# Introduction to Algebraic Geometry. Jilong Tong

Size: px
Start display at page:

Transcription

1 Introduction to Algebraic Geometry Jilong Tong December 6, 2012

2 2

3 Contents 1 Algebraic sets and morphisms Affine algebraic sets Some definitions Hilbert s Nullstellensatz Zariski topology on an affine algebraic set Coordinate ring of an affine algebraic set Projective algebraic sets Definitions Homogeneous Nullstellensatz Homogeneous coordinate ring Exercise: plane curves Morphisms of algebraic sets Affine case Quasi-projective case The Language of schemes Sheaves and locally ringed spaces Sheaves on a topological spaces Ringed space Schemes Definition of schemes Morphisms of schemes Projective schemes First properties of schemes and morphisms of schemes Topological properties Noetherian schemes Reduced and integral schemes Finiteness conditions Dimension Dimension of a topological space Dimension of schemes and rings The noetherian case Dimension of schemes over a field Fiber products and base change Sum of schemes Fiber products of schemes Base change; fibers Some global properties of morphisms

4 4 CONTENTS Separatedness of schemes Proper morphisms Projective morphisms Some local properties of schemes and morphisms of schemes Zariski tangent spaces Regular schemes Flatness and smoothness Normal schemes An introduction to algebraic curves Divisors and invertible sheaves Divisors Invertible sheaves Čech cohomology of a topological space Riemann-Roch theorem for a curve (first form) Serre s duality theorem Differentials Residues Classes of répartitions Duality theorem Riemann-Roch theorem Definitive form Riemann-Roch theorem Some applications

5 Introduction The starting point of the algebraic geometry is trying to study the solutions of systems of polynomials: for simplicity, let k be a field, let P 1,, P m k[x 1,, X n ] to be m polynomials with n variables. Then we want to study the solutions of the following system: One could ask the following questions: P 1 (X 1,, X n ) = 0 P 2 (X,, X n ) = 0 P m (X 1,, X n ) = 0 Does this system have at least a solution in k? If we have a positive answer, then what can we say about these solutions (i.e., are there just finitely many or there are infinite?). Or further more, can we know exactly what are the solutions? Of course, when each polynomial P i is of degree 1, we know in the course of linear algebra that, at least theoretically, there is a way to find explicitly all the solutions (if there is any) or the show the contrast. But in the general case, these become very difficult questions. One can consider the following famous example Fermet s last theorem We consider k = Q, and we want to understand the solutions of the following equations: for n 3 an integer, what can way say about the solutions in Q of the following single equation with only two variables X n + Y n = 1. Sure, one can easily find some solutions (x, y) {(1, 0), (0, 1)} or several more like ( 1, 0), (0, 1) when the integer n is even. After these more or less trivial solutions, it becomes difficult to find some new one (maybe you can try with the computers). In fact, one has the following Conjecture (Fermat s last theorem, 1637). There is no non trivial solutions for n 3. 5

6 6 CONTENTS Fermat himself solved this conjecture when n = 4. In fact, Fermet claimed also that he had found a proof for this conjecture for all n 3. But today we believe that Fermet must make a mistake in his proof. In fact, the human had to spend the next 358 years to completely prove this statement! 1 And the proof involves in fact very deep mathematical subjects such as algebraic geometry and number theory, and it uses in a very impressed way the mathematical tools and methods that are developed in the recent years. Elliptic curves Let again k = Q (or more generally a number field), and a, b k. We consider the following equation (1) y 2 = x 3 + ax + b. We suppose that 4a b 2 0. Let E aff the set of solutions in k of the previous equation. Of course, there are three possibilities: (1) has no solutions in Q, i.e., E aff =. (1) has finitely many solutions in Q, i.e., E aff is a nonempty finite set. (1) has infinitely many solutions in Q, i.e., E aff is an infinite set. Of course, by saying this, we get nothing. To get something interesting, we add an extra point, denoted by, to E aff, we let E = E aff { }. The equation (1) together the point at infinite give an elliptic curve, which amounts to consider the projective version of the equation (1) above. The set E is then the k-points of this curve. Now the miracle does happen: we have the following result Theorem One can endow a natural group structure to E so that E becomes an abelian group, and is the neutral element of this group. 2 Moreover, we have the following Mordell-Weil theorem Theorem (Mordell-Weil). With the group structure above, E is a finitely generated abelian group. Hence, by the structure theorem of abelian groups of finite type, we find E Z r E tor with E tor the torsion part of E. The torsion part is more or less known according to the famous result of Mazur, in fact, there are only finitely many possibilities be the torsion part of E. The free part of E, or equivalently the rank of E, is still very mysterious. If we could completely understand the free part of E, we can earn one millon dollars by proving the so-called Birch and Swinnerton-Dyer conjecture, which relates the rank of E with some analytic invariant attached to the equation (1). 1 This is finally done by Andrew Wiles in 1995, with the help of his former student Richard Taylor. 2 The more correct statement should be the following: the elliptic curve has a natural structure of group scheme over the field of definition (in the example, this is Q), so that gives the neutral element of this group scheme. In fact, what we get is an abelian variety of dimension 1.

7 CONTENTS 7 Remark Note that, an elliptic curve is not the same as an ellipse. Here the latter is given in general by equation of the follow form x 2 a 2 + y2 b 2 = 1. In fact, we will see later that an elliptic curve is a projective smooth curve of genus 1, while an ellipse is a projective smooth curve of genus 0, these two kinds of curves are of completely different nature. 2. The solution of Fermat s Last Theorem is built on a very sophisticate study of the elliptic curves. In fact, elliptic curve is a very important object both in algebraic geometry and modern number theory. This course Since a direct way is impossible, one have to find another way. We will put some extra structures on the set of solutions (Zariski topology, variety, sheaf, cohomology), and then try to understand this set together with the structure. That is what we do in algebraic geometry. In this course, we will try to follow the approach of Grothendieck by using the language of schemes. The main technique that we will use is then the commutative algebra. Hence, we will first review some basic notions in commutative algebra. Review of some basic notions in commutative algebra The commutative algebra is the basic tool to study algebraic geometry. In this section, we will fix some notations by reviewing some well-known things in commutative algebra. Recall first the following definition Definition A commutative unitary ring is the given of the following datum (R, +,, 0, 1), where R is a non empty set, +, are two binary operations defined on R, and 1, 0 R are two distinguish elements (maybe coincide) such that (R, +, 0) is an abelian group, while (R,, 1) is a commutative unitary monoid; The two binary operations are compatible: for any x, y, z R, we have (x + y) z = x z + y z. A commutative unitary ring R will be called trivial if 1 = 0, or equivalently, if R has only one element. Convention Unless specifically stated to the contrast, in this course, 1. the word ring means a commutative unitary ring. Same thing for the subrings etc; 2. the word field means a commutative field. Recall the following definition Definition Let A be a ring.

8 8 CONTENTS 1. An ideal I of A is a subgroup (for the addition) of A such that a A, x I, a x I. An ideal I A is called of finite type if it can be generated as an ideal by finitely many elements. 2. An ideal I is called prime if I A, and whenever x y I for some x, y A, we must have either x I or y I. The set of prime ideals of A will be denoted by Spec(A). 3. An ideal I is called maximal if the following condition holds: if I A, and for any ideal J A such that I J A, then we have either I = J or A = J. The set of maximal ideal will be denoted by Max(A). The proof of the following lemma uses Zorn s lemma, or equivalently, the axiom of choice. Lemma We have Max(A) Spec(A). Moreover, if A is non trivial, then Max(A). Lemma Let A be a ring such that 0 1, and I A be an ideal. Then I is a prime ideal if and only if the quotient A/I is an integral domain. 3 I is a maximal ideal if and only if the quotient A/I is a field. Definition A ring A is call noetherian if it satisfies the following increasing chain condition: for any increasing family of ideals of A: I 0 I 1 I r I r+1 there exists some sufficiently large integer r 0 0 such that I r = I r0 for any r r 0. In an equivalent way, a ring A is noetherian if and only if any ideal I A is finitely generated. Lemma Let A be a ring. 1. If A is a field, or a principal ideal domain, then it s noetherian. 2. Let I be an ideal of A. Then if A is noetherian, so is the quotient A/I. Exercise In general, a subring of a noetherian ring is not necessarily still noetherian. Could you find an example? Lemma Suppose A noetherian, and M an A-module of finite type. noetherian as A-module, namely, any A-submodule of M is finitely generated. Then M is Theorem (Hilbert s basis theorem). If A is noetherian, so is A[X]. Proof. Let (0) J A[X] be an ideal, and let I be the set of leading coefficient of polynomials in J. 4 Then I A is an ideal. As A is noetherian, I is finitely generated, say by LC(P 1 ),, LC(P n ). Let J 1 = (P 1,, P n ) A[X] 3 By definition, a domain is always non trivial. 4 By convention, the leading coefficient of zero polynomial is 0. For f A[X], we denote by LC(f) the leading coefficient of f.

