Rings and groups. Ya. Sysak

Rings and groups. Ya. Sysak 1 Noetherian rings Let R be a ring. A (right) R -module M is called noetherian if it satisfies the maximum condition for its submodules. In other words, if M 1... M i M i+1...

### ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ.

ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ. ANDREW SALCH 1. Hilbert s Nullstellensatz. The last lecture left off with the claim that, if J k[x 1,..., x n ] is an ideal, then

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

MATH 22 NOTES BRENT HO Date: January 3, 2009. 0 Table of Contents. Localizations......................................................................... 2 2. Zariski Topology......................................................................

### NOTES FOR COMMUTATIVE ALGEBRA M5P55

NOTES FOR COMMUTATIVE ALGEBRA M5P55 AMBRUS PÁL 1. Rings and ideals Definition 1.1. A quintuple (A, +,, 0, 1) is a commutative ring with identity, if A is a set, equipped with two binary operations; addition

### Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35

Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35 1. Let R be a commutative ring with 1 0. (a) Prove that the nilradical of R is equal to the intersection of the prime

### ADVANCED COMMUTATIVE ALGEBRA: PROBLEM SETS

ADVANCED COMMUTATIVE ALGEBRA: PROBLEM SETS UZI VISHNE The 11 problem sets below were composed by Michael Schein, according to his course. Take into account that we are covering slightly different material.

### ALGEBRA EXERCISES, PhD EXAMINATION LEVEL

ALGEBRA EXERCISES, PhD EXAMINATION LEVEL 1. Suppose that G is a finite group. (a) Prove that if G is nilpotent, and H is any proper subgroup, then H is a proper subgroup of its normalizer. (b) Use (a)

### Homological Methods in Commutative Algebra

Homological Methods in Commutative Algebra Olivier Haution Ludwig-Maximilians-Universität München Sommersemester 2017 1 Contents Chapter 1. Associated primes 3 1. Support of a module 3 2. Associated primes

### Assigned homework problems S. L. Kleiman, fall 2008

18.705 Assigned homework problems S. L. Kleiman, fall 2008 Problem Set 1. Due 9/11 Problem R 1.5 Let ϕ: A B be a ring homomorphism. Prove that ϕ 1 takes prime ideals P of B to prime ideals of A. Prove

### 11. Dimension. 96 Andreas Gathmann

96 Andreas Gathmann 11. Dimension We have already met several situations in this course in which it seemed to be desirable to have a notion of dimension (of a variety, or more generally of a ring): for

### 4.4 Noetherian Rings

4.4 Noetherian Rings Recall that a ring A is Noetherian if it satisfies the following three equivalent conditions: (1) Every nonempty set of ideals of A has a maximal element (the maximal condition); (2)

### Injective Modules and Matlis Duality

Appendix A Injective Modules and Matlis Duality Notes on 24 Hours of Local Cohomology William D. Taylor We take R to be a commutative ring, and will discuss the theory of injective R-modules. The following

### n P say, then (X A Y ) P

COMMUTATIVE ALGEBRA 35 7.2. The Picard group of a ring. Definition. A line bundle over a ring A is a finitely generated projective A-module such that the rank function Spec A N is constant with value 1.

### The most important result in this section is undoubtedly the following theorem.

28 COMMUTATIVE ALGEBRA 6.4. Examples of Noetherian rings. So far the only rings we can easily prove are Noetherian are principal ideal domains, like Z and k[x], or finite. Our goal now is to develop theorems

### Course 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra

Course 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra D. R. Wilkins Contents 3 Topics in Commutative Algebra 2 3.1 Rings and Fields......................... 2 3.2 Ideals...............................

