Spring 2016, lecture notes by Maksym Fedorchuk 51


 Shannon Sharp
 4 years ago
 Views:
Transcription
1 Spring 2016, lecture notes by Maksym Fedorchuk 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) The radical of an ideal I is the intersection of all prime ideals of R containing I. Proof. For the quotient homomorphism : R! R/I, we have 1 nilrad(r/i) = {f 2 R (f) n = 0 for some n 2 N} = {f 2 R f n 2 I for some n 2 N} = rad(i). Therefore, the second claim follows from the first one, which we proceed to prove. The inclusion nilrad(r) \ p p2spec R is obvious, for f n = 0 implies that f 2 p for every prime p. Suppose f is not nilpotent. Then S = {f n } 1 n=0 is a multiplicative set not containing 0. Hence by Problem 2.6, there exists a prime ideal p disjoint from S. (See below for details.) In particular, f/2 p. This finishes the proof. Problem. Let k be a field. Describe the contraction of the maximal ideal (x 1,y 1) under the ring homomorphism f : k[z 1,z 2,z 3 ]! k[x, y] given by z 1 7! x 2, z 2 7! xy, z 3 7! y 2. We will do this problem by proving a much more general statement: Proposition Let R = k[z 1,...,z m ]/I and S = k[x 1,...,x n ]/J be finitely generated algebras over a field k. Suppose f : R! S is a kalgebra homomorphism defined by f(z i )= i (x 1,...,x n ) for some polynomials i 2 k[x 1,...,x n ]. (Note that 1,..., m have to satisfy F ( 1,..., m) 2 J for every F 2 I.) Then for every (a 1,...,a n ) 2 k n such that J (x 1 a 1,...,x n a n ), the contraction of the induced maximal ideal (x 1 a 1,...,x n a n ) S is a maximal ideal z 1 1 (a 1,...,a n ),...,z n n (a 1,...,a n ) R. Lemma Every kalgebra homomorphism ev a : k[x 1,...,x n ]! k, is defined by ev a (x i )=a i for some a =(a 1,...,a n ) 2 k n and we have ker(ev a )= (x 1 a 1,...,x n a n ). In particular, (x 1 a 1,...,x n a n ) is a maximal ideal for every (a 1,...,a n ) 2 k n.
2 52 Algebra II Commutative and Homological Algebra Proof. This follows from two observations. The first is that ev a F (x 1,...,x n ) = F (a 1,...,a n ) for every F 2 k[x 1,...,x n ]. The second is that F (x 1,...,x n )=F (a 1,...,a n ) mod (x 1 a 1,...,x n a n ). N.B. The above follows from the existence of the Taylor expansion of F (x 1,...,x n ) around (a 1,...,a n ). N.B. A kalgebra homomorphism from a finitely generated kalgebras to k is called an evaluation homomorphism. Lemma completely describes such homomorphisms. Proof. By Lemma , we have that (x 1 a 1,...,x n a n ) S is the kernel of some evaluation homomorphism ev a : S! k defined by ev a (x i )=a i. Consider now the composition homomorphism R f! S ev! k. Then ev a f is the evaluation homomorphism from R to k defined by z i 7! i(a 1,...,a n ). In particular, z 1 1 (a 1,...,a n ),...,z n n (a 1,...,a n ) = ker(ev f), again by Lemma This finishes the proof. Problem. Suppose k is a field (not necessarily algebraically closed; the case of algebraically closed field was done in class). For a homomorphism of finitely generated kalgebras R! S, prove that the contraction of a maximal ideal in S is a maximal ideal in R. Proof. Let n S be a maximal ideal and K = S/n be the corresponding quotient field. Then k K is a finitely generated field extension, hence k K is a finite field extension by Weak Nullstellensatz (Proposition 1.4.1). Let m := n \ R be the contraction of n in R. Then we have a sequence of ring extensions: k R/m K. Since k K is a finite ring extension, we must have that R/m K is a finite ring extension. But both of these rings are domains and K is a field. It follows that R/m is a field by Proposition Problem. Finish the proof of Proposition from the lecture notes by showing that for an irreducible a ne variety X k n, the ideal of polynomials vanishing on X is prime. Proof. Suppose that X is an irreducible a ne variety but the ideal I := I(X) is not prime. Then there exist f,g 2 k[x 1,...,x n ] such that f,g /2 I(X), but fg 2 I(X). Let X 1 = Z(I,f) Z(I(X)) = X and X 2 = Z(I,g) Z(I(X)) = X be algebraic sets. But X 1 [ X 2 = Z (I,f)(I,g) Z(I) =X. It follows that X 1 [ X 2 = X. But because X is irreducible, we conclude that either X 1 = X and X 2 = X. Suppose X 1 = X. Then rad(i,f) = rad(i) =I. (Note that I is automatically radical.) But then f 2 I. A contradiction!
