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

Size: px
Start display at page:

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

Transcription

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 integral domains that are used in the classification of finitely generated modules over a principal ideal domain in Samuel s book on algebraic number theory. The assertions are very easy. They were prepared for Basic Algebra, 60210, discussion on modules in December Rank and basis Let R be a ring and let M be a R-module with submodules M 1 and M 2. Consider the R-module homomorphism, f : M 1 M 2 M, given by f(x,y) = x+y, for x M 1, y M 2. We set M 1 + M 2 = f(m 1 M 2 ) = {x 1 + x 2 : x 1 M 1,x 2 M 2 }. Since the image of R-module homomorphism is a submodule, it follows that M 1 +M 2 is a submodule of M. By Theorem in Ash, f : M 1 M 2 M is a R-module isomorphism if and only if: (1) M = M 1 +M 2, and (2) M 1 M 2 = 0. When f is an isomorphism, we say M = M 1 M 2, or M = M 1 +M 2 is direct. Recall that a R-module M is called free if M has a basis. The next result asserts that if M is a direct sum of two free modules, then M is free. Lemma 2.1. Let M = M 1 +M 2 be direct. Suppose S 1 = {v 1,...,v t } is a basis of M 1 and S 2 = {w 1,...,w u } is a basis of M 2. Then T := S 1 S 2 is a basis of M. In particular, if M 1 and M 2 are free, then M 1 + M 2 is free, and the number of elements in a basis of M 1 +M 2 is t+u. 1

2 2 Proof : We first show that T is linearly independent, so suppose that t i=1 a iv i + u j=1 b jw j = 0, a i,b j R. Set v = t i=1 a iv i and set w = u j=1 b jw j = 0. Note that v M 1 and w M 2. Then by assumption, v +w = 0, so v = w M 1 M 2 = 0. Since v = 0, t i=1 a iv i = 0, so a i = 0,i = 1,...,t, since S 1 is a basis, and hence linearly independent. Since w = 0, u j=1 b jw j = 0, so b j = 0,j = 1,...,u, since S 2 is linearly independent. Hence T is linearly independent. We now show that T generates M. Since M = M 1 +M 2, if m M, then m = v +w, with v M 1, and w M 2. Since S 1 generates M 1, v = t i=1 a iv i for some a i R. Since S 2 generates M 2, w = u j=1 b jw j, for some b j R. Then m = v +w = t i=1 a iv i + u j=1 b jw j, so T generates M. The last assertion follows from the definition of a free R-module. Definition 2.2. We say a R-module M is finitely generated if there exists a finite generating set S of M, or equivalently, if there is a finite subset S M such that if v M, then v = v i S a iv i, with a i R. By convention, the 0-submodule {0} is generated by the empty set, and an empty sum 0 i=1 r iv i = 0.. Let R be an integral domain and let F = Frac(R) be its fraction field. We regard R F via the injective map a a. Then it follows that 1 R n = {(a 1,...,a n ) : a i R} F n = {(α 1,...,α n ) : α i F}. Let V R n be a R-submodule. Let FV := {αv : α F,v V} F n. Lemma 2.3. FV is a F-subspace of F n. Proof : It suffices to show that if u 1,u 2 FV and λ F, then u 1 + u 2 FV and λu 1 FV. We set u 1 = α 1 v 1 and u 2 = α 2 v 2 FV, with v 1,v 2 V, and α 1 = a 1 b 1,α 2 = a 2 b 2 F, so a 1,a 2,b 1,b 2 R, and b 1,b 2 are both nonzero. Then α 1 v 1 +α 2 v 2 = a 1 v 1 + a 2 v 2 = 1 (a 1 b 2 v 1 +b 1 a 2 v 2 ) FV, since a 1 b 2 v 1 +b 1 a 2 v 2 V by definition of submodule. If λ F, then λ α 1 v 1 = (λ α 1 ) v 1 FV. Thus, FV is a F-subspace. Definition 2.4. Let V R n be a R-submodule. Then rk R (V) = dim F (FV).

3 Proposition 2.5. Let R be an integral domain, and let V R n be a R-submodule. (1) 0 rk R (V) n. (2) rk R (R n ) = n. (3) Let M 1,M 2 be R-submodules of R n and suppose M = M 1 + M 2 is direct. Then FM 1 FM 2 = 0, FM 1 +FM 2 = F(M 1 +M 2 ), and rk R (M) = rk R (M 1 )+rk R (M 2 ). (4) Let v 1,...,v r be a basis of a submodule M of R n. Then v 1,...,v r is a basis of FM, so rk R (M) = r. (5) Let v M be nonzero. Then {v} is a basis of R v, so rk R (R v) = 1. 3 Proof : (1) and (2): Since V R n, it follows from definitions that FV FR n = F n. Hence, by linear algebra over fields, 0 rk R (V) = dim F (FV) dim F (F n ) = n. For (3), we first show that FM 1 FM 2 = 0. For this, let α 1 v 1 FM 1, and α 2 v 2 FM 2, with (*) α 1 = a 1 b 1, α 2 = a 2 b 2, for v i M 1, v 2 M 2, a 1,a 2 R, and b 1,b 2 nonzero elements of R. If α 1 v 1 = α 2 v 2 FM 1 FM 2, then a 1 b 1 v 1 = a 2 b 2 v 2, so a 1b 2 v 1 = a 2b 1 v 2. Since 0, then a 1 b 2 v 1 = a 2 b 1 v 2 M 1 M 2 = 0. Since b 2 0, a 1 v 1 = b 2 1 a 1 b 2 v 2 = 0, so α 1 v 1 = 0. It follows that FM 1 FM 2 = 0. We next show that FM 1 + FM 2 = F(M 1 + M 2 ). Indeed, let v 1 M 1 and v 2 M 2. Then if α F, then α(v 1 +v 2 ) = αv 1 +αv 2 FM 1 +FM 2, so F(M 1 +M 2 ) FM 1 +FM 2. Conversely, let α 1 v 1 FM 1 and α 2 v 2 FM 2 with α 1,α 2 as in (*) above. Then α 1 v 1 +α 2 v 2 = 1 1 (a 1 b 2 v 1 +a 2 b 1 v 2 ) F(M 1 +M 2 ), since F and a 1 b 2 v 1 +a 2 b 1 v 2 M 1 +M 2. Given these observations, rk R (M) = rk R (M 1 +M 2 ) = dim F (F(M 1 +M 2 )) = dim F (FM 1 +FM 2 ). But from linear algebra over fields, if V 1 and V 2 are F-subspaces of F n, then dim F (V 1 +V 2 ) = dim F (V 1 )+dim F (V 2 ) dim F (V 1 V 2 ). Applying this with V 1 = FM 1 and V 2 = FM 2 gives assertion (3), since FM 1 FM 2 = 0. We now prove assertion (5). It is clear from the definition of R v that {v} generates R v. Let v R n is nonzero, then v = r 1 e r n e n for some r 1,...,r n R (here e 1,...,e n are standard basis vectors of R n.) Since v 0, some r i 0. If a v = 0 for a R, then ar 1 e 1 + +ar n e n = 0, so since {e 1,...,e n } is a basis of R n, ar i = 0. Since r i 0, it follows that a = 0. Hence, the set {v} is linearly independent. It follows easily that {v} is a basis of F v = FR v, so dim F (F v) = 1, and this completes the proof of (5).

4 4 We prove assertion (4) by induction on r, and note that the case r = 0 is trivial and the case r = 1 was proved as part of assertion (5). Let M 1 = Rv 1 + +Rv r 1. Then it follows easily that S 1 = {v 1,...,v r 1 } is a basis of M 1. By induction, S 1 is a basis of FM 1 and rk R (M 1 ) = r 1. Let M 2 = R v r. By (5), {v r } is a basis of M 2, and rk R (M 2 ) = 1. By Lemma 2.1, it follows that v 1,...,v r is a basis of M = M 1 +M 2. Note that M 1 M 2 = 0 since {v 1,...,v r } is a basis. Hence, FM 1 FM 2 = 0 by assertion (3), and FM = F(M 1 +M 2 ) = FM 1 +FM 2 = F v i by induction. This proves (4). Definition 2.6. Let M be a free finitely generated R-module with R-module isomorphism φ : M R n. Let M 1 M be a R-submodule. Then rk R (M 1 ) = rk R (φ(m 1 )). Proposition 2.7. Let M be a free finitely generated R-module with R-module isomorphism φ : M R n. Then (1) Let M 1,M 2 M be submodules. If M 1 M 2 = 0, then rk R (M 1 ) + rk R (M 2 ) = rk R (M 1 +M 2 ). (2) Let M 1 M be a free submodule with basis v 1,...,v r. Then rk R (M 1 ) = r. (3) If M 1 M is a submodule, then 0 rk R (M 1 ) n. (4) If v M is nonzero, then rk R (R v) = 1. Proof : For(1), notethatφ(m 1 ) φ(m 2 ) = φ(m 1 M 2 ) = 0. Thus, by(3)ofproposition 2.5, rk R (φ(m 1 +M 2 )) = rk R (φ(m 1 )+φ(m 2 )) = rk R (φ(m 1 ))+rk R (φ(m 2 )), and this implies (1). For (2), it follows easily from definitions that φ(v 1 ),...,φ(v r ) is a basis of φ(m 1 ), and now (2) follows from (4) of Proposition 2.5. Assertion (3) is clear by (1) of Proposition 2.5, and Assertion (4) is clear by (5) of Proposition 2.5. Note that a R-submodule M 1 of a free module M may not be generated by one element, but may still have rk R (M 1 ) = 1. For example, let R = F[x,y], the polynomial ring in two variables, and let M = R, which is free with basis {1}. Let M 1 = (x,y). Then M 1 R v for any v M 1 since (x,y) is not a principal ideal. But FM 1 = Fx+Fy = F, so rk R (M 1 ) = The set of linear maps We discuss the R-linear maps from a R-module M to N. As before, R is a commutative ring. Definition 3.1. Let R be a ring and let M,N be R-modules. Then Hom R (M,N) = {φ : M N : φ is a R module homomorphism}.

5 Remark 3.2. Let f Hom R (M,N). Let M 1 M be a R-submodule, and N 1 N be a R-submodule. (i) f(m 1 ) := {f(x) : x M 1 } is a R-submodule of N. (2) f 1 (N 1 ) := {x M : f(x) N 1 } is a R-submodule of M. (3) Let P M be a R-submodule. Then P M 1 is a R-submodule of M. All these assertions are routine to prove, and we leave them to the reader, aside from noting that some work can be saved by citing in Ash. We can deduce (3) from (2) using the inclusion map i : P M, and realizing that P M 1 = i 1 (M 1 ). Forf 1,f 2 Hom R (M,N), letf 1 +f 2 : M N bedefinedby(f 1 +f 2 )(x) = f 1 (x)+f 2 (x) for x M. For r R, let r f 1 : M N be defined by (r f 1 )(x) = r (f 1 (x)). Note that f 1 +f 2,r f 1 Hom R (M,N). Indeed, for the second assertion, let a R and x M, and compute (r f 1 )(a x) = r (f 1 (a x)) = f 1 (ra x) = f 1 (ar x) = ar (f 1 (x)) = a (r f 1 )(x). This requires that R is commutative, but f 1 + f 2 Hom R (M,N) even when R is noncommutative, and we leave the easy verification to the reader. Lemma 3.3. Hom R (M,N) is a R-module. We leave these assertions to the reader. They are quite easy. We consider the special case when N = R, viewed as a R-module using multiplication in R. We let M have basis x 1,...,x n, and recall that if m M, then we can write m = r 1 x 1 + +r n x n. For i = 1,...,n, we define p i : M R by the formula p i (r 1 x 1 + +r n x n ) = r i. Remark 3.4. For i = 1,...,n, then p i Hom R (M,R). Indeed, if m = n i=1 r ix i and n sum n i=1s i x i, then m+n = n i=1 (r i +s i )x i, so p i (m+n) = r i +s i = p i (m)+p i (n). We leave to the reader the verification that if s M, then p i (s m) = s p i (m). Remark 3.5. In the above, let M be the free module R n, with standard basis e 1,...,e n. Then p i (r 1,...,r n ) = r i is projection on the rth factor. Exercise 3.6. If R is a commutative ring and M is a free R-module with basis x 1,...,x n, then p 1,...,p n is a basis for the R-module Hom R (M,R) has basis p 1,...,p n as an R- module. It follows that Hom R (M,R) = R n as a R-module. We establish one further result concerning principal ideal domains. Lemma 3.7. Let R be a principal ideal domain, and let S = {I j } j J be a collection of ideals of R. Then S has a maximal element N; i.e., N S is an ideal, and if I j S, and N I j, then N = I j. Proof : We suppose that there does not exist a maximal element N. Then for every K S, there exists L S such that K L and K L. Pick an ideal I 1 S. By 5

6 6 assumption, there exists an ideal I 2 S such that I 1 I 2, but I 1 I 2. Similarly, for each j 1, there exists an ideal I j+1 S such that I j I j+1, but I j I j+1. Thus, I 1 I 2 I n... is an infinite ascending chain of proper inclusions of ideals. But R is a PID, so it satisfies the ascending chain condition on ideals by the PID theorem proved in class (or Ash, Theorem and Theorem 2.6.8).

### 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

### 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

### Injective Modules and Matlis Duality

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

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

### 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)

### and this makes M into an R-module by (1.2). 2

1. Modules Definition 1.1. Let R be a commutative ring. A module over R is set M together with a binary operation, denoted +, which makes M into an abelian group, with 0 as the identity element, together

### IDEAL CLASSES AND RELATIVE INTEGERS

IDEAL CLASSES AND RELATIVE INTEGERS KEITH CONRAD The ring of integers of a number field is free as a Z-module. It is a module not just over Z, but also over any intermediate ring of integers. That is,

### ALGEBRA HW 3 CLAY SHONKWILER

ALGEBRA HW 3 CLAY SHONKWILER (a): Show that R[x] is a flat R-module. 1 Proof. Consider the set A = {1, x, x 2,...}. Then certainly A generates R[x] as an R-module. Suppose there is some finite linear combination

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

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

### Ring Theory Problems. A σ

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

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

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

### 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

### NOTES ON SPLITTING FIELDS

NOTES ON SPLITTING FIELDS CİHAN BAHRAN I will try to define the notion of a splitting field of an algebra over a field using my words, to understand it better. The sources I use are Peter Webb s and T.Y

### 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

### 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

### (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

### 5 Dedekind extensions

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

### Review of Linear Algebra

Review of Linear Algebra Throughout these notes, F denotes a field (often called the scalars in this context). 1 Definition of a vector space Definition 1.1. A F -vector space or simply a vector space

### Noetherian property of infinite EI categories

Noetherian property of infinite EI categories Wee Liang Gan and Liping Li Abstract. It is known that finitely generated FI-modules over a field of characteristic 0 are Noetherian. We generalize this result

### 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

### which is a group homomorphism, such that if W V U, then

4. Sheaves Definition 4.1. Let X be a topological space. A presheaf of groups F on X is a a function which assigns to every open set U X a group F(U) and to every inclusion V U a restriction map, ρ UV

### 10. Noether Normalization and Hilbert s Nullstellensatz

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

### CHAPTER 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

### Math 121 Homework 4: Notes on Selected Problems

Math 121 Homework 4: Notes on Selected Problems 11.2.9. If W is a subspace of the vector space V stable under the linear transformation (i.e., (W ) W ), show that induces linear transformations W on W

### NONCOMMUTATIVE POLYNOMIAL EQUATIONS. Edward S. Letzter. Introduction

NONCOMMUTATIVE POLYNOMIAL EQUATIONS Edward S Letzter Introduction My aim in these notes is twofold: First, to briefly review some linear algebra Second, to provide you with some new tools and techniques

### 4.2 Chain Conditions

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

### The Fitting Submodule

The Fitting Submodule U. Meierfrankenfeld B. Stellmacