# 10. Noether Normalization and Hilbert s Nullstellensatz

Size: px
Start display at page:

Transcription

1 10. Noether Normalization and Hilbert s Nullstellensatz Noether Normalization and Hilbert s Nullstellensatz In the last chapter we have gained much understanding for integral and finite ring extensions. We now want to prove an elementary but powerful theorem stating that every finitely generated algebra R over a field K (so in particular every coordinate ring of a variety by Remark 1.31) is a finite extension ring of a polynomial ring K[z 1,...,z r ] and hence of a very simple K-algebra that is easy to deal with. Let us start by giving the geometric idea behind this so-called Noether Normalization theorem, which is in fact very simple. Example 10.1 (Idea of Noether Normalization). Let R = C[x 1,x 2 ]/(x 1 x 2 1) be the coordinate ring of the variety X = V (x 1 x 2 1) A 2 C as in Example 9.4 (b). We know already that R is not integral (and hence not finite) over C[x 1 ]; this is easily seen geometrically in the picture below on the left since this map does not satisfy the Lying Over property for the origin as in Example R coordinate change x 1 = y 2 + y 1 x 2 = y 2 y 1 R C[x 1 ] x 1 x 2 1 = 0 not finite C[y 1 ] y 2 2 y2 1 1 = 0 finite It is easy to change this however by a linear coordinate transformation: if we set e. g. x 1 = y 2 +y 1 and x 2 = y 2 y 1 then we can write R also as R = C[y 1,y 2 ]/(y 2 2 y2 1 1), and this is now finite over C[y 1] by Proposition 9.5 since the polynomial y 2 2 y2 1 1 is monic in y 2. Geometrically, the coordinate transformation has tilted the space X as in the picture above on the right so that e. g. the Lying Over property now obviously holds. Note that this is not special to the particular transformation that we have chosen; in fact, almost any linear coordinate change would have worked to achieve this goal. In terms of geometry, we are therefore looking for a change of coordinates so that a suitable coordinate projection to some affine space A r K then corresponds to a finite ring extension of a polynomial ring over K in r variables. Note that this number r can already be thought of as the dimension of X (a concept that we will introduce in Chapter 11) as finite ring extensions correspond to surjective geometric maps with finite fibers by Example 9.19, and thus should not change the dimension (we will prove this in Lemma 11.8). As we have seen above already, the strategy to achieve our goal is to find a suitable change of coordinates so that the given relations among the variables become monic. The first thing we have to do is therefore to prove that such a change of coordinates is always possible. It turns out that a linear change of coordinates works in general only for infinite fields, whereas for arbitrary fields one has to allow more general coordinate transformations. Lemma Let f K[x 1,...,x n ] be a non-zero polynomial over an infinite field K. Assume that f is homogeneous, i. e. every monomial of f has the same degree (in the sense of Exercise 0.16). Then there are a 1,...,a n 1 K such that f (a 1,...,a n 1,1) Proof. We will prove the lemma by induction on n. The case n = 1 is trivial, since a homogeneous polynomial in one variable is just a constant multiple of a monomial.

2 92 Andreas Gathmann So assume now that n > 1, and write f as f = d i=0 f i x1 i where the f i K[x 2,...,x n ] are homogeneous of degree d i. As f is non-zero, at least one f i has to be non-zero. By induction we can therefore choose a 2,...,a n 1 such that f i (a 2,...,a n 1,1) 0 for this i. But then f (,a 2,...,a n 1,1) K[x 1 ] is a non-zero polynomial, so it has only finitely many zeroes. As K is infinite, we can therefore find a 1 K such that f (a 1,...,a n 1,1) 0. Lemma Let f K[x 1,...,x n ] be a non-zero polynomial over an infinite field K. Then there are λ K and a 1,...,a n 1 K such that λ f (y 1 + a 1 y n,y 2 + a 2 y n,...,y n 1 + a n 1 y n,y n ) K[y 1,...,y n ] is monic in y n (i. e. as an element of R[y n ] with R = K[y 1,...,y n 1 ]). Proof. Let d be the degree of f in the sense of Exercise 0.16, and write f = k1,...,k n c k1,...,k n x k 1 1 xk n n with c k1,...,k n K. Then the leading term of λ f (y 1 + a 1 y n,y 2 + a 2 y n,...,y n 1 + a n 1 y n,y n ) = λ c k1,...,k n (y 1 + a 1 y n ) k1 (y n 1 + a n 1 y n ) k n 1 y k n n k 1,...,k n in y n is obtained by always taking the second summand in the brackets and only keeping the degree-d terms, i. e. it is equal to λ c k1,...,k n a k 1 1 ak n 1 n 1 yk 1+ +k n n = λ f d (a 1,...,a n 1,1)y d n, k 1,...,k n k 1 + +k n =d where f d is the (homogeneous) degree-d part of f. Now pick a 1,...,a n 1 by Lemma 10.2 such that f d (a 1,...,a n 1,1) 0, and set λ = f d (a 1,...,a n 1,1) 1. Exercise Let f K[x 1,...,x n ] be a non-zero polynomial over an arbitrary field K. Prove that there are λ K and a 1,...,a n 1 N such that is monic in y n. λ f (y 1 + y a 1 n,y 2 + y a 2 n,...,y n 1 + y a n 1 n,y n ) K[y 1,...,y n ] Proposition 10.5 (Noether Normalization). Let R be a finitely generated algebra over a field K, with generators x 1,...,x n R. Then there is an injective K-algebra homomorphism K[z 1,...,z r ] R from a polynomial ring over K to R that makes R into a finite extension ring of K[z 1,...,z r ]. Moreover, if K is an infinite field the images of z 1,...,z r in R can be chosen to be K-linear combinations of x 1,...,x n. Proof. We will prove the statement by induction on the number n of generators of R. The case n = 0 is trivial, as we can then choose r = 0 as well. So assume now that n > 0. We have to distinguish two cases: (a) There is no algebraic relation among the x 1,...,x n R, i. e. there is no non-zero polynomial f over K such that f (x 1,...,x n ) = 0 in R. Then we can choose r = n and the map K[z 1,...,z n ] R given by z i x i for all i, which is even an isomorphism in this case. (b) There is a non-zero polynomial f over K such that f (x 1,...,x n ) = 0 in R. Then we choose λ and a 1,...,a n 1 as in Lemma 10.3 (if K is infinite) or Exercise 10.4 (for any K) and set y 1 := x 1 a 1 x n,..., y n 1 := x n 1 a n 1 x n, y n := x n (so that x 1 = y 1 + a 1 y n,..., x n 1 = y n 1 + a n 1 y n, x n = y n ) or y 1 := x 1 x a 1 n,..., y n 1 := x n 1 x a n 1 n, y n := x n (so that x 1 = y 1 + y a 1 n,..., x n 1 = y n 1 + y a n 1 n, x n = y n ),

3 10. Noether Normalization and Hilbert s Nullstellensatz 93 respectively. Note that in both cases these relations show that the K-subalgebra K[y 1,...,y n ] of R generated by y 1,...,y n R is the same as that generated by x 1,...,x n, i. e. all of R. Moreover, y n is integral over the K-subalgebra K[y 1,...,y n 1 ] of R, since λ f (y 1 + a 1 y n,...,y n 1 + a n 1 y n,y n ) or λ f (y 1 + y a 1 n,...,y n 1 + y a n 1 n,y n ), respectively, is monic in y n and equal to λ f (x 1,...,x n ) = 0. Hence R = K[y 1,...,y n ] is finite over K[y 1,...,y n 1 ] by Proposition 9.5. In addition, the subalgebra K[y 1,...,y n 1 ] of R is finite over a polynomial ring K[z 1,...,z r ] by the induction hypothesis, and thus R is finite over K[z 1,...,z r ] by Lemma 9.6 (a). Moreover, if K is infinite we can always choose the coordinate transformation of Lemma 10.3, and thus y 1,...,y n (i. e. also the images of z 1,...,z r by induction) are linear combinations of x 1,...,x n. Remark Let R = A(X) be the coordinate ring of a variety X over a field K. (a) In the Noether Normalization of Proposition 10.5, the (images of) z 1,...,z r in R are algebraically independent functions on X in the sense that there is no polynomial relation among them with coefficients in K. On the other hand, every other element of R is algebraically dependent on z 1,...,z r, i. e. it satisfies a (monic) polynomial relation with coefficients in K[z 1,...,z r ]. We can therefore think of r as the number of parameters needed to describe X, i. e. as the dimension of X as already mentioned in Example In fact, we will see in Remark that the number r in Proposition 10.5 is uniquely determined to be the dimension of X in the sense of Chapter 11. (b) As one would have guessed already from the geometric picture in Example 10.1, the proof of Lemma 10.2 shows that most choices of linear coordinate transformations are suitable to obtain a Noether normalization: in each application of this lemma, only finitely many values of a 1 K have to be avoided. Hence we can translate Proposition 10.5 into geometry by saying that a sufficiently general projection to an r-dimensional linear subspace corresponds to a finite ring extension, and hence to a surjective map with finite fibers (where r is the dimension of X as in (a)). Exercise Find a Noether normalization of the C-algebra C[x,y,z]/(xy+z 2,x 2 y xy 3 +z 4 1). Exercise Let R R be an integral ring extension, and assume that R is a finitely generated algebra over some field K. Moreover, let P 1 P 3 be prime ideals in R and P 1 P 3 be prime ideals in R such that P 1 R = P 1 and P 3 R = P 3. (a) Prove: If there is a prime ideal P 2 in R with P 1 P 2 P 3, then there is also a prime ideal P 2 in R with P 1 P 2 P 3. R : P 1 P 2 P 3 (b) Can we always find P 2 in (a) such that in addition P 2 R = P 2 holds? R: P 1 P 2 P 3 As an important application of Noether Normalization we can now give rigorous proofs of some statements in our dictionary between algebra and geometry, namely of the correspondence between (maximal) ideals in the coordinate ring A(X) of a variety X over an algebraically closed field and subvarieties (resp. points) of X. There are various related statements along these lines, and they are all known in the literature by the German name Hilbert s Nullstellensatz ( theorem of the zeroes ). Let us start with the simplest instance of this family of propositions. Still very algebraic in nature, it is the statement most closely related to Noether Normalization, from which the geometric results will then follow easily. Corollary 10.9 (Hilbert s Nullstellensatz, version 1). Let K be a field, and let R be a finitely generated K-algebra which is also a field. Then K R is a finite field extension. In particular, if in addition K is algebraically closed then R = K.