### CHAPTER 1. AFFINE ALGEBRAIC VARIETIES

CHAPTER 1. AFFINE ALGEBRAIC VARIETIES During this first part of the course, we will establish a correspondence between various geometric notions and algebraic ones. Some references for this part of the

### 5 Dedekind extensions

18.785 Number theory I Fall 2016 Lecture #5 09/22/2016 5 Dedekind extensions In this lecture we prove that the integral closure of a Dedekind domain in a finite extension of its fraction field is also

### Extended Index. 89f depth (of a prime ideal) 121f Artin-Rees Lemma. 107f descending chain condition 74f Artinian module

Extended Index cokernel 19f for Atiyah and MacDonald's Introduction to Commutative Algebra colon operator 8f Key: comaximal ideals 7f - listings ending in f give the page where the term is defined commutative

### CRing Project, Chapter 5

Contents 5 Noetherian rings and modules 3 1 Basics........................................... 3 1.1 The noetherian condition............................ 3 1.2 Stability properties................................

### Commutative Algebra. Contents. B Totaro. Michaelmas Basics Rings & homomorphisms Modules Prime & maximal ideals...

Commutative Algebra B Totaro Michaelmas 2011 Contents 1 Basics 4 1.1 Rings & homomorphisms.............................. 4 1.2 Modules........................................ 6 1.3 Prime & maximal ideals...............................

### 5 Dedekind extensions

18.785 Number theory I Fall 2017 Lecture #5 09/20/2017 5 Dedekind extensions In this lecture we prove that the integral closure of a Dedekind domain in a finite extension of its fraction field is also

### Structure of rings. Chapter Algebras

Chapter 5 Structure of rings 5.1 Algebras It is time to introduce the notion of an algebra over a commutative ring. So let R be a commutative ring. An R-algebra is a ring A (unital as always) together

Algebra Qualifying Exam August 2001 Do all 5 problems. 1. Let G be afinite group of order 504 = 23 32 7. a. Show that G cannot be isomorphic to a subgroup of the alternating group Alt 7. (5 points) b.

### Dedekind Domains. Mathematics 601

Dedekind Domains Mathematics 601 In this note we prove several facts about Dedekind domains that we will use in the course of proving the Riemann-Roch theorem. The main theorem shows that if K/F is a finite

### INJECTIVE MODULES: PREPARATORY MATERIAL FOR THE SNOWBIRD SUMMER SCHOOL ON COMMUTATIVE ALGEBRA

INJECTIVE MODULES: PREPARATORY MATERIAL FOR THE SNOWBIRD SUMMER SCHOOL ON COMMUTATIVE ALGEBRA These notes are intended to give the reader an idea what injective modules are, where they show up, and, to

### Commutative Algebra. B Totaro. Michaelmas Basics Rings & homomorphisms Modules Prime & maximal ideals...

Commutative Algebra B Totaro Michaelmas 2011 Contents 1 Basics 2 1.1 Rings & homomorphisms................... 2 1.2 Modules............................. 4 1.3 Prime & maximal ideals....................

### COURSE SUMMARY FOR MATH 504, FALL QUARTER : MODERN ALGEBRA

COURSE SUMMARY FOR MATH 504, FALL QUARTER 2017-8: MODERN ALGEBRA JAROD ALPER Week 1, Sept 27, 29: Introduction to Groups Lecture 1: Introduction to groups. Defined a group and discussed basic properties

### Algebra Homework, Edition 2 9 September 2010

Algebra Homework, Edition 2 9 September 2010 Problem 6. (1) Let I and J be ideals of a commutative ring R with I + J = R. Prove that IJ = I J. (2) Let I, J, and K be ideals of a principal ideal domain.

### Dimension Theory. Mathematics 683, Fall 2013

Dimension Theory Mathematics 683, Fall 2013 In this note we prove some of the standard results of commutative ring theory that lead up to proofs of the main theorem of dimension theory and of the Nullstellensatz.

### Summer Algebraic Geometry Seminar

Summer Algebraic Geometry Seminar Lectures by Bart Snapp About This Document These lectures are based on Chapters 1 and 2 of An Invitation to Algebraic Geometry by Karen Smith et al. 1 Affine Varieties

### Math 418 Algebraic Geometry Notes

Math 418 Algebraic Geometry Notes 1 Affine Schemes Let R be a commutative ring with 1. Definition 1.1. The prime spectrum of R, denoted Spec(R), is the set of prime ideals of the ring R. Spec(R) = {P R

### Integral Extensions. Chapter Integral Elements Definitions and Comments Lemma

Chapter 2 Integral Extensions 2.1 Integral Elements 2.1.1 Definitions and Comments Let R be a subring of the ring S, and let α S. We say that α is integral over R if α isarootofamonic polynomial with coefficients

### Math 120 HW 9 Solutions

Math 120 HW 9 Solutions June 8, 2018 Question 1 Write down a ring homomorphism (no proof required) f from R = Z[ 11] = {a + b 11 a, b Z} to S = Z/35Z. The main difficulty is to find an element x Z/35Z

### Commutative Algebra and Algebraic Geometry. Robert Friedman

Commutative Algebra and Algebraic Geometry Robert Friedman August 1, 2006 2 Disclaimer: These are rough notes for a course on commutative algebra and algebraic geometry. I would appreciate all suggestions

### MATH 8253 ALGEBRAIC GEOMETRY WEEK 12

MATH 8253 ALGEBRAIC GEOMETRY WEEK 2 CİHAN BAHRAN 3.2.. Let Y be a Noetherian scheme. Show that any Y -scheme X of finite type is Noetherian. Moreover, if Y is of finite dimension, then so is X. Write f

### (1) A frac = b : a, b A, b 0. We can define addition and multiplication of fractions as we normally would. a b + c d

The Algebraic Method 0.1. Integral Domains. Emmy Noether and others quickly realized that the classical algebraic number theory of Dedekind could be abstracted completely. In particular, rings of integers

### Exploring the Exotic Setting for Algebraic Geometry

Exploring the Exotic Setting for Algebraic Geometry Victor I. Piercey University of Arizona Integration Workshop Project August 6-10, 2010 1 Introduction In this project, we will describe the basic topology

### Math 210B. Artin Rees and completions

Math 210B. Artin Rees and completions 1. Definitions and an example Let A be a ring, I an ideal, and M an A-module. In class we defined the I-adic completion of M to be M = lim M/I n M. We will soon show

### CRing Project, Chapter 7

Contents 7 Integrality and valuation rings 3 1 Integrality......................................... 3 1.1 Fundamentals................................... 3 1.2 Le sorite for integral extensions.........................

### ALGEBRAIC GROUPS. Disclaimer: There are millions of errors in these notes!

ALGEBRAIC GROUPS Disclaimer: There are millions of errors in these notes! 1. Some algebraic geometry The subject of algebraic groups depends on the interaction between algebraic geometry and group theory.

### Yuriy Drozd. Intriduction to Algebraic Geometry. Kaiserslautern 1998/99

Yuriy Drozd Intriduction to Algebraic Geometry Kaiserslautern 1998/99 CHAPTER 1 Affine Varieties 1.1. Ideals and varieties. Hilbert s Basis Theorem Let K be an algebraically closed field. We denote by

### Introduction to Arithmetic Geometry Fall 2013 Lecture #18 11/07/2013

18.782 Introduction to Arithmetic Geometry Fall 2013 Lecture #18 11/07/2013 As usual, all the rings we consider are commutative rings with an identity element. 18.1 Regular local rings Consider a local

### Bachelorarbeit. Zur Erlangung des akademischen Grades. Bachelor of Science. Dimension theory for commutative rings. Universität Regensburg

Bachelorarbeit Zur Erlangung des akademischen Grades Bachelor of Science Dimension theory for commutative rings Eingereicht von: Eingereicht bei: Daniel Heiß Prof. Dr. Niko Naumann Universität Regensburg

### 10. Noether Normalization and Hilbert s Nullstellensatz

10. Noether Normalization and Hilbert s Nullstellensatz 91 10. Noether Normalization and Hilbert s Nullstellensatz In the last chapter we have gained much understanding for integral and finite ring extensions.