9 CONTENTS 9 the ideal generated by these P i, and d = max i (deg(p i )). Then one verifies easily that J = J M d + J 1 with M d the set of polynomials in A[X] with degree d 1. As A is noetherian, and as M d is a A-module of finite type, according to the previous lemma, we find that M d is noetherian as A-module. In particular, the A-submodule J M d M d is of finite type as A-module. Let Q 1,, Q r J M d J be a family of generators, then we have J = J M + J 1 (Q 1,, Q r ) + (P 1,, P n ) = (Q 1,, Q r, P 1,, P n ) J. As a result, J = (Q 1,, Q r, P 1,, P n ) is finitely generated. This finishes the proof. 5 Corollary For k a field. The polynomial rings k[x 1,, X n ] are all noetherian. Hence any k-algebra of finite type is noetherian. Remark Let k be a field. Then k[x 1,, X n ] are the so-called unique factorization domain (UFD for short). But in general, it is not a principal domain. In fact, it s principal if and only if n = 1. Definition Let k be a field, then k is called algebraically closed, if any polynomial P k[x] of degree 1 has at least one solution in k (hence has exactly deg(p ) solutions). Example C is algebraically closed by the fundamental theorem of algebra. 6 Moreover, for any field k, one can always find an algebraically closed field K containing k, the smallest one is called the algebraic closure of k. 5 An alternative way to avoiding the previous lemma: after we get the equality J = J M +J 1, let I 1 be the set of leading coefficients of polynomials in J M. Then I 1 A is again an ideal, hence there is Q 1,, Q r J M such that I 1 = (LC(Q 1),, LC(Q r)). Then similarly, we have J M d J M d1 + (Q 1,, Q r) where d 1 is the maximum of deg(q i) for 1 i r. Clearly, d 1 < d, and we have J = J M d1 + (P 1,, P n, Q 1,, Q r). Now, we continue with this process, and we can conclude. 6 Despite the name of this theorem, there exists no purely algebraic proof of this result. The reason is that the construction of C is not completely algebraic.

10 10 CONTENTS

11 Chapter 1 Algebraic sets and morphisms The aim of this section is trying to gives some geometric background of the algebraic geometry. So here k = k is an algebraically closed field. 1.1 Affine algebraic sets Some definitions Definition Let k be an algebraically closed field as before. The affine space A n k of dimension n is just k n = {(x 1,, x n ): x i k}. 1 Let S k[x 1,, X n ] be a subset of polynomials of n variables. Define V (S) := {(x 1,, x n ) A n k : P (x 1,, x n ) = 0 for any P S} Such a subset of A n k is called an algebraic set of An k. Remark Keeping the notations as before. If S 1 S 2 k[x 1,, X n ] are two subsets, then V (S 2 ) V (S 1 ). Recall that for A a ring and S A a subset. The ideal generated by S (denoted by < S >) consists of all finite sums a i s i, a i A, s i S. i Hence V (< S >) V (S) V (< S >). As a result, we find V (S) = V (< S >). So for algebraic sets, we only need to consider those of the form V (a) with a k[x 1,, X n ] an ideal. Recall also that since k is a field, k[x 1,, X n ] is noetherian. Hence any ideal of k[x 1,, X n ] is finitely generated, say by P 1,, P m A. So to define an algebraic set, we only need finitely many equations V (a) = {(x 1,, x n ) k n : P 1 (x 1,, x n ) = = P m (x 1,, x n ) = 0}. 1 The more correct notation should be A n k (k), but here we just write A n k for simplicity. 11

12 12 CHAPTER 1. ALGEBRAIC SETS AND MORPHISMS Proposition The union of two algebraic sets (of A n k ) is again algebraic; the intersection of any family of algebraic sets is again algebraic. Moreover, empty set and the total space A n k are algebraic. Proof. In fact, we have (i) V (a) V (b) = V (a b); (ii) i V (a i ) = V ( i a i); (iii) V ((1)) =, and V ((0)) = A n k. Remark We remark that we have always V (a) = V (a 2 ) = V ( a), here a := {a A: r Z 1 s.t. a r a} is the radical of the ideal a. Hence one can not expect to recover the ideal a from its corresponding algebraic set V (a) Hilbert s Nullstellensatz Definition Let Y A n k be a subset. Define the ideal of Y to be I(Y ) := {P k[x 1,, X n ]: P (x) = 0 x Y }. Lemma Keeping the notations as before: For any subset Y A n k, we have I(Y ) = I(Y ). 2 For any subsets Y 1 Y 2 A n k, we have I(Y 2) I(Y 1 ). For any subsets Y 1, Y 2 A n k, we have I(Y 1 Y 2 ) = I(Y 1 ) I(Y 2 ). For any ideal a k[x 1,, X n ], a I(Z(a)). Now, we may state and prove the first fundamental result in this course Theorem (Hilbert s Nullstellensatz). For any ideal a k[x 1,, X n ], I(V (a)) = a. To prove this theorem, we begin with the following lemma: Lemma Let K be a field, L/K be a field extension which is finitely generated as K-algebra. Then L/K is an algebraic extension. Proof. 3 Up to replace K by its algebraic closure in L, we may assume that L/K is an transcendant field extension, i.e., any non zero element of L is transcendant over K. Suppose first that L/K is of transcendant degree 1, namely L contains a copy of K(x) (:=the fraction field of the polynomial ring K[x]) such that L is algebraic over K(x). Since L is of finite type as K-algebra, it s also of finite type as K(x)-algebra. In particular, L is of finite dimension 2 Such an ideal is called radical. 3 When K is a uncountable field, this lemma can also be proved as follows: remark first that since L/K is finitely generated as K-algebra, then L is of countable dimension over K as a K-vector space. On the other hand, once the field K is uncountable, and once one can find x L which is transcendant over K, then the following family {1/(x a): a K} is linearly independent and of cardinality uncountable. From this, we find that the K-space L must has uncountable dimension, whence a contradiction. This gives the lemma when K is uncountable.

13 1.1. AFFINE ALGEBRAIC SETS 13 over K(x). Choose e 1, e n a basis of L over K(x), and write down the multiplication table for L: (1.1) e i e j = k a ijk (x) b ijk (x) e k with a ijk (x), b ijk (x) K[x]. Now suppose L = K[f 1,, f m ], and we write each f j under the basis {e i }: f j = c ij (x) d i ij (x) e i Since for any element x L, x can be written as a polynomial in the f j s with coefficients in K, it follows that x can also be written as a K(x)-combination of 1 and the product of e i s such that the denominator of each coefficient involves only the products of the polynomials d ij s. Now, by applying the multiplication table (1.1), we find that x can be written as a K(x)-combination of 1 and the e i s such that the denominator of each coefficient involves only the products of the polynomials d ij s and b ijk s. But there exists infinitely many irreducible polynomials in K[x], there exists some polynomial p(x) K[x] which does not divide any d ij nor b ijk. As a result, the fraction 1/p(x) cannot be written in the form described above, which means 1/p(x) / K[f 1,, f m ] = L, a contradiction since we suppose that L is a field, hence à priori, 1/p(x) L. This finishes the case when L/K is of transcendant degree 1. For the general case (i.e. tr.deg(l/k) 1), one can always find a subfield K of L containing K such that L/K is of transcendant degree 1. In particular, the previous argument shows that L cannot be of finite type as K -algebra. A priori, L cannot be of finite type as K-algebra. This gives a contradiction. From this lemma, we get the Weak Nullstellensatz: Theorem (Weak Nullstellensatz). Let k be an algebraically closed field, then every maximal ideal in A = k[x 1,, X n ] has the form (X 1 x 1,, X n x n ) for some (x 1,, x n ) k n. As a consequence, a family of polynomials functions on k n with no common zero generates the unit ideal of A. Proof. Let m A be a maximal ideal, then the corresponding quotient K := A/m is a field extension of k. Moreover, since A is finitely generated as k-algebra, so is K. Hence the previous lemma tells us that K/k is in fact an algebraic extension. But since the base field k is algebraically closed, the inclusion k K must be an isomorphism. Let now x i k be the image of X i in the quotient k K = A/m, then X i x i m for all i. As a result, we get (X 1 x 1,, X n x n ) m A. On the other hand, we can verify directly that the ideal (X 1 x 1,, X n x n ) is itself maximal, hence the first inclusion above is an equality: (X 1 x 1,, X n x n ) = m. In this way, we get the first assertion. For the last assertion, let J be the ideal generated by this family. Since this family of polynomials functions has no common zero on k n, J is not contained in any maximal ideal of A. As a result, we must have J = A. This finishes the proof. Exercise Prove directly (i.e., without using Weak Nullstellensatz) that I(A n k ) = (0). Proof of Hilbert s Nullstellensatz. Suppose a = (f 1,, f m ), and let g I(V (a)) {0}. We consider now the ring of polynomials in n+1 variables k[x 1,, X n, X n+1 ], and the polynomials

14 14 CHAPTER 1. ALGEBRAIC SETS AND MORPHISMS f i, g can be viewed naturally as polynomials in n + 1 variables. Now, consider following family of polynomials {f 1,, f m, X n+1 g 1}, it has no common zero in k n+1. Hence by the Weak Nullstellensatz, we get (f 1,, f m, X n+1 g 1) = k[x 1,, X n, X n+1 ]. Hence, there are polynomials Q 1,, Q n+1 k[x 1,, X n+1 ] such that 1 = Q 1 f 1 + Q n f n + Q n+1 (X n+1 g 1) Now, take the image of this equality by the morphism k[x 1,, X n, X n+1 ] k(x 1,, X n ), X n+1 1/g. We get the following equality in k(x 1,, X n ) 1 = P 1 f 1 + P n f n g s with P i k[x 1,, X n ]. In particular, g s = P 1 f P n f n a. Hence g a, this finishes the proof. Exercise Show that V (I(Y )) = Y for any subset Y A n k. Corollary There is one-to-one inclusion-reversing correspondence between algebraic sets in A n k and radical ideals a k[x 1,, X n ]. Proof. The correspondence is given by Y I(Y ), and conversely by a V (a) Zariski topology on an affine algebraic set Definition (Topology). A topology on a set X is defined by giving a family T of subsets of X such that, X T ; T is stable by finite intersection; T is stable by any union. In this case, a subset U X is called open iff U T, and F X is called closed iff its complement F c = X F is open in X. As a corollary of the Proposition , we have the following Definition (Zariski topology). We define the Zariski topology on A n k by taking the open subsets to be the complements of algebraic sets. For an algebraic subset V A n k, then the Zariski topology on V is the subspace topology induced from A n k. Example A subset F A 1 k is closed iff F is a finite set. As a result, a subset U A 1 k is open iff its complement An k U is a finite set. The Zariski topology on A n k is not Hausdorff. It s just a T 1 space: for any two different points x, y A n k, one can find an open U An k such that x U and y / U.

15 1.1. AFFINE ALGEBRAIC SETS 15 Exercise Determine the closed subsets of A 2 k, and then prove that the topology of A2 k is not the product topology on A 2 k = A1 k A1 k (here, this is an equality as sets). Definition A non empty subset Y of a topological space X is called irreducible if it cannot be expressed as the union Y = Y 1 Y 2 of two proper subsets such that each Y i is closed in Y. Remark Thought the empty set satisfies vacuously the condition of , in this course, we don t consider it as an irreducible set. Example V (X 1 X 2 ) = V (X 1 ) V (X 2 ) A 2 k is not irreducible, while A1 k is irreducible. Proposition A closed subset F A n k is irreducible iff its ideal I(F ) k[x 1,, X n ] is a prime ideal. Proof. Let F A n k be irreducible, and let f, g k[x 1,, X n ] be two elements such that f g I(F ). We have in particular (f g) I(F ), hence F = V (I(F )) V (f g) = V (f) V (g), and we get F = V (f g) F = (V (f) F ) (V (g) F ). As V (f) and V (g) are both closed, and F is irreducible, we have either V (f) F = F or V (g) F = F. Suppose for example V (f) F = F, in particular, we get F V (f), hence f I(V (f)) I(F ). This shows that I(F ) k[x 1,, X n ] is prime. Conversely, suppose I(F ) is a prime ideal, and show that F is irreducible. Let F = F 1 F 2 with F i F two closed subsets. Then we have I(F ) I(F i ). Now suppose F 1 F, in particular, by Hilbert s Nullstellensatz, I(F ) I(F 1 ). Let f I(F 1 ) I(F ), then for any g I(F 2 ), then f g I(F 1 F 2 ) = I(F ). As I(F ) is a prime ideal, we get g I(F ). In particular, I(F ) = I(F 2 ), hence F = I(V (F )) = I(V (F 2 )) = F 2. This shows that F is irreducible. In this way, we get the proposition. Example The affine space A n k prime. is irreducible since its ideal is (0) which is of course Definition A topological space X is called noetherian, if it satisfies the descending chain condition for closed subsets: for any sequence Y 0 Y 1 Y i Y i+1 of closed subsets, there exists some integer i 0 0 such that Y i = Y i0 for any i i 0. Proposition Any algebraic subset together with the Zariski topology is a noetherian topological space. Proof. Since any closed subset of a noetherian space is again noetherian, we only need to show that A n k is noetherian. Recall that the set of closed subsets of An k is in inclusion-reversing oneto-one correspondence with the set of radical ideal of k[x 1,, X n ]. Hence to show that A n k is noetherian, we only need to show that k[x 1,, X n ] satisfies the increasing chain condition, or equivalently, k[x 1,, X n ] is noetherian. This is exactly Hilbert s basis theorem. As a result, A n k is noetherian.

16 16 CHAPTER 1. ALGEBRAIC SETS AND MORPHISMS Lemma Let X be a neotherian topological space, and Y X be a closed subset. Then Y can be expressed as a finite union Y = Y 1 Y 2 Y r of irreducible closed subsets Y i. If we require moreover Y i Y j whenever i j, then the family {Y i : 1 i r} is uniquely determined. They are called the irreducible components of Y. Proof. We will first show the existence of such decomposition. Let S be the set of closed subset Y X which can not be written as a finite union of irreducible closed subsets, we need to show that S =. If not, let Y 0 S be an arbitrary element of this set, then Y 0 is not irreducible, hence we have Y 0 = Y 0 Y 0, with Y 0, Y 0 Y 0 two proper closed subsets of Y 0. By the choice of Y 0, either Y 0 S or Y 0 S. For simplicity, suppose Y 0 S, and we note Y 1 = Y 0. Then we continue with this construction, and we get in this way a infinite sequence of closed subsets of X: Y 0 Y 1 Y 2 Y r Y r+1. This gives a contradiction with the assumption that X is noetherian. As a result, S =, and any closed Y X can be written as a finite union of irreducible closed subsets Y = Y 1 Y 2 Y r By throwing away a few if necessary, we may assume that Y i Y j whenever i j. This gives the existence. For the uniqueness, suppose Y = r Y i = i=1 such that Y i Y i and Y j Y j whenever i i and j j. Then Y 1 iy i, hence there exists some i such that Y 1 Y i. Similarly, we have Y i Y j for some j, hence Y 1 Y j. We must have 1 = j, and Y 1 = Y i. After renumbering the index i, we may assume i = 1. Now to finish the proof, it remains remark that r i=2 s j=1 Y j Y i = Y Y 1 = Y Y 1 = s Hence the uniqueness follows from an induction on min(r, s). Corollary Any closed subset F A n k following form s F = i=1 of irreducible closed subsets, no one containing another. F i j=2 Y j. can be expressed as a finite union of the Exercise Show that a noetherian topological space X is always quasi-compact. If moreover it s Hausdorff, then X is a finite set with the discrete topology Coordinate ring of an affine algebraic set Let V A n k be an algebraic set, we define its coordinate ring k[v ] := k[x 1,, X n ]/I(V ).

17 1.1. AFFINE ALGEBRAIC SETS 17 Lemma An element f k[x 1,, X n ] defines a function A n k k, x f(x), whose restriction to V A n k depends only on the coset f + I(V ). In particular, any element of its coordinate ring k[v ] can be viewed naturally as a function defined on V, which is called a regular function of V. We call also k[v ] the ring of regular functions. Moreover, we endow V with the induced topology as V is a subset of the topological space A n k. As a corollary of Hilbert s Nullstellensatz, we have the following Proposition ideals of k[v ]. (a) The points of V are in one-to-one correspondence with the maximal (b) The closed subsets of V are in one-to-one correspondence with the radical ideals of k[v ], such that a closed subset F V is irreducible iff the corresponding radical ideal under this correspondence is a prime ideal of k[v ]. 4 (c) For any f k[v ], let D(f) := {x V : f(x) }. Then the family {D(f): f k[v ]} forms a basis for the topology of V : i.e., each D(f) is open, and any open U V is a union of D(f) s. Proof. (a) Let P = (x 1,, x n ) V A n k be an element of V, then I({P }) = m P = (X 1 x 1,, X n x n ) k[x 1,, X n ] is a maximal ideal which contains I(V ). Hence its image m P k[v ] = k[x 1,, X n ]/I(V ) is again a maximal ideal. Now we claim that the correspondence V Max(k[V ]), P m P is bijective. Indeed, let n be a maximal ideal of k[v ], its inverse image m in k[x 1,, X n ] is again maximal, hence of the form m Pn for a unique point P n A n k by Nullstellensatz. One verifies easily that P n is contained in V (since I(V ) m Pn ), and the map Max(k[V ]) V, n P n gives an inverse of the previous correspondence. (b) Similar proof as (a). (c) First of all D(f) V is open. Indeed, let f be an arbitrary lifting of f in k[x 1,, X n ], and consider D( f) := {x A n k : f(x) 0}. Then D(f) = D( f) V, hence, we only need to show D( f) A n k complement can be described in the following way, A n k D( f) = {x A n k : f(x) = 0} = V ({ f}) is open. But its hence this is closed. As a result, D( f) is open. This gives the openness of D(f). For the last assertion, suppose U = Ũ V be an open of V with Ũ is open in An k. Hence, there is some ideal a k[x 1,, X n ] such that A n k Ũ = V (a) = g av (g). Hence Ũ = g ad(g). As a result, U = Ũ V = g a(d(g) V ) = g a D(ḡ) with ḡ the image of g in k[v ]. This finishes the proof of (c). 4 Note that by our convention , the empty set is not irreducible.

18 18 CHAPTER 1. ALGEBRAIC SETS AND MORPHISMS Exercise Consider the affine twisted cubic curve: C = {(t, t 2, t 3 ): t k} A 3 k. Show that C is an irreducible closed subset of A 3 k, and find generators of the ideal I(C) k[x, Y, Z]. 2. Let V = V (X 2 Y Z, XZ X) A 3 k. Show that V consists of three irreducible components, and determine the corresponding prime ideals. 3. We identify the space M 2 (k) of 2 2-matrices over k with A 4 k with coordinates a, b, c, d: ( ) M 2 (k) A 4 a b k, (a, b, c, d). c d Show that the set of nilpotent matrices is an algebraic subset of A 4 k, and determine its ideal. Exercise Consider C = {(t 3, t 4, t 5 ): t k} A 3 k. Show that C is an irreducible algebraic set, et determine I(C). Can I(C) be generated by 2 elements? Solution. Consider the ideal J generated by the following three elements XZ Y 2, X 2 Y Z 2, Y Z X 3. Then J I(C), and V (J) = C. In particular, C is an algebraic set. Next, we claim that J is a prime ideal. Indeed, consider the following morphism φ: k[x, Y, Z] k[t ], X T 3, Y T 4, Z T 5. Its kernel contains J, and we only need to show that ker(φ) = J. Let f k[x, Y, Z], using the three generators of J as above, we have f(x, Y, Z) a 0 (X) + a 1 (X) Y + a 2 (X) Z mod J with a i (X) k[x]. Now f ker(φ) means that a 0 (T 3 ) + a 1 (T 3 ) T 4 + a 2 (T 3 ) T 5 = 0, which is possible only when a i (X) = 0, that is f J. This proves that J = ker(φ) is prime. In particular, J = I(C) by Hilbert s Nullstellensatz. Moreover, I(C) cannot be generated by two elements. In fact, if so, the k-vector space J := J k[x,y,z] k[x, Y, Z]/(X, Y, Z) = J/(X, Y, Z) J must have dimension at most 2. On the other hands, we claim that the image of the three generators above in J is k-linearly independent: let a, b, c k such that a (XZ Y 2 ) + b (X 2 Y Z 2 ) + c (Y Z X 3 ) (X, Y, Z) J Since each term of all non zero element of (X, Y, Z) J has degree at least 3, we must have a (XZ Y 2 ) b Z 2 + c Y Z = 0 As a result, a = b = c = 0, this gives the claim. In particular, J = I(C) cannot be generated by 2 elements.

19 1.2. PROJECTIVE ALGEBRAIC SETS Projective algebraic sets Definitions Let n 1 be an integer, and we consider the following equivalent relation on k n+1 {0}: for x = (x 0,, x n ) and y = (y 0,, y n ), we say x y iff there exists some λ k (must be non zero) such that x i = λ y i for each i. Definition We define the projective space of dimension n to be the set of equivalent classes: P n k := ( k n+1 {0} ) / 5 An element of P n k is called a point of the projective space Pn k. For each element (x 0,, x n ) k n+1 {0}, its equivalent class in P n k will be denoted by (x 0 : x 1 : : x n ). The n + 1-tuple (x 0,, x n ) is called the homogeneous coordinate of this equivalent class. Recall that for a polynomial P k[x 0,, X n ], it s called homogeneous if there exists some integer d 0, such that P can be written under the following form P (X 0,, X n ) = a i0,,i n X i 0 0 Xin n In particular, i 0,,i n Z 0,i 0 + +i n=d P (λ X 0, λ X 1,, λ X n ) = λ d P (X 0, X 1,, X n ). for any λ k. Let now P be such a homogeneous polynomial, and let x = (x 0,, x n ) be a zero of P. For any λ k, λ x = (λx 0,, λx n ) is again a zero of P. Hence it makes sense to say that for a point x P n k, whether the value P (x) is zero or not. Moreover, an ideal a k[x 0,, X n ] is called homogeneous if it can be generated as an ideal by homogeneous elements, or equivalently, if a = a k[x 0,, X n ] d. d 0 Definition Let P k[x 0,, X n ] be a homogeneous polynomial. Let a = (f 1,, f m ) be a homogeneous ideal with {f 1,, f m } a family of homogeneous generators. We define the corresponding projective algebraic set to be V + (a) := {(x 0 : x n ) P n k : f i(x 0,, x n ) = 0 i } One can verify that this definition is independent of the choice of the homogeneous generators {f 1,, f m } Example Let (a 0 : : a n ) P n k be a point of the projective space, we consider the ideal p generated by (X i a j X j a i ) i,j. Then it s homogeneous, and V + (p) = {(a 0 : : a n )} P n k. Note that this ideal is prime but not maximal. Moreover, V (p) A n+1 k is the line passing through the origin of A n+1 k and also the point (a 0,, a n ) A n+1 k. Example (Hypersurfaces). Let f(x 0,, X n ) be an irreducible homogeneous polynomial. Then f = 0 defines an irreducible algebraic set in P n k, called a hypersurface. It may happen that V + (a) = for a homogeneous ideal a even if a k[x 0,, X n ]. In fact, we have 5 Similar to the affine case, the more correct notation here is P n k (k), here we write P n k just for simplicity.

20 20 CHAPTER 1. ALGEBRAIC SETS AND MORPHISMS Proposition Let a k[x 0,, X n ] be a homogeneous ideal. The following two assertions are equivalent 1. V + (a) = ; 2. a contains k[x 0,, X n ] d for some d > 0. Proof. 6 Suppose V + (a) =, which implies that V (a) A n+1 k has at most the single point {0}, that is V (a) {0}. As a result, we find (X 0,, X n ) = I({0}) I(V (a)) = a. In particular, there exists some s > 0 such that X s i a, hence k[x 0,, X n ] d a with d = (n + 1) s. This gives (2). Conversely, suppose a contains k[x 0,, X n ] d for some d > 0, then V (a) {0}, hence V + (a) =. Proposition The union of two algebraic sets is algebraic. The intersection of any family of algebraic sets is algebraic. The empty set and the whole space P n k are both algebraic. Definition We define the Zariski topology on P n k by taking open subsets to be the complements of algebraic sets. For a subset Y P n k, the Zariski topology is the subspace topology induced from P n k. Exercise Describe the closed subsets of P 1 k. Definition A subset V P n k is called a quasi-projective algebraic set, if it can be realized as an open of a projective algebraic set of P n k. Example For each integer i [0, n], let H i = Z(X i ) = {(x 0 : : x n ): x i = 0} P n k. Then it s a projective algebraic set, and we will denote by U i its complement. There is a canonical map of sets ( U i A n k, (x x0 0 : : x n ),, x i 1, x i+1,, x ) n x i x i x i x i In fact, this is a homoeomorphism of topological spaces. Note that i H i =, hence i U i = P n k. Example [Any affine algebraic set is quasi-projective] Let V = V (a) A n k be an affine algebraic set. We identify A n k with the open subset U 0 of P n k of elements with homogeneous coordinate (x 0 : x 1 : : x n ) such that x 0 0. In this way, V can be seen as a subset of P n k. Since V U 0 is closed, we have V = V U 0 with V P n k the closure of V in Pn k. Exercise Keeping the notations of Example Describe explicitly from a the closure V of V. Solution. We consider the following construction: for each polynomial f k[x 1,, X n ], et define β(f) k[x 0,, x n ] the homogeneous polynomial given by ( β(f) = x deg(f) x1 0 f,, x ) n. x 0 x 0 Now, we consider b k[x 0,, x n ] the homogeneous ideal generated by β(a) k[x 0,, x n ], then clearly V V + (b), hence V V + (b). Conversely, for G k[x 0,, x n ] a homogeneous polynomial such that V V + (G), or equivalently V V (g) with g = G(1, X 1,, X n ), we have g a by Hilbert s Nullstellensatz. In particular, there exists some integer r > 0 such that g r a. In particular, β(g) r = β(g r ) (β(a)) = b. Moreover, we have G = x deg(g) deg(g) 0 β(g), hence G r b, so G b. As a result, we find V + (b) V + (G). In this way, we find V + (b) is contained in any closed subset of P n k which contains V. From this, we find V = V +(b). 6 In the course, it was said that V +(a) = iff V (a) = {0}, but of course this equivalence works only when a k[x 0,, X n]. Hence here the more correct statement is that V +(a) = iff V (a) {0}.

21 1.2. PROJECTIVE ALGEBRAIC SETS 21 Example (Twisted cubic in P 3 k ). Recall that the affine twisted cubic is the affine algebraic set given by C = {(t, t 2, t 3 ) : t k} A 3 k In terms of equations, this is V (X 2 Y, X 3 Z). Now, we identify A 3 k with an open of P3 k by A 3 k P3 k, (x, y, z) (1 : x : y : z). Then the closure C of C is an projective algebraic set of P 3 k, called the (projective) twisted cubic. Now we want to give an explicitly description of C in terms of equations. Note that (X 2 Y, X 3 Y ) = (X 2 Y, XY Z) = (X 2 Y, XY Z, XZ Y 2 ) Hence, if we consider I = (X 2 Y W, XY ZW, XZ Y 2 ) k[w, X, Y, Z], then V + (I) A 3 k = C. Moreover, this ideal is prime: indeed, we can consider the following morphism of rings α: k[w, X, Y, Z] k[s, T ], W S 3, X S 2 T, Y ST 2, Z T 3. Then I α, hence to show that I is prime, we only need to show that I = ker(α). Now, any polynomial f k[w, X, Y, Z] can be written as f a 0 (W, Z) + a 1 (W, Z) X + a 2 (W, Z) Y mod I with a i k[w, Z]. If f ker(α), we then have 0 = a 0 (S 3, T 3 ) + a 1 (S 3, T 3 ) S 2 T + a 2 (S 3, T 3 ) ST 2 which is possible iff a i = 0 for i {0, 1, 2}. Hence f I. This shows I is prime, hence V + (I) is irreducible, and it contains V as an open subset, hence V + (I) = V. One shows also that this ideal I cannot be generated by two elements. Exercise Consider the map ν : P 1 k P3 k, (x; y) (x3 : x 2 y : xy 2 : y 3 ). Prove that the image of this map is a projective algebraic set of P 3 k. Can you recognize this algebraic set? Homogeneous Nullstellensatz Definition Let Y P n k be a subset. We define its homogeneous ideal I(Y ) to be the ideal generated by the homogeneous polynomial P k[x 0,, X n ] such that P (x) = 0 for any x Y. Proposition Keeping the notations as before. 1. We have I(V + (a)) = a, and V + (I(Y )) = Y for Y P n k a subset. 2. The mappings a V (a) and Y I(Y ) set up a bijection between the set of closed algebraic subsets of P n k, and the set of all radical homogeneous ideal a k[x 0,, X n ] such that a (X 0,, X n ) Under the previous correspondence, the irreducible closed subsets (resp. the points) of P n k correspond to the prime homogeneous ideals not equal to (X 0,, X n ) (resp. the prime homogeneous ideals p (X 0,, X n ) which is maximal with respect to this property). 4. P n k is neotherian with the Zariski topology. 7 Note that the radical of a homogeneous ideal is still homogeneous.

22 22 CHAPTER 1. ALGEBRAIC SETS AND MORPHISMS Exercises: affine cone over a projective subset Let U = A n+1 k {0}, and p : U P n k be the canonical projection. Let Y Pn k be a projective algebraic subset, let Ỹ := p 1 (Y ) {0} A n+1 k. Exercise Show that Ỹ An+1 k is an algebraic subset of A n+1 k, whose ideal is I(Y ) considered as an ordinary ideal of k[x 0,, X n ]. Moreover show that Y is irreducible if and only if Ỹ is irreducible. The affine algebraic set Ỹ is called the affine cone over Y. Exercise Use the notion of affine cone to establish Homogeneous coordinate ring Recall first that a graded ring is a ring A, together with a decomposition of A into a direct sum of its subgroups A = d 0 A d such that A d A d A d+d. The decomposition above is called a gradation of A. Let V = V + (a) P n k be a projective algebraic set, with a k[x 0,, X n ] a homogeneous ideal. Its homogeneous coordinate ring is then define to be the quotient k[x 0,, X n ]/I(V ), with I(V ) the ideal of V. Since I(V ) is homogeneous, the quotient A := k[x 0,, X n ]/I(V ) is naturally graded: A = d 0 A d where A d := k[x 0,, X n ] d /k[x 0,, X n ] d I(V ) with k[x 0,, X n ] = d 0 k[x 0,, X n ] d the usual gradation of k[x 0,, X n ]. Let A + := d>0 A d. Let f A be a homogeneous element, which is the image of a homogeneous element f k[x 0,, X n ]. It s easy to see that the set {x V P n k : f(x) 0} is independent of the lifting f of f, hence we will denote it by D + (f). For the future use, we record the following proposition, which is a projective analogue of Proposition There is a one-to-one correspondence between the set of closed subsets of V, and the set of homogeneous radical ideals not equal to A Under the previous correspondence, the irreducible closed subsets (resp. points) correspond to the homogeneous prime ideals of A not equal to A + (resp. prime homogeneous ideals p A + which are maximal with respect to this property). 3. For f A = k[x 0,, X n ]/I(V ), the subset D + (f) is open in V. Moreover, the family {D + (f): f A homogeneous} forms a basis for the Zariski topology of V. Such an open subset will be called principal Exercise: plane curves We begin with a general ressult. Exercise Let a k[x 1,, X n ] be an ideal. Show that V (a) is a finite set if and only if the quotient k[x 1,, X n ]/a is of finite dimensional over k.

23 1.2. PROJECTIVE ALGEBRAIC SETS 23 Now, an affine plane curve, is the set of solutions in A 2 k k[x, Y ]. of a nonconstant polynomial f Exercise Let C = V (f), D = V (g) A 2 k be two plane curves with f, g two irreducible polynomial of degree respectively m and n, such that f g. We want to show that C D is a finite set in the following way. 1. For each integers d 0, let P d be the set of polynomials k[x, Y ] of total degree d. Compute the dimension of this k-vector space P d. 2. Show that for d Max{m, n}, the k-vector space P d /(f, g) P d is of dimension mn. For this, one could use the following sequence of maps (1.2) P d m P d n α P d β P d /(f, g) P d where α(u, v) = uf + vg, and β is the natural projection, and note that β α = Show that dim k (k[x, Y ]/(f, g)) mn and then conclude. But it might happen that C D =. To remedy this, we consider the projective plane curves. By definition, a projective plane curve, or just a plane curve for short, is the set of solutions in P 2 k of a nonconstant homogeneous polynomial F k[x, Y, Z]. Exercise Show that for any two plane curves C, D P 2 k, their intersection C D is not empty. One could try to use the sequence (1.2) for help. Now, we will examiner the case where C = V + (F ) is a plane curve with F k[x, Y, Z] a homogenenous irreducible polynomial, and D = L is a line in P 2 k. Let s first define the intersction multiplicity of L and C at a point P P 2 k. We may assume that P = (x 0 : y 0 : 1) A 2 k = D + (Z) P 2 k. Hence L A2 k is given by a polynomial f in x, y of degree 1: g(x, y) = a (x x 0 ) + b (y y 0 ) and C A 2 k is given by a polynomial f(x, y) = F (x, y, 1) k[x, y] such that f(x 0, y 0 ) = 0. Without loss of generality, we assume b 0, then replace y by a b (x x 0) + y 0 in the expression of f, and we get a polynomial f(x) = f(x, a b (x x 0)+y 0 ) k[x]. By definition, x 0 is a root of this polynomial. Now, we define the intersection multiplicity i(p ; L, C) to be the multiplicity of the root x = x 0 of the polynomial f. Moreover, by convention, for P / L C, we define i(p ; L, C) = 0. Exercise (A special case of Bézout s theorem). 8 Let L be a line of P 2 k and C = V +(F ) be a plane curve given by an irreducible homogeneous polynomial of degree n. Suppose C L, then the following equality holds n = i(p ; L, C) P P 2 k The actual Bézout s theorem is the following 8 Of course, the affine analogue of this proposition is not true.

24 24 CHAPTER 1. ALGEBRAIC SETS AND MORPHISMS Theorem Let C and C be two plane curves of degree respectively m and n. Suppose that C or C doesnot contain any irreducible component of the other curve as its irreducible component, then the following equality holds: m n = i(p ; C, C ) P P 2 k 1.3 Morphisms of algebraic sets We will need to know when 2 algebraic sets are to be considered isomorphic. More generally, we will need to define not just the set of all algebraic sets, but the category of algebraic sets Affine case Let U A n k and V Am k be two affine algebraic sets. Definition A map σ : U V is called a morphism if there exist m polynomials in n variables P 1,, P m k[x 1,, X n ] such that for any x = (x 1,, x n ), we have σ(x) = (P 1 (x),, P m (x)). A morphism σ : X Y is called an isomorphism if it s bijective, 9 with inverse σ 1 again a morphism. In this case, we say that U and V are isomorphic. Remark If we replace in the previous definition polynomials by regular functions on U, we get the same notion. Hence a morphism σ : U V A m k is given by m regular functions f 1,, f m k[u] such that for any x U, σ(x) = (f 1 (x),, f m (x)) V. Example Look at A 1 k the affine line, and the parabola C = V (y x2 ) A 2 k. The projection π : (x, y) x of the parabola onto the x-axis should surely be an isomorphism between these algebraic sets. Indeed, we consider the following map Then we have σ π = id and π σ = id. σ : A 1 k C, x (x, x2 ). Proposition Let f : U V be a morphism of affine algebraic sets. Then f induces a continuous map between the underlying topological spaces. Proof. Suppose U A n k, and V Am k. By the definition, f is just the restriction to U of a regular function F : A n k Am k. Since U and V are endow with the subspace topology, hence we are reduced to show that F is continuous. We write F = (P 1,, P m ) with P i k[x 1,, X n ]. Let Y = V (I) A m k be a closed subset with I = (g 1,, g r ) k[y 1,, Y m ] an ideal. Then F 1 (V (I)) = V (f 1,, f r ) with f i = g i (P 1,, P m ) k[x 1,, X n ]. In particular, F 1 (V (I)) is closed. Hence, F is continuous. This finishes the proof. Let σ : U V be a morphism of affine algebraic sets. For any regular function f : V U, we claim the composition f σ : U k gives a regular function on U. Indeed, as f is regular, we may assume that f(y) = F (y) for y V with F k[y 1,, Y m ] a polynomial. Moreover, let P 1,, P m k[x 1,, X n ] such that σ(x) = (P 1 (x),, P m (x)) for x U. Then f σ(x) = F (P 1,, P n )(x) for x U. As F (P 1,, P m ) k[x 1,, X n ] is a polynomial, hence the composition f σ is a regular function on U. In this way, we get a map, denoted by σ : σ : k[v ] k[u] Note that k = k. 10 The proof given here is slicely different the proof give in the lecture.

25 1.3. MORPHISMS OF ALGEBRAIC SETS 25 Proposition The map is bijective. ι: Hom(U, V ) Hom k (k[v ], k[u]), σ σ Proof. We remark first that, for any x U, if we note m x = {f k[u]: f(x) = 0} k[u] the maximal ideal corresponding to x X, and n σ(x) k[v ] the maximal ideal corresponding to σ(x). Then we have σ 1 (m x ) = n σ(x). This remark implies immediately that the map ι is injective. To establish the surjectivity, let τ : k[v ] k[u] be a morphism of k-algebras. Let P i k[x 1,, X n ] be a lifting of τ(y i ) for each 1 i m, we have then the following commutative diagram k[y 1,, Y m ] Y i P i k[x 1,, X n ] Y i y i k[v ] τ k[u]. X i x i Let σ = (P 1,, P m ): A n k Am k be the morphism given by the polynomials P i s. Then σ sends U to V, which means that for any x U, (P 1 (x),, P m (x)) V. Indeed, we need to show that f(p 1 (x),, P m (x)) = 0 for any f I(V ). But since the polynomial P i is a lifting of τ(y i ), we have f(p 1 (X),, P m (X)) I(U), hence f(p 1 (x),, P m (x)) = 0 for any x U. In this way, we get a morphism from U to V, still denoted by σ. Finally, it remains to show that σ = τ. We only need to show σ (y i ) = τ(y i ) for each i: in fact, we have This finishes then the proof. σ (y i ) = P i (x 1,, x n ) = P i (X 1,, X n ) = τ(y i ) Corollary A morphism of affine algebraic sets σ : U V is an isomorphism iff the induced maps between the coordinate rings is an isomorphism. In particular, two affine algebraic sets are isomorphic if and only if there coordinate rings are isomorphic as k-algebras. But note that a bijective morphism σ : U V is not an isomorphism in general. Example Suppose k of characteristic p. Define the morphism f : A 1 k A1 k, t tp which is bijective. Then in the level of coordinate rings, f induces the map f : k[x] k[x], X X p. Clearly, this map is not an isomorphism. Hence, f is not an isomorphism of affine algebraic sets. Example We consider the following map σ : A 1 k A2 k, t (t2, t 3 ). Let C be its image, then C is an affine algebraic set, and σ induces a bijection between A 1 k and C. But this morphism is not an isomorphism. In fact, these two affine algebraic sets are not isomorphic. Exercise Verify the assertions in the previous exercise.

### 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...............................

### 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

### 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

### 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.

### 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

### 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

### 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

### 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

### 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

### 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

### 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...

### 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

### 1. Algebraic vector bundles. Affine Varieties

0. Brief overview Cycles and bundles are intrinsic invariants of algebraic varieties Close connections going back to Grothendieck Work with quasi-projective varieties over a field k Affine Varieties 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

### Projective Varieties. Chapter Projective Space and Algebraic Sets

Chapter 1 Projective Varieties 1.1 Projective Space and Algebraic Sets 1.1.1 Definition. Consider A n+1 = A n+1 (k). The set of all lines in A n+1 passing through the origin 0 = (0,..., 0) is called the

### where m is the maximal ideal of O X,p. Note that m/m 2 is a vector space. Suppose that we are given a morphism

8. Smoothness and the Zariski tangent space We want to give an algebraic notion of the tangent space. In differential geometry, tangent vectors are equivalence classes of maps of intervals in R into the

### 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 ) :

### 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

### Math 203A - Solution Set 1

Math 203A - Solution Set 1 Problem 1. Show that the Zariski topology on A 2 is not the product of the Zariski topologies on A 1 A 1. Answer: Clearly, the diagonal Z = {(x, y) : x y = 0} A 2 is closed in

### 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

### 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

### Pure Math 764, Winter 2014

Compact course notes Pure Math 764, Winter 2014 Introduction to Algebraic Geometry Lecturer: R. Moraru transcribed by: J. Lazovskis University of Waterloo April 20, 2014 Contents 1 Basic geometric objects

### π X : X Y X and π Y : X Y Y

Math 6130 Notes. Fall 2002. 6. Hausdorffness and Compactness. We would like to be able to say that all quasi-projective varieties are Hausdorff and that projective varieties are the only compact varieties.

### Institutionen för matematik, KTH.

Institutionen för matematik, KTH. Contents 7 Affine Varieties 1 7.1 The polynomial ring....................... 1 7.2 Hypersurfaces........................... 1 7.3 Ideals...............................

### 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

### MATH32062 Notes. 1 Affine algebraic varieties. 1.1 Definition of affine algebraic varieties

MATH32062 Notes 1 Affine algebraic varieties 1.1 Definition of affine algebraic varieties We want to define an algebraic variety as the solution set of a collection of polynomial equations, or equivalently,

### Algebraic Geometry. Andreas Gathmann. Notes for a class. taught at the University of Kaiserslautern 2002/2003

Algebraic Geometry Andreas Gathmann Notes for a class taught at the University of Kaiserslautern 2002/2003 CONTENTS 0. Introduction 1 0.1. What is algebraic geometry? 1 0.2. Exercises 6 1. Affine varieties

### 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

### 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

### Math 203A - Solution Set 1

Math 203A - Solution Set 1 Problem 1. Show that the Zariski topology on A 2 is not the product of the Zariski topologies on A 1 A 1. Answer: Clearly, the diagonal Z = {(x, y) : x y = 0} A 2 is closed in

### Arithmetic Algebraic Geometry

Arithmetic Algebraic Geometry 2 Arithmetic Algebraic Geometry Travis Dirle December 4, 2016 2 Contents 1 Preliminaries 1 1.1 Affine Varieties.......................... 1 1.2 Projective Varieties........................

### 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

### 1 Notations and Statement of the Main Results

An introduction to algebraic fundamental groups 1 Notations and Statement of the Main Results Throughout the talk, all schemes are locally Noetherian. All maps are of locally finite type. There two main

### 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)

### 3. The Sheaf of Regular Functions

24 Andreas Gathmann 3. The Sheaf of Regular Functions After having defined affine varieties, our next goal must be to say what kind of maps between them we want to consider as morphisms, i. e. as nice

### Locally Free Sheaves

Locally Free Sheaves Patrick Morandi Algebra Seminar, Spring 2002 In these talks we will discuss several important examples of locally free sheaves and see the connection between locally free sheaves and

### 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

### Algebraic Geometry. Andreas Gathmann. Class Notes TU Kaiserslautern 2014

Algebraic Geometry Andreas Gathmann Class Notes TU Kaiserslautern 2014 Contents 0. Introduction......................... 3 1. Affine Varieties........................ 9 2. The Zariski Topology......................

### AN EXPOSITION OF THE RIEMANN ROCH THEOREM FOR CURVES

AN EXPOSITION OF THE RIEMANN ROCH THEOREM FOR CURVES DOMINIC L. WYNTER Abstract. We introduce the concepts of divisors on nonsingular irreducible projective algebraic curves, the genus of such a curve,

### (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

### div(f ) = D and deg(d) = deg(f ) = d i deg(f i ) (compare this with the definitions for smooth curves). Let:

Algebraic Curves/Fall 015 Aaron Bertram 4. Projective Plane Curves are hypersurfaces in the plane CP. When nonsingular, they are Riemann surfaces, but we will also consider plane curves with singularities.

### ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 4: MORE ABOUT VARIETIES AND REGULAR FUNCTIONS.

ALGERAIC GEOMETRY COURSE NOTES, LECTURE 4: MORE AOUT VARIETIES AND REGULAR FUNCTIONS. ANDREW SALCH. More about some claims from the last lecture. Perhaps you have noticed by now that the Zariski topology

### PROBLEMS, MATH 214A. Affine and quasi-affine varieties

PROBLEMS, MATH 214A k is an algebraically closed field Basic notions Affine and quasi-affine varieties 1. Let X A 2 be defined by x 2 + y 2 = 1 and x = 1. Find the ideal I(X). 2. Prove that the subset

### 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.............

### 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......................................................................

### Adic Spaces. Torsten Wedhorn. June 19, 2012

Adic Spaces Torsten Wedhorn June 19, 2012 This script is highly preliminary and unfinished. It is online only to give the audience of our lecture easy access to it. Therefore usage is at your own risk.

### 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)

### TWO IDEAS FROM INTERSECTION THEORY

TWO IDEAS FROM INTERSECTION THEORY ALEX PETROV This is an expository paper based on Serre s Local Algebra (denoted throughout by [Ser]). The goal is to describe simple cases of two powerful ideas in intersection

### 0.1 Spec of a monoid

These notes were prepared to accompany the first lecture in a seminar on logarithmic geometry. As we shall see in later lectures, logarithmic geometry offers a natural approach to study semistable schemes.

### DIVISORS ON NONSINGULAR CURVES

DIVISORS ON NONSINGULAR CURVES BRIAN OSSERMAN We now begin a closer study of the behavior of projective nonsingular curves, and morphisms between them, as well as to projective space. To this end, we introduce

### ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 9: SCHEMES AND THEIR MODULES.

ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 9: SCHEMES AND THEIR MODULES. ANDREW SALCH 1. Affine schemes. About notation: I am in the habit of writing f (U) instead of f 1 (U) for the preimage of a subset

### Synopsis of material from EGA Chapter II, 4. Proposition (4.1.6). The canonical homomorphism ( ) is surjective [(3.2.4)].

Synopsis of material from EGA Chapter II, 4 4.1. Definition of projective bundles. 4. Projective bundles. Ample sheaves Definition (4.1.1). Let S(E) be the symmetric algebra of a quasi-coherent O Y -module.

### 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.

### COMPLEX VARIETIES AND THE ANALYTIC TOPOLOGY

COMPLEX VARIETIES AND THE ANALYTIC TOPOLOGY BRIAN OSSERMAN Classical algebraic geometers studied algebraic varieties over the complex numbers. In this setting, they didn t have to worry about the Zariski

### Algebraic Geometry Spring 2009

MIT OpenCourseWare http://ocw.mit.edu 18.726 Algebraic Geometry Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 18.726: Algebraic Geometry

### A GLIMPSE OF ALGEBRAIC K-THEORY: Eric M. Friedlander

A GLIMPSE OF ALGEBRAIC K-THEORY: Eric M. Friedlander During the first three days of September, 1997, I had the privilege of giving a series of five lectures at the beginning of the School on Algebraic

### 9. Birational Maps and Blowing Up

72 Andreas Gathmann 9. Birational Maps and Blowing Up In the course of this class we have already seen many examples of varieties that are almost the same in the sense that they contain isomorphic dense

### Algebraic varieties and schemes over any scheme. Non singular varieties

Algebraic varieties and schemes over any scheme. Non singular varieties Trang June 16, 2010 1 Lecture 1 Let k be a field and k[x 1,..., x n ] the polynomial ring with coefficients in k. Then we have two

### CHAPTER 0 PRELIMINARY MATERIAL. Paul Vojta. University of California, Berkeley. 18 February 1998

CHAPTER 0 PRELIMINARY MATERIAL Paul Vojta University of California, Berkeley 18 February 1998 This chapter gives some preliminary material on number theory and algebraic geometry. Section 1 gives basic

### Math 248B. Applications of base change for coherent cohomology

Math 248B. Applications of base change for coherent cohomology 1. Motivation Recall the following fundamental general theorem, the so-called cohomology and base change theorem: Theorem 1.1 (Grothendieck).