3 Spring 2016, lecture notes by Maksym Fedorchuk 53 Problem. Let k be a field. For integers 1 apple m<n, consider the matrix x 1 x 2 x m 1 x m (10.2.3) x 1 x 2 x 3 x m x m+1 M m,n := B A. x n m x n 1 x n Let I n,m be the ideal in k[x 1,...,x n ] generated by the 2 2 minors of M m,n.prove that Z(I n,m )={(t, t 2,t 3,...,t n ) t 2 k} k n. Proof. Note that for i apple m and j apple n m, the polynomial x i+j x i x j belong to I n,m. Suppose (a 1,...,a n ) 2 Z(I n,m ). Then we must have a i+j = a i a j for all i apple m and j apple n m. It follows that a r = a r 1 for every 1 apple r apple n. Problem. For any ideal I R and a multiplicative set S disjoint from I, prove that there exists a prime ideal p I disjoint from S. Proof. Let R := R/I and let S be the image of S in R. ThenS is a multiplicative set in R that does not contain zero. In particular, the ring S 1 R is not a zero ring. Let q be a prime ideal in S 1 R. Then the contraction of q to R is a prime ideal avoiding S and containing I. (Alternatively, one can apply Zorn s lemma to the set of the ideals containing I and disjoint from S.) Problem. Let p be a prime ideal of a ring R. Prove that the residue field of R p is the fraction field of R/p: R p /pr p = Frac(R/p). Proof. Let : R! R/p be the quotient homomorphism. There is an obvious ring homomorphism : R p! Frac(R/p) that sends a/s 2 R p to (a)/ (s). It is easy to see that is welldefined (do this explicitly if you are still shaky with localization.) Clearly, is surjective: Every element of R/p can be written as (a) for a 2 R and every nonzero element of R/p can be written as (s) for s/2 p. Finally, (a/s) = 0 if and only if (a) = 0 if and only if a 2 p if and only if a/s 2 pr p. Hence ker = pr p. We are done. Problem. For an Rmodule M and a multiplicative set S R, prove that S 1 R R M ' S 1 M. Proof. We have an obvious Rbilinear map S 1 R M! S 1 M given by r s,m 7! rm s. By the universal property of the tensor product, this gives rise to an Rlinear homomorphism f : S 1 R R M ' S 1 M,
4 54 Algebra II Commutative and Homological Algebra r that satisfies f s m = rm s. In fact, it is easy to see that f is also S 1 Rlinear. It is a tedious but straightforward exercise to check that m g : S 1! S 1 R R M, g = 1 s s m is a welldefined Rmodule homomorphism (which is also an S 1 Rmodule homomorphism). Clearly, f and g are inverses of each other and so define an S 1 Rmodule isomorphism S 1 R R M ' S 1 M. Problem. For a finitely generated Rmodule M and a multiplicative set S R, prove that S 1 M = 0 if and only if there exists s 2 S such that sm = 0. Proof. If sm = 0 for some s 2 S, then for every element x = m/t 2 S 1 M,we have x = m t = sm st = 0 st =02 S 1 M. Conversely, suppose S 1 M = 0. Then for every m 2 M, there exists an element s m 2 S such that s m m = 0. (This follows by unwinding what it means for m = m/1 tobezeroins 1 M.) By the assumption, we can choose finitely many generators m 1,...,m n of M. Then it is easy to see that s := s m1 s mn satisfies sm = 0. Problem. Let R be a ring. Suppose that for each prime p, the local ring R p has no nonzero nilpotent elements. Prove that R has no nonzero nilpotent elements. If each R p is an integral domain, is R necessarily an integral domain? Proof. Suppose r 2 R is a nilpotent element. Then r = r remains nilpotent in 1 R p.thuswemusthave r 1 =02 R p, for every prime p. Thus for each primes p, there exists an element in R \ p that kills r. In other words, the annihilator of r Ann(r) :={a 2 R ar =0} is not entirely contained in any of the prime ideals. Since Ann(r) is an ideal itself, we conclude that Ann(r) =R = (1) and so r = 0.