### CHAPTER I. Rings. Definition A ring R is a set with two binary operations, addition + and

CHAPTER I Rings 1.1 Definitions and Examples Definition 1.1.1. A ring R is a set with two binary operations, addition + and multiplication satisfying the following conditions for all a, b, c in R : (i)

### Homework 2 - Math 603 Fall 05 Solutions

Homework 2 - Math 603 Fall 05 Solutions 1. (a): In the notation of Atiyah-Macdonald, Prop. 5.17, we have B n j=1 Av j. Since A is Noetherian, this implies that B is f.g. as an A-module. (b): By Noether

### Commutative Algebra. Andreas Gathmann. Class Notes TU Kaiserslautern 2013/14

Commutative Algebra Andreas Gathmann Class Notes TU Kaiserslautern 2013/14 Contents 0. Introduction......................... 3 1. Ideals........................... 9 2. Prime and Maximal Ideals.....................

### FILTERED RINGS AND MODULES. GRADINGS AND COMPLETIONS.

FILTERED RINGS AND MODULES. GRADINGS AND COMPLETIONS. Let A be a ring, for simplicity assumed commutative. A filtering, or filtration, of an A module M means a descending sequence of submodules M = M 0

### ALGEBRA QUALIFYING EXAM SPRING 2012

ALGEBRA QUALIFYING EXAM SPRING 2012 Work all of the problems. Justify the statements in your solutions by reference to specific results, as appropriate. Partial credit is awarded for partial solutions.

### Lecture 2. (1) Every P L A (M) has a maximal element, (2) Every ascending chain of submodules stabilizes (ACC).

Lecture 2 1. Noetherian and Artinian rings and modules Let A be a commutative ring with identity, A M a module, and φ : M N an A-linear map. Then ker φ = {m M : φ(m) = 0} is a submodule of M and im φ is

### SUMMARY ALGEBRA I LOUIS-PHILIPPE THIBAULT

SUMMARY ALGEBRA I LOUIS-PHILIPPE THIBAULT Contents 1. Group Theory 1 1.1. Basic Notions 1 1.2. Isomorphism Theorems 2 1.3. Jordan- Holder Theorem 2 1.4. Symmetric Group 3 1.5. Group action on Sets 3 1.6.

### Primary Decomposition and Associated Primes

Chapter 1 Primary Decomposition and Associated Primes 1.1 Primary Submodules and Ideals 1.1.1 Definitions and Comments If N is a submodule of the R-module M, and a R, let λ a : M/N M/N be multiplication

### Algebraic Varieties. Notes by Mateusz Micha lek for the lecture on April 17, 2018, in the IMPRS Ringvorlesung Introduction to Nonlinear Algebra

Algebraic Varieties Notes by Mateusz Micha lek for the lecture on April 17, 2018, in the IMPRS Ringvorlesung Introduction to Nonlinear Algebra Algebraic varieties represent solutions of a system of polynomial

### 7 Orders in Dedekind domains, primes in Galois extensions

18.785 Number theory I Lecture #7 Fall 2015 10/01/2015 7 Orders in Dedekind domains, primes in Galois extensions 7.1 Orders in Dedekind domains Let S/R be an extension of rings. The conductor c of R (in

### HARTSHORNE EXERCISES

HARTSHORNE EXERCISES J. WARNER Hartshorne, Exercise I.5.6. Blowing Up Curve Singularities (a) Let Y be the cusp x 3 = y 2 + x 4 + y 4 or the node xy = x 6 + y 6. Show that the curve Ỹ obtained by blowing

### 10. Smooth Varieties. 82 Andreas Gathmann

82 Andreas Gathmann 10. Smooth Varieties Let a be a point on a variety X. In the last chapter we have introduced the tangent cone C a X as a way to study X locally around a (see Construction 9.20). It

### RINGS: SUMMARY OF MATERIAL

RINGS: SUMMARY OF MATERIAL BRIAN OSSERMAN This is a summary of terms used and main results proved in the subject of rings, from Chapters 11-13 of Artin. Definitions not included here may be considered

### SCHEMES. David Harari. Tsinghua, February-March 2005

SCHEMES David Harari Tsinghua, February-March 2005 Contents 1. Basic notions on schemes 2 1.1. First definitions and examples.................. 2 1.2. Morphisms of schemes : first properties.............

### 4.2 Chain Conditions

4.2 Chain Conditions Imposing chain conditions on the or on the poset of submodules of a module, poset of ideals of a ring, makes a module or ring more tractable and facilitates the proofs of deep theorems.

### ALGEBRA HW 4. M 0 is an exact sequence of R-modules, then M is Noetherian if and only if M and M are.

ALGEBRA HW 4 CLAY SHONKWILER (a): Show that if 0 M f M g M 0 is an exact sequence of R-modules, then M is Noetherian if and only if M and M are. Proof. ( ) Suppose M is Noetherian. Then M injects into

### COMMUNICATIONS IN ALGEBRA, 15(3), (1987) A NOTE ON PRIME IDEALS WHICH TEST INJECTIVITY. John A. Beachy and William D.

COMMUNICATIONS IN ALGEBRA, 15(3), 471 478 (1987) A NOTE ON PRIME IDEALS WHICH TEST INJECTIVITY John A. Beachy and William D. Weakley Department of Mathematical Sciences Northern Illinois University DeKalb,

### Math 145. Codimension

Math 145. Codimension 1. Main result and some interesting examples In class we have seen that the dimension theory of an affine variety (irreducible!) is linked to the structure of the function field in

### 2. Prime and Maximal Ideals

18 Andreas Gathmann 2. Prime and Maximal Ideals There are two special kinds of ideals that are of particular importance, both algebraically and geometrically: the so-called prime and maximal ideals. Let

### Topics in Module Theory

Chapter 7 Topics in Module Theory This chapter will be concerned with collecting a number of results and constructions concerning modules over (primarily) noncommutative rings that will be needed to study

### Ring Theory Problems. A σ

Ring Theory Problems 1. Given the commutative diagram α A σ B β A σ B show that α: ker σ ker σ and that β : coker σ coker σ. Here coker σ = B/σ(A). 2. Let K be a field, let V be an infinite dimensional

### MATH 326: RINGS AND MODULES STEFAN GILLE

MATH 326: RINGS AND MODULES STEFAN GILLE 1 2 STEFAN GILLE 1. Rings We recall first the definition of a group. 1.1. Definition. Let G be a non empty set. The set G is called a group if there is a map called

### COMMUTATIVE ALGEBRA, LECTURE NOTES

COMMUTATIVE ALGEBRA, LECTURE NOTES P. SOSNA Contents 1. Very brief introduction 2 2. Rings and Ideals 2 3. Modules 10 3.1. Tensor product of modules 15 3.2. Flatness 18 3.3. Algebras 21 4. Localisation

### CRing Project, Chapter 16

Contents 16 Homological theory of local rings 3 1 Depth........................................... 3 1.1 Depth over local rings.............................. 3 1.2 Regular sequences................................

### Tensor Product of modules. MA499 Project II

Tensor Product of modules A Project Report Submitted for the Course MA499 Project II by Subhash Atal (Roll No. 07012321) to the DEPARTMENT OF MATHEMATICS INDIAN INSTITUTE OF TECHNOLOGY GUWAHATI GUWAHATI

### Introduction Non-uniqueness of factorization in A[x]... 66

Abstract In this work, we study the factorization in A[x], where A is an Artinian local principal ideal ring (briefly SPIR), whose maximal ideal, (t), has nilpotency h: this is not a Unique Factorization

### Algebra Exam Syllabus

Algebra Exam Syllabus The Algebra comprehensive exam covers four broad areas of algebra: (1) Groups; (2) Rings; (3) Modules; and (4) Linear Algebra. These topics are all covered in the first semester graduate

### Math 121 Homework 5: Notes on Selected Problems

Math 121 Homework 5: Notes on Selected Problems 12.1.2. Let M be a module over the integral domain R. (a) Assume that M has rank n and that x 1,..., x n is any maximal set of linearly independent elements

### Honors Algebra 4, MATH 371 Winter 2010 Assignment 3 Due Friday, February 5 at 08:35

Honors Algebra 4, MATH 371 Winter 2010 Assignment 3 Due Friday, February 5 at 08:35 1. Let R 0 be a commutative ring with 1 and let S R be the subset of nonzero elements which are not zero divisors. (a)

### Q N id β. 2. Let I and J be ideals in a commutative ring A. Give a simple description of

Additional Problems 1. Let A be a commutative ring and let 0 M α N β P 0 be a short exact sequence of A-modules. Let Q be an A-module. i) Show that the naturally induced sequence is exact, but that 0 Hom(P,

### 11. Finitely-generated modules

11. Finitely-generated modules 11.1 Free modules 11.2 Finitely-generated modules over domains 11.3 PIDs are UFDs 11.4 Structure theorem, again 11.5 Recovering the earlier structure theorem 11.6 Submodules

### REPRESENTATION THEORY WEEK 9

REPRESENTATION THEORY WEEK 9 1. Jordan-Hölder theorem and indecomposable modules Let M be a module satisfying ascending and descending chain conditions (ACC and DCC). In other words every increasing sequence

### Algebraic Number Theory

TIFR VSRP Programme Project Report Algebraic Number Theory Milind Hegde Under the guidance of Prof. Sandeep Varma July 4, 2015 A C K N O W L E D G M E N T S I would like to express my thanks to TIFR for

### HILBERT FUNCTIONS. 1. Introduction

HILBERT FUCTIOS JORDA SCHETTLER 1. Introduction A Hilbert function (so far as we will discuss) is a map from the nonnegative integers to themselves which records the lengths of composition series of each

### Algebraic Varieties. Chapter Algebraic Varieties

Chapter 12 Algebraic Varieties 12.1 Algebraic Varieties Let K be a field, n 1 a natural number, and let f 1,..., f m K[X 1,..., X n ] be polynomials with coefficients in K. Then V = {(a 1,..., a n ) :

### MAS 6396 Algebraic Curves Spring Semester 2016 Notes based on Algebraic Curves by Fulton. Timothy J. Ford April 4, 2016

MAS 6396 Algebraic Curves Spring Semester 2016 Notes based on Algebraic Curves by Fulton Timothy J. Ford April 4, 2016 FLORIDA ATLANTIC UNIVERSITY, BOCA RATON, FLORIDA 33431 E-mail address: ford@fau.edu

### 214A HOMEWORK KIM, SUNGJIN

214A HOMEWORK KIM, SUNGJIN 1.1 Let A = k[[t ]] be the ring of formal power series with coefficients in a field k. Determine SpecA. Proof. We begin with a claim that A = { a i T i A : a i k, and a 0 k }.

### Polynomial Rings. i=0. i=0. n+m. i=0. k=0

Polynomial Rings 1. Definitions and Basic Properties For convenience, the ring will always be a commutative ring with identity. Basic Properties The polynomial ring R[x] in the indeterminate x with coefficients

### NOTES ON FINITE FIELDS

NOTES ON FINITE FIELDS AARON LANDESMAN CONTENTS 1. Introduction to finite fields 2 2. Definition and constructions of fields 3 2.1. The definition of a field 3 2.2. Constructing field extensions by adjoining

### Spring 2016, lecture notes by Maksym Fedorchuk 51

Spring 2016, lecture notes by Maksym Fedorchuk 51 10.2. Problem Set 2 Solution Problem. Prove the following statements. (1) The nilradical of a ring R is the intersection of all prime ideals of R. (2)

### 12. Hilbert Polynomials and Bézout s Theorem

12. Hilbert Polynomials and Bézout s Theorem 95 12. Hilbert Polynomials and Bézout s Theorem After our study of smooth cubic surfaces in the last chapter, let us now come back to the general theory of

### 8. Prime Factorization and Primary Decompositions

70 Andreas Gathmann 8. Prime Factorization and Primary Decompositions 13 When it comes to actual computations, Euclidean domains (or more generally principal ideal domains) are probably the nicest rings

### Algebra Prelim Notes

Algebra Prelim Notes Eric Staron Summer 2007 1 Groups Define C G (A) = {g G gag 1 = a for all a A} to be the centralizer of A in G. In particular, this is the subset of G which commuted with every element

### MATH 8254 ALGEBRAIC GEOMETRY HOMEWORK 1

MATH 8254 ALGEBRAIC GEOMETRY HOMEWORK 1 CİHAN BAHRAN I discussed several of the problems here with Cheuk Yu Mak and Chen Wan. 4.1.12. Let X be a normal and proper algebraic variety over a field k. Show

### INTRODUCTION TO COMMUTATIVE ALGEBRA MAT6608. References

INTRODUCTION TO COMMUTATIVE ALGEBRA MAT6608 ABRAHAM BROER References [1] Atiyah, M. F.; Macdonald, I. G. Introduction to commutative algebra. Addison-Wesley Publishing Co., Reading, Mass.-London-Don Mills,

### Commutative Algebra. Timothy J. Ford

Commutative Algebra Timothy J. Ford DEPARTMENT OF MATHEMATICS, FLORIDA ATLANTIC UNIVERSITY, BOCA RA- TON, FL 33431 Email address: ford@fau.edu URL: http://math.fau.edu/ford Last modified January 9, 2018.

### Commutative Algebra. Gabor Wiese. Winter Term 2013/2014. Université du Luxembourg.

Commutative Algebra Winter Term 2013/2014 Université du Luxembourg Gabor Wiese gabor.wiese@uni.lu Version of 16th December 2013 2 Preface In number theory one is naturally led to study more general numbers

### 38 Irreducibility criteria in rings of polynomials

38 Irreducibility criteria in rings of polynomials 38.1 Theorem. Let p(x), q(x) R[x] be polynomials such that p(x) = a 0 + a 1 x +... + a n x n, q(x) = b 0 + b 1 x +... + b m x m and a n, b m 0. If b m

### Introduction to Algebraic Geometry. Jilong Tong

Introduction to Algebraic Geometry Jilong Tong December 6, 2012 2 Contents 1 Algebraic sets and morphisms 11 1.1 Affine algebraic sets.................................. 11 1.1.1 Some definitions................................

### Math 711: Lecture of September 7, Symbolic powers

Math 711: Lecture of September 7, 2007 Symbolic powers We want to make a number of comments about the behavior of symbolic powers of prime ideals in Noetherian rings, and to give at least one example of

### Commutative Algebra Lecture Notes

Marco Andrea Garuti Commutative Algebra Lecture Notes Version of January 17, 2017 This text consists of the notes of a course in Commutative Algebra taught in Padova from 2014-15 to 2016-17. Some topics

### CHEVALLEY S THEOREM AND COMPLETE VARIETIES

CHEVALLEY S THEOREM AND COMPLETE VARIETIES BRIAN OSSERMAN In this note, we introduce the concept which plays the role of compactness for varieties completeness. We prove that completeness can be characterized

### Commutative Algebra I

Commutative Algebra I Craig Huneke 1 June 27, 2012 1 A compilation of two sets of notes at the University of Kansas; one in the Spring of 2002 by?? and the other in the Spring of 2007 by Branden Stone.