4 94 Andreas Gathmann Proof. By Noether Normalization as in Proposition 10.5 we know that R is finite over a polynomial ring K[z 1,...,z r ], and thus also integral over K[z 1,...,z r ] by Proposition 9.5. But R is a field, hence K[z 1,...,z r ] must be a field as well by Corollary 9.21 (a). This is only the case for r = 0, and so R is finite over K. In particular, if K is algebraically closed then there are no algebraic extension fields of K since all zeroes of polynomials over K lie already in K. Hence by Proposition 9.5 there are no finite extensions either in this case, and we must have R = K. Corollary (Hilbert s Nullstellensatz, version 2). Let K be an algebraically closed field. Then all maximal ideals of the polynomial ring K[x 1,...,x n ] are of the form for some a = (a 1,...,a n ) A n K. I(a) = (x 1 a 1,...,x n a n ) 19 Proof. Let P K[x 1,...,x n ] be a maximal ideal. Then K[x 1,...,x n ]/P is a field by Lemma 2.3 (b), and in addition certainly a finitely generated K-algebra. Hence K[x 1,...,x n ]/P = K by Corollary 10.9, i. e. the natural map K K[x 1,...,x n ]/P, c c is an isomorphism. Choosing inverse images a 1,...,a n of x 1,...,x n we get x i = a i for all i, and thus (x 1 a 1,...,x n a n ) P. But (x 1 a 1,...,x n a n ) is a maximal ideal by Example 2.6 (c), and so we must already have P = (x 1 a 1,...,x n a n ) = I(a). Remark (Points of a variety correspond to maximal ideals). It is easy to extend Corollary to a statement about an arbitrary variety X A n K over an algebraically closed field K: if R = A(X) is the coordinate ring of X we claim that there is a one-to-one correspondence {points of X} 1:1 a I(a) V (I) I. {maximal ideals in A(X)} In fact, the maximal ideals of A(X) = K[x 1,...,x n ]/I(X) are in one-to-one correspondence with maximal ideals I K[x 1,...,x n ] such that I I(X) by Lemma 1.21 and Corollary 2.4. By Corollary this is the same as ideals of the form I(a) = (x 1 a 1,...,x n a n ) containing I(X). But I(a) I(X) is equivalent to a X by Lemma 0.9 (a) and (c), so the result follows. Remark (Zeroes of ideals in K[x 1,...,x n ]). Another common reformulation of Hilbert s Nullstellensatz is that every proper ideal I K[x 1,...,x n ] in the polynomial ring over an algebraically closed field K has a zero: by Corollary 2.17 we know that I is contained in a maximal ideal, which must be of the form I(a) by Corollary But I I(a) implies a V (I) by Lemma 0.9 (a) and (c), and hence V (I) /0. Note that this statement is clearly false over fields that are not algebraically closed, as e. g. (x 2 + 1) is a proper ideal in R[x] with empty zero locus in A 1 R. In order to extend the correspondence between points and maximal ideals to arbitrary subvarieties we need another algebraic preliminary result first: recall that in any ring R the radical of an ideal I equals the intersection of all prime ideals containing I by Lemma We will show now that it is in fact sufficient to intersect all maximal ideals containing I if R is a finitely generated algebra over a field. Corollary (Hilbert s Nullstellensatz, version 3). For every ideal I in a finitely generated algebra R over a field K we have I = P. P maximal P I Proof. The inclusion follows immediately from Lemma 2.21, since every maximal ideal is prime.

5 10. Noether Normalization and Hilbert s Nullstellensatz 95 For the opposite inclusion, let f R with f / I; we have to find a maximal ideal P I with f / P. Consider the multiplicatively closed set S = { f n : n N}. As f / I implies I S = /0, we get by Exercise 6.14 (a) a prime ideal P R with P I and P S = /0, in particular with f / P. Moreover, we can assume by Exercise 6.14 (a) that S 1 P is maximal. It only remains to show that P is maximal. To do this, consider the ring extension K R/P (R/P) f = R f /P f, where the subscript f denotes localization at S as in Example 6.5 (c). Note that the second map is in fact an inclusion since R/P is an integral domain, and the stated equality holds by Corollary 6.22 (b). Moreover, R f /P f is a field since P f is maximal, and finitely generated as a K-algebra (as generators we can choose the classes of generators for R together with 1 f ). So K R f /P f is a finite field extension by Corollary 10.9, and hence integral by Proposition 9.5. But then R/P R f /P f is integral as well, which means by Corollary 9.21 (a) that R/P is a field since R f /P f is. Hence P is maximal by Lemma 2.3 (b). Corollary (Hilbert s Nullstellensatz, version 4). Let X A n K be a variety over an algebraically closed field K. Then for every ideal I A(X) we have I(V (I)) = I. In particular, there is a one-to-one correspondence {subvarieties of X} 1:1 Y I(Y ) V (I) I. Proof. Let us first prove the equality I(V (I)) = I. {radical ideals in A(X)} : Assume that f / I. By Corollary there is then a maximal ideal P A(X) with P I and f / P. But by Remark this maximal ideal has to be of the form I(a) = (x 1 a 1,...,x n a n ) for some point a X. Now I(a) I implies a V (I) by Lemma 0.9 (a) and (c), and f / I(a) means f (a) 0. Hence f / I(V (I)). : Let f I, i. e. f n I for some n N. Then ( f (a)) n = 0, and hence f (a) = 0, for all a V (I). This means that f I(V (I)). The one-to-one correspondence now follows immediately from what we already know: the two maps are well-defined since I(Y ) is always radical by Remark 1.10, and they are inverse to each other by Lemma 0.9 (c) and the statement I(V (I)) = I proven above.

### 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

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

### 9. Integral Ring Extensions

80 Andreas Gathmann 9. Integral ing Extensions In this chapter we want to discuss a concept in commutative algebra that has its original motivation in algebra, but turns out to have surprisingly many applications

### 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

### 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

### 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

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

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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.

### 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

### 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

### 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

### 2. Intersection Multiplicities

2. Intersection Multiplicities 11 2. Intersection Multiplicities Let us start our study of curves by introducing the concept of intersection multiplicity, which will be central throughout these notes.

### 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

### 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 ) :

### 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

### 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

### (dim Z j dim Z j 1 ) 1 j i

Math 210B. Codimension 1. Main result and some interesting examples Let k be a field, and A a domain finitely generated k-algebra. In class we have seen that the dimension theory of A is linked to the

### 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

### 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

### AN INTRODUCTION TO AFFINE SCHEMES

AN INTRODUCTION TO AFFINE SCHEMES BROOKE ULLERY Abstract. This paper gives a basic introduction to modern algebraic geometry. The goal of this paper is to present the basic concepts of algebraic geometry,

### 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

### 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

### Extension theorems for homomorphisms

Algebraic Geometry Fall 2009 Extension theorems for homomorphisms In this note, we prove some extension theorems for homomorphisms from rings to algebraically closed fields. The prototype is the following