### 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.

### ADVANCED TOPICS IN ALGEBRAIC GEOMETRY

ADVANCED TOPICS IN ALGEBRAIC GEOMETRY DAVID WHITE Outline of talk: My goal is to introduce a few more advanced topics in algebraic geometry but not to go into too much detail. This will be a survey of

### Exercises of the Algebraic Geometry course held by Prof. Ugo Bruzzo. Alex Massarenti

Exercises of the Algebraic Geometry course held by Prof. Ugo Bruzzo Alex Massarenti SISSA, VIA BONOMEA 265, 34136 TRIESTE, ITALY E-mail address: alex.massarenti@sissa.it These notes collect a series of

### A finite universal SAGBI basis for the kernel of a derivation. Osaka Journal of Mathematics. 41(4) P.759-P.792

Title Author(s) A finite universal SAGBI basis for the kernel of a derivation Kuroda, Shigeru Citation Osaka Journal of Mathematics. 4(4) P.759-P.792 Issue Date 2004-2 Text Version publisher URL https://doi.org/0.890/838

### 3. Lecture 3. Y Z[1/p]Hom (Sch/k) (Y, X).

3. Lecture 3 3.1. Freely generate qfh-sheaves. We recall that if F is a homotopy invariant presheaf with transfers in the sense of the last lecture, then we have a well defined pairing F(X) H 0 (X/S) F(S)

### ALGEBRAIC GEOMETRY (NMAG401) Contents. 2. Polynomial and rational maps 9 3. Hilbert s Nullstellensatz and consequences 23 References 30

ALGEBRAIC GEOMETRY (NMAG401) JAN ŠŤOVÍČEK Contents 1. Affine varieties 1 2. Polynomial and rational maps 9 3. Hilbert s Nullstellensatz and consequences 23 References 30 1. Affine varieties The basic objects

### 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

### Notes on p-divisible Groups

Notes on p-divisible Groups March 24, 2006 This is a note for the talk in STAGE in MIT. The content is basically following the paper [T]. 1 Preliminaries and Notations Notation 1.1. Let R be a complete

### MATH 233B, FLATNESS AND SMOOTHNESS.

MATH 233B, FLATNESS AND SMOOTHNESS. The discussion of smooth morphisms is one place were Hartshorne doesn t do a very good job. Here s a summary of this week s material. I ll also insert some (optional)

### 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

### Introduction to Arithmetic Geometry Fall 2013 Lecture #23 11/26/2013

18.782 Introduction to Arithmetic Geometry Fall 2013 Lecture #23 11/26/2013 As usual, a curve is a smooth projective (geometrically irreducible) variety of dimension one and k is a perfect field. 23.1

### Algebraic varieties. Chapter A ne varieties

Chapter 4 Algebraic varieties 4.1 A ne varieties Let k be a field. A ne n-space A n = A n k = kn. It s coordinate ring is simply the ring R = k[x 1,...,x n ]. Any polynomial can be evaluated at a point

### Algebraic Geometry Spring 2009

MIT OpenCourseWare http://ocw.mit.edu 18.726 Algebraic Geometry Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 18.726: Algebraic Geometry

### NOTES FOR DRAGOS: MATH 210 CLASS 12, THURS. FEB. 22

NOTES FOR DRAGOS: MATH 210 CLASS 12, THURS. FEB. 22 RAVI VAKIL Hi Dragos The class is in 381-T, 1:15 2:30. This is the very end of Galois theory; you ll also start commutative ring theory. Tell them: midterm

### 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.

### Math 429/581 (Advanced) Group Theory. Summary of Definitions, Examples, and Theorems by Stefan Gille

Math 429/581 (Advanced) Group Theory Summary of Definitions, Examples, and Theorems by Stefan Gille 1 2 0. Group Operations 0.1. Definition. Let G be a group and X a set. A (left) operation of G on X is

### INTRODUCTION TO ALGEBRAIC GEOMETRY. Throughout these notes all rings will be commutative with identity. k will be an algebraically

INTRODUCTION TO ALGEBRAIC GEOMETRY STEVEN DALE CUTKOSKY Throughout these notes all rings will be commutative with identity. k will be an algebraically closed field. 1. Preliminaries on Ring Homomorphisms

### Math 762 Spring h Y (Z 1 ) (1) h X (Z 2 ) h X (Z 1 ) Φ Z 1. h Y (Z 2 )

Math 762 Spring 2016 Homework 3 Drew Armstrong Problem 1. Yoneda s Lemma. We have seen that the bifunctor Hom C (, ) : C C Set is analogous to a bilinear form on a K-vector space, : V V K. Recall that

### Algebraic Geometry. Question: What regular polygons can be inscribed in an ellipse?

Algebraic Geometry Question: What regular polygons can be inscribed in an ellipse? 1. Varieties, Ideals, Nullstellensatz Let K be a field. We shall work over K, meaning, our coefficients of polynomials

### NONSINGULAR CURVES BRIAN OSSERMAN

NONSINGULAR CURVES BRIAN OSSERMAN The primary goal of this note is to prove that every abstract nonsingular curve can be realized as an open subset of a (unique) nonsingular projective curve. Note that

### Outline of the Seminar Topics on elliptic curves Saarbrücken,

Outline of the Seminar Topics on elliptic curves Saarbrücken, 11.09.2017 Contents A Number theory and algebraic geometry 2 B Elliptic curves 2 1 Rational points on elliptic curves (Mordell s Theorem) 5

### FOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 43

FOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 43 RAVI VAKIL CONTENTS 1. Facts we ll soon know about curves 1 1. FACTS WE LL SOON KNOW ABOUT CURVES We almost know enough to say a lot of interesting things about

### Topics in Algebraic Geometry

Topics in Algebraic Geometry Nikitas Nikandros, 3928675, Utrecht University n.nikandros@students.uu.nl March 2, 2016 1 Introduction and motivation In this talk i will give an incomplete and at sometimes

### 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

### This is a closed subset of X Y, by Proposition 6.5(b), since it is equal to the inverse image of the diagonal under the regular map:

Math 6130 Notes. Fall 2002. 7. Basic Maps. Recall from 3 that a regular map of affine varieties is the same as a homomorphism of coordinate rings (going the other way). Here, we look at how algebraic properties

### Matsumura: Commutative Algebra Part 2

Matsumura: Commutative Algebra Part 2 Daniel Murfet October 5, 2006 This note closely follows Matsumura s book [Mat80] on commutative algebra. Proofs are the ones given there, sometimes with slightly more

### The Proj Construction

The Proj Construction Daniel Murfet May 16, 2006 Contents 1 Basic Properties 1 2 Functorial Properties 2 3 Products 6 4 Linear Morphisms 9 5 Projective Morphisms 9 6 Dimensions of Schemes 11 7 Points of

### 12. Linear systems Theorem Let X be a scheme over a ring A. (1) If φ: X P n A is an A-morphism then L = φ O P n

12. Linear systems Theorem 12.1. Let X be a scheme over a ring A. (1) If φ: X P n A is an A-morphism then L = φ O P n A (1) is an invertible sheaf on X, which is generated by the global sections s 0, s

### 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

### 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...............................

### THROUGH THE FIELDS AND FAR AWAY

THROUGH THE FIELDS AND FAR AWAY JONATHAN TAYLOR I d like to thank Prof. Stephen Donkin for helping me come up with the topic of my project and also guiding me through its various complications. Contents

### 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

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

Advanced Algebra II Mar. 2, 2007 (Fri.) 1. commutative ring theory In this chapter, rings are assume to be commutative with identity. 1.1. basic definitions. We recall some basic definitions in the section.

### 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

### Math 797W Homework 4

Math 797W Homework 4 Paul Hacking December 5, 2016 We work over an algebraically closed field k. (1) Let F be a sheaf of abelian groups on a topological space X, and p X a point. Recall the definition

### Introduction to Arithmetic Geometry Fall 2013 Lecture #17 11/05/2013

18.782 Introduction to Arithmetic Geometry Fall 2013 Lecture #17 11/05/2013 Throughout this lecture k denotes an algebraically closed field. 17.1 Tangent spaces and hypersurfaces For any polynomial f k[x

### k k would be reducible. But the zero locus of f in A n+1

Math 145. Bezout s Theorem Let be an algebraically closed field. The purpose of this handout is to prove Bezout s Theorem and some related facts of general interest in projective geometry that arise along

### 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

### Smooth morphisms. Peter Bruin 21 February 2007

Smooth morphisms Peter Bruin 21 February 2007 Introduction The goal of this talk is to define smooth morphisms of schemes, which are one of the main ingredients in Néron s fundamental theorem [BLR, 1.3,

### 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