5 Spring 2016, lecture notes by Maksym Fedorchuk 55 Example: Consider a product of two domains, say R = S T. Then a prime ideal in S T is of the form either S q, whereq T is prime, or p T, where p S is prime. It is easy to see that (S T ) S q ' T q. (Indeed, the complement of S q contains 0 1, which allows to kill all nonzero elements of S.) It follows that the localization of R is a domain for all primes of R, butr is clearly not a domain. Definition A ring R with no nonzero nilpotent elements (that is, satisfying nilrad(r) = (0)) is called reduced. Problem. Let R be a ring and S R a multiplicative set. For a ring homomorphism : R! S 1 R, prove that ] (Spec S 1 R)= \ Spec R f. Is the above set necessarily open? Proof. By definition, ] (Spec S 1 R) is the set of prime ideals in R that are contractions of prime ideals in S 1 R. The distinguished open Spec R f Spec R is defined to be the complement of V(f), and so consists of those prime ideals in R that do not contain f. A prime ideal p R is a contraction of a prime ideal in S 1 R if and only if S \ p = ; if and only if f /2 p for every f 2 S if and only if p 2 T f2s Spec R f. This finishes the proof. If we take p to be a prime ideal and S = R\p, then T f2s Spec R f is the set of all prime ideals contained in p. Such a set is generally not open in Zariski topology. For example, if R = Z and p =(p) for a prime p, thenspecr p = {(0), (p)}, which is not open in Spec Z because its complement is not closed. Another example would be R = k[x] and p =(x), etc. f2s
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
More informationALGEBRAIC 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
More informationMath 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
More informationCommutative 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...............................
More information4.4 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)
More informationCommutative Algebra. B Totaro. Michaelmas Basics Rings & homomorphisms Modules Prime & maximal ideals...
Commutative Algebra B Totaro Michaelmas 2011 Contents 1 Basics 2 1.1 Rings & homomorphisms................... 2 1.2 Modules............................. 4 1.3 Prime & maximal ideals....................
More informationCourse 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...............................
More informationIntegral 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
More informationExtended Index. 89f depth (of a prime ideal) 121f ArtinRees 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
More informationNOTES IN COMMUTATIVE ALGEBRA: PART 2
NOTES IN COMMUTATIVE ALGEBRA: PART 2 KELLER VANDEBOGERT 1. Completion of a Ring/Module Here we shall consider two seemingly different constructions for the completion of a module and show that indeed they
More information1.5 The Nil and Jacobson Radicals
1.5 The Nil and Jacobson Radicals The idea of a radical of a ring A is an ideal I comprising some nasty piece of A such that A/I is wellbehaved or tractable. Two types considered here are the nil and
More informationCHAPTER 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
More informationThe 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
More informationAlgebraic 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
More information12. 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
More informationALGEBRA HW 4. M 0 is an exact sequence of Rmodules, 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 Rmodules, then M is Noetherian if and only if M and M are. Proof. ( ) Suppose M is Noetherian. Then M injects into
More informationMATH 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
More informationReid 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
More informationIn Theorem 2.2.4, we generalized a result about field extensions to rings. Here is another variation.
Chapter 3 Valuation Rings The results of this chapter come into play when analyzing the behavior of a rational function defined in the neighborhood of a point on an algebraic curve. 3.1 Extension Theorems
More informationAssigned homework problems S. L. Kleiman, fall 2008
18.705 Assigned homework problems S. L. Kleiman, fall 2008 Problem Set 1. Due 9/11 Problem R 1.5 Let ϕ: A B be a ring homomorphism. Prove that ϕ 1 takes prime ideals P of B to prime ideals of A. Prove
More informationADVANCED 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.
More informationCommutative 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.....................
More informationALGEBRAIC 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.
More informationAlgebraic 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 ) :
More informationHARTSHORNE 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
More information3. 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
More informationCOURSE SUMMARY FOR MATH 504, FALL QUARTER : MODERN ALGEBRA
COURSE SUMMARY FOR MATH 504, FALL QUARTER 20178: MODERN ALGEBRA JAROD ALPER Week 1, Sept 27, 29: Introduction to Groups Lecture 1: Introduction to groups. Defined a group and discussed basic properties
More informationRings 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...
More informationV (f) :={[x] 2 P n ( ) f(x) =0}. If (x) ( x) thenf( x) =
20 KIYOSHI IGUSA BRANDEIS UNIVERSITY 2. Projective varieties For any field F, the standard definition of projective space P n (F ) is that it is the set of one dimensional F vector subspaces of F n+.
More informationMATH 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......................................................................
More informationLecture 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
More informationInjective Modules and Matlis Duality
Appendix A Injective Modules and Matlis Duality Notes on 24 Hours of Local Cohomology William D. Taylor We take R to be a commutative ring, and will discuss the theory of injective Rmodules. The following
More informationWritten Homework # 2 Solution
Math 517 Spring 2007 Radford Written Homework # 2 Solution 02/23/07 Throughout R and S are rings with unity; Z denotes the ring of integers and Q, R, and C denote the rings of rational, real, and complex
More informationMath 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
More informationMATH 101B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA
MATH 101B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA These are notes for our first unit on the algebraic side of homological algebra. While this is the last topic (Chap XX) in the book, it makes sense to
More information10. 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
More informationABSTRACT NONSINGULAR CURVES
ABSTRACT NONSINGULAR CURVES Affine Varieties Notation. Let k be a field, such as the rational numbers Q or the complex numbers C. We call affine nspace the collection A n k of points P = a 1, a,..., a
More informationLecture 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 Alinear map. Then ker φ = {m M : φ(m) = 0} is a submodule of M and im φ is
More informationREPRESENTATION THEORY, LECTURE 0. BASICS
REPRESENTATION THEORY, LECTURE 0. BASICS IVAN LOSEV Introduction The aim of this lecture is to recall some standard basic things about the representation theory of finite dimensional algebras and finite
More information10. 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.
More informationALGEBRAIC 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
More informationWEAK NULLSTELLENSATZ
WEAK NULLSTELLENSATZ YIFAN WU, wuyifan@umich.edu Abstract. We prove weak Nullstellensatz which states if a finitely generated k algebra is a field, then it is a finite algebraic field extension of k. We
More informationFormal power series rings, inverse limits, and Iadic completions of rings
Formal power series rings, inverse limits, and Iadic completions of rings Formal semigroup rings and formal power series rings We next want to explore the notion of a (formal) power series ring in finitely
More informationPrimary Decomposition and Associated Primes
Chapter 1 Primary Decomposition and Associated Primes 1.1 Primary Submodules and Ideals 1.1.1 Definitions and Comments If N is a submodule of the Rmodule M, and a R, let λ a : M/N M/N be multiplication
More informationChapter 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
More informationCRing Project, Chapter 5
Contents 5 Noetherian rings and modules 3 1 Basics........................................... 3 1.1 The noetherian condition............................ 3 1.2 Stability properties................................
More informationCOMMUNICATIONS IN ALGEBRA, 15(3), (1987) A NOTE ON PRIME IDEALS WHICH TEST INJECTIVITY. John A. Beachy and William D.
COMMUNICATIONS IN ALGEBRA, 15(3), 471 478 (1987) A NOTE ON PRIME IDEALS WHICH TEST INJECTIVITY John A. Beachy and William D. Weakley Department of Mathematical Sciences Northern Illinois University DeKalb,
More informationHomological Methods in Commutative Algebra
Homological Methods in Commutative Algebra Olivier Haution LudwigMaximiliansUniversität München Sommersemester 2017 1 Contents Chapter 1. Associated primes 3 1. Support of a module 3 2. Associated primes
More informationMath 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
More informationA Primer on Homological Algebra
A Primer on Homological Algebra Henry Y Chan July 12, 213 1 Modules For people who have taken the algebra sequence, you can pretty much skip the first section Before telling you what a module is, you probably
More information(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 kalgebra. In class we have seen that the dimension theory of A is linked to the
More information11. 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
More informationLecture 7.3: Ring homomorphisms
Lecture 7.3: Ring homomorphisms Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4120, Modern Algebra M. Macauley (Clemson) Lecture 7.3:
More informationPROBLEMS, MATH 214A. Affine and quasiaffine varieties
PROBLEMS, MATH 214A k is an algebraically closed field Basic notions Affine and quasiaffine 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
More informationExercises MAT2200 spring 2014 Ark 5 Rings and fields and factorization of polynomials
Exercises MAT2200 spring 2014 Ark 5 Rings and fields and factorization of polynomials This Ark concerns the weeks No. (Mar ) andno. (Mar ). Status for this week: On Monday Mar : Finished section 23(Factorization
More informationSummer 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