### 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)

### 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

### ne varieties (continued)

Chapter 2 A ne varieties (continued) 2.1 Products For some problems its not very natural to restrict to irreducible varieties. So we broaden the previous story. Given an a ne algebraic set X A n k, we

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

### 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.

### 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

### 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

### 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

### 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

### 4.5 Hilbert s Nullstellensatz (Zeros Theorem)

4.5 Hilbert s Nullstellensatz (Zeros Theorem) We develop a deep result of Hilbert s, relating solutions of polynomial equations to ideals of polynomial rings in many variables. Notation: Put A = F[x 1,...,x

### 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

### 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.

### 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

### 4. Noether normalisation

4. Noether normalisation We shall say that a ring R is an affine ring (or affine k-algebra) if R is isomorphic to a polynomial ring over a field k with finitely many indeterminates modulo an ideal, i.e.,

### 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

### Chapter 1. Affine algebraic geometry. 1.1 The Zariski topology on A n

Chapter 1 Affine algebraic geometry We shall restrict our attention to affine algebraic geometry, meaning that the algebraic varieties we consider are precisely the closed subvarieties of affine n- space

### Introduction to Arithmetic Geometry Fall 2013 Lecture #15 10/29/2013

18.782 Introduction to Arithmetic Geometry Fall 2013 Lecture #15 10/29/2013 As usual, k is a perfect field and k is a fixed algebraic closure of k. Recall that an affine (resp. projective) variety is an

### 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,

### (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

### Algebraic Geometry: MIDTERM SOLUTIONS

Algebraic Geometry: MIDTERM SOLUTIONS C.P. Anil Kumar Abstract. Algebraic Geometry: MIDTERM 6 th March 2013. We give terse solutions to this Midterm Exam. 1. Problem 1: Problem 1 (Geometry 1). When is

### 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.

### LECTURE Affine Space & the Zariski Topology. It is easy to check that Z(S)=Z((S)) with (S) denoting the ideal generated by elements of S.

LECTURE 10 1. Affine Space & the Zariski Topology Definition 1.1. Let k a field. Take S a set of polynomials in k[t 1,..., T n ]. Then Z(S) ={x k n f(x) =0, f S}. It is easy to check that Z(S)=Z((S)) with

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

### Math 40510, Algebraic Geometry

Math 40510, Algebraic Geometry Problem Set 1, due February 10, 2016 1. Let k = Z p, the field with p elements, where p is a prime. Find a polynomial f k[x, y] that vanishes at every point of k 2. [Hint:

### 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

### 3.1. Derivations. Let A be a commutative k-algebra. Let M be a left A-module. A derivation of A in M is a linear map D : A M such that

ALGEBRAIC GROUPS 33 3. Lie algebras Now we introduce the Lie algebra of an algebraic group. First, we need to do some more algebraic geometry to understand the tangent space to an algebraic variety at

### NOTES ON FIBER DIMENSION

NOTES ON FIBER DIMENSION SAM EVENS Let φ : X Y be a morphism of affine algebraic sets, defined over an algebraically closed field k. For y Y, the set φ 1 (y) is called the fiber over y. In these notes,

### THE ALGEBRAIC GEOMETRY DICTIONARY FOR BEGINNERS. Contents

THE ALGEBRAIC GEOMETRY DICTIONARY FOR BEGINNERS ALICE MARK Abstract. This paper is a simple summary of the first most basic definitions in Algebraic Geometry as they are presented in Dummit and Foote ([1]),

### 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

### 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

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

### NOTES ON LINEAR ALGEBRA OVER INTEGRAL DOMAINS. Contents. 1. Introduction 1 2. Rank and basis 1 3. The set of linear maps 4. 1.

NOTES ON LINEAR ALGEBRA OVER INTEGRAL DOMAINS Contents 1. Introduction 1 2. Rank and basis 1 3. The set of linear maps 4 1. Introduction These notes establish some basic results about linear algebra over

### Lecture 11 The Radical and Semisimple Lie Algebras

18.745 Introduction to Lie Algebras October 14, 2010 Lecture 11 The Radical and Semisimple Lie Algebras Prof. Victor Kac Scribe: Scott Kovach and Qinxuan Pan Exercise 11.1. Let g be a Lie algebra. Then

### 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.

### 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

### The Geometry-Algebra Dictionary

Chapter 1 The Geometry-Algebra Dictionary This chapter is an introduction to affine algebraic geometry. Working over a field k, we will write A n (k) for the affine n-space over k and k[x 1,..., x n ]

### A MODEL-THEORETIC PROOF OF HILBERT S NULLSTELLENSATZ

A MODEL-THEORETIC PROOF OF HILBERT S NULLSTELLENSATZ NICOLAS FORD Abstract. The goal of this paper is to present a proof of the Nullstellensatz using tools from a branch of logic called model theory. In

### TROPICAL SCHEME THEORY

TROPICAL SCHEME THEORY 5. Commutative algebra over idempotent semirings II Quotients of semirings When we work with rings, a quotient object is specified by an ideal. When dealing with semirings (and lattices),

### 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

### 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

### 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

### Algebraic geometry of the ring of continuous functions

Algebraic geometry of the ring of continuous functions Nicolas Addington October 27 Abstract Maximal ideals of the ring of continuous functions on a compact space correspond to points of the space. For

### 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

### 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

### 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

### 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

### MATH 631: ALGEBRAIC GEOMETRY: HOMEWORK 1 SOLUTIONS

MATH 63: ALGEBRAIC GEOMETRY: HOMEWORK SOLUTIONS Problem. (a.) The (t + ) (t + ) minors m (A),..., m k (A) of an n m matrix A are polynomials in the entries of A, and m i (A) = 0 for all i =,..., k if and

### 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.

### 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 6.

Math 203A, Solution Set 6. Problem 1. (Finite maps.) Let f 0,..., f m be homogeneous polynomials of degree d > 0 without common zeros on X P n. Show that gives a finite morphism onto its image. f : X P

### Lecture 6. s S} is a ring.

Lecture 6 1 Localization Definition 1.1. Let A be a ring. A set S A is called multiplicative if x, y S implies xy S. We will assume that 1 S and 0 / S. (If 1 / S, then one can use Ŝ = {1} S instead of

### Polynomials, Ideals, and Gröbner Bases

Polynomials, Ideals, and Gröbner Bases Notes by Bernd Sturmfels for the lecture on April 10, 2018, in the IMPRS Ringvorlesung Introduction to Nonlinear Algebra We fix a field K. Some examples of fields

### 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...

### Basic facts and definitions

Synopsis Thursday, September 27 Basic facts and definitions We have one one hand ideals I in the polynomial ring k[x 1,... x n ] and subsets V of k n. There is a natural correspondence. I V (I) = {(k 1,

### Plane Algebraic Curves

Plane Algebraic Curves Andreas Gathmann Class Notes TU Kaiserslautern 2018 Contents 0. Introduction......................... 3 1. Affine Curves......................... 6 2. Intersection Multiplicities.....................

### 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

### FOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 37

FOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 37 RAVI VAKIL CONTENTS 1. Motivation and game plan 1 2. The affine case: three definitions 2 Welcome back to the third quarter! The theme for this quarter, insofar

### 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)

### POLYNOMIAL IDENTITY RINGS AS RINGS OF FUNCTIONS

POLYNOMIAL IDENTITY RINGS AS RINGS OF FUNCTIONS Z. REICHSTEIN AND N. VONESSEN Abstract. We generalize the usual relationship between irreducible Zariski closed subsets of the affine space, their defining

### Pacific Journal of Mathematics

Pacific Journal of Mathematics GROUP ACTIONS ON POLYNOMIAL AND POWER SERIES RINGS Peter Symonds Volume 195 No. 1 September 2000 PACIFIC JOURNAL OF MATHEMATICS Vol. 195, No. 1, 2000 GROUP ACTIONS ON POLYNOMIAL

### Definitions. Notations. Injective, Surjective and Bijective. Divides. Cartesian Product. Relations. Equivalence Relations

Page 1 Definitions Tuesday, May 8, 2018 12:23 AM Notations " " means "equals, by definition" the set of all real numbers the set of integers Denote a function from a set to a set by Denote the image of

### Math 249B. Nilpotence of connected solvable groups

Math 249B. Nilpotence of connected solvable groups 1. Motivation and examples In abstract group theory, the descending central series {C i (G)} of a group G is defined recursively by C 0 (G) = G and C

### 7. Localization. (d 1, m 1 ) (d 2, m 2 ) d 3 D : d 3 d 1 m 2 = d 3 d 2 m 1. (ii) If (d 1, m 1 ) (d 1, m 1 ) and (d 2, m 2 ) (d 2, m 2 ) then

7. Localization To prove Theorem 6.1 it becomes necessary to be able to a enominators to rings (an to moules), even when the rings have zero-ivisors. It is a tool use all the time in commutative algebra,

### 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)

### 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