More informationALGEBRAIC 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
More informationCohomology and Base Change
Cohomology and Base Change Let A and B be abelian categories and T : A B and additive functor. We say T is halfexact if whenever 0 M M M 0 is an exact sequence of Amodules, the sequence T (M ) T (M)
More informationON STRONGLY PRIME IDEALS AND STRONGLY ZERODIMENSIONAL RINGS. Christian Gottlieb
ON STRONGLY PRIME IDEALS AND STRONGLY ZERODIMENSIONAL RINGS Christian Gottlieb Department of Mathematics, University of Stockholm SE106 91 Stockholm, Sweden gottlieb@math.su.se Abstract A prime ideal
More informationNOTES ON ALGEBRAIC GEOMETRY MATH 202A. Contents Introduction Affine varieties 22
NOTES ON ALGEBRAIC GEOMETRY MATH 202A KIYOSHI IGUSA BRANDEIS UNIVERSITY Contents Introduction 1 1. Affine varieties 2 1.1. Weak Nullstellensatz 2 1.2. Noether s normalization theorem 2 1.3. Nullstellensatz
More informationLECTURES ON ALGEBRAIC GEOMETRY MATH 202A
LECTURES ON ALGEBRAIC GEOMETRY MATH 202A KIYOSHI IGUSA BRANDEIS UNIVERSITY Contents Introduction 1 Overfield 1 1. A ne varieties 2 1.1. Lecture 1: Weak Nullstellensatz 2 1.2. Lecture 2: Noether s normalization
More informationTROPICAL 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),
More information2. 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 socalled prime and maximal ideals. Let
More informationNOTES FOR COMMUTATIVE ALGEBRA M5P55
NOTES FOR COMMUTATIVE ALGEBRA M5P55 AMBRUS PÁL 1. Rings and ideals Definition 1.1. A quintuple (A, +,, 0, 1) is a commutative ring with identity, if A is a set, equipped with two binary operations; addition
More informationne 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
More informationExploring the Exotic Setting for Algebraic Geometry
Exploring the Exotic Setting for Algebraic Geometry Victor I. Piercey University of Arizona Integration Workshop Project August 610, 2010 1 Introduction In this project, we will describe the basic topology
More informationRings With Topologies Induced by Spaces of Functions
Rings With Topologies Induced by Spaces of Functions Răzvan Gelca April 7, 2006 Abstract: By considering topologies on Noetherian rings that carry the properties of those induced by spaces of functions,
More informationAlgebraic 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
More informationLecture 4. Corollary 1.2. If the set of all nonunits is an ideal in A, then A is local and this ideal is the maximal one.
Lecture 4 1. General facts Proposition 1.1. Let A be a commutative ring, and m a maximal ideal. Then TFAE: (1) A has only one maximal ideal (i.e., A is local); (2) A \ m consists of units in A; (3) For
More informationPolynomials, 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
More informationCRing Project, Chapter 7
Contents 7 Integrality and valuation rings 3 1 Integrality......................................... 3 1.1 Fundamentals................................... 3 1.2 Le sorite for integral extensions.........................
More informationMath 203A  Solution Set 3
Math 03A  Solution Set 3 Problem 1 Which of the following algebraic sets are isomorphic: (i) A 1 (ii) Z(xy) A (iii) Z(x + y ) A (iv) Z(x y 5 ) A (v) Z(y x, z x 3 ) A Answer: We claim that (i) and (v)
More informationLECTURE 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
More informationCommutative Algebra. Instructor: Graham Leuschke & Typist: Caleb McWhorter. Spring 2016
Commutative Algebra Instructor: Graham Leuschke & Typist: Caleb McWhorter Spring 2016 Contents 1 General Remarks 3 2 Localization 3 2.1 Localization......................................... 3 2.6 The Construction
More information8. 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
More informationREPRESENTATION THEORY WEEK 9
REPRESENTATION THEORY WEEK 9 1. JordanHölder theorem and indecomposable modules Let M be a module satisfying ascending and descending chain conditions (ACC and DCC). In other words every increasing sequence
More informationExtension 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
More informationAlgebraic varieties. Chapter A ne varieties
Chapter 4 Algebraic varieties 4.1 A ne varieties Let k be a field. A ne nspace 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
More informationMath 120 HW 9 Solutions
Math 120 HW 9 Solutions June 8, 2018 Question 1 Write down a ring homomorphism (no proof required) f from R = Z[ 11] = {a + b 11 a, b Z} to S = Z/35Z. The main difficulty is to find an element x Z/35Z
More informationRepresentations of algebraic groups and their Lie algebras Jens Carsten Jantzen Lecture III
Representations of algebraic groups and their Lie algebras Jens Carsten Jantzen Lecture III Lie algebras. Let K be again an algebraically closed field. For the moment let G be an arbitrary algebraic group
More informationCommutative Algebra I
Commutative Algebra I Craig Huneke 1 June 27, 2012 1 A compilation of two sets of notes at the University of Kansas; one in the Spring of 2002 by?? and the other in the Spring of 2007 by Branden Stone.
More information5 Dedekind extensions
18.785 Number theory I Fall 2016 Lecture #5 09/22/2016 5 Dedekind extensions In this lecture we prove that the integral closure of a Dedekind domain in a finite extension of its fraction field is also
More informationφ(a + b) = φ(a) + φ(b) φ(a b) = φ(a) φ(b),
16. Ring Homomorphisms and Ideals efinition 16.1. Let φ: R S be a function between two rings. We say that φ is a ring homomorphism if for every a and b R, and in addition φ(1) = 1. φ(a + b) = φ(a) + φ(b)
More informationNOTES 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 381T, 1:15 2:30. This is the very end of Galois theory; you ll also start commutative ring theory. Tell them: midterm
More informationTensor Product of modules. MA499 Project II
Tensor Product of modules A Project Report Submitted for the Course MA499 Project II by Subhash Atal (Roll No. 07012321) to the DEPARTMENT OF MATHEMATICS INDIAN INSTITUTE OF TECHNOLOGY GUWAHATI GUWAHATI
More informationHomework 2  Math 603 Fall 05 Solutions
Homework 2  Math 603 Fall 05 Solutions 1. (a): In the notation of AtiyahMacdonald, Prop. 5.17, we have B n j=1 Av j. Since A is Noetherian, this implies that B is f.g. as an Amodule. (b): By Noether
More informationCommutative Algebra. Gabor Wiese. Winter Term 2013/2014. Université du Luxembourg.
Commutative Algebra Winter Term 2013/2014 Université du Luxembourg Gabor Wiese gabor.wiese@uni.lu Version of 16th December 2013 2 Preface In number theory one is naturally led to study more general numbers
More informationAdvanced 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.
More informationCOURSE SUMMARY FOR MATH 508, WINTER QUARTER 2017: ADVANCED COMMUTATIVE ALGEBRA
COURSE SUMMARY FOR MATH 508, WINTER QUARTER 2017: ADVANCED COMMUTATIVE ALGEBRA JAROD ALPER WEEK 1, JAN 4, 6: DIMENSION Lecture 1: Introduction to dimension. Define Krull dimension of a ring A. Discuss
More informationTWO 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
More informationOn the vanishing of Tor of the absolute integral closure
On the vanishing of Tor of the absolute integral closure Hans Schoutens Department of Mathematics NYC College of Technology City University of New York NY, NY 11201 (USA) Abstract Let R be an excellent
More informationAlgebraic 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
More informationAlgebra Qualifying Exam Solutions January 18, 2008 Nick Gurski 0 A B C 0
1. Show that if B, C are flat and Algebra Qualifying Exam Solutions January 18, 2008 Nick Gurski 0 A B C 0 is exact, then A is flat as well. Show that the same holds for projectivity, but not for injectivity.
More informationMATH 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
More informationFormal Groups. Niki Myrto Mavraki
Formal Groups Niki Myrto Mavraki Contents 1. Introduction 1 2. Some preliminaries 2 3. Formal Groups (1 dimensional) 2 4. Groups associated to formal groups 9 5. The Invariant Differential 11 6. The Formal
More informationCOMMUTATIVE ALGEBRA, LECTURE NOTES
COMMUTATIVE ALGEBRA, LECTURE NOTES P. SOSNA Contents 1. Very brief introduction 2 2. Rings and Ideals 2 3. Modules 10 3.1. Tensor product of modules 15 3.2. Flatness 18 3.3. Algebras 21 4. Localisation
More informationSolutions Ark3. 1 It is also an exercise in the aikido toho iai of writing diagrams in L A TEX
Solutions Ark3 From the book: Number 1, 2, 3, 5 and 6 on page 43 and 44. Number 1: Let S be a multiplicatively closed subset of a ring A, and let M be a finitely generated Amodule. Then S 1 M = 0 if and
More informationAlgebraic 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
More informationHonors 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)
More information4.1 Primary Decompositions
4.1 Primary Decompositions generalization of factorization of an integer as a product of prime powers. unique factorization of ideals in a large class of rings. In Z, a prime number p gives rise to a prime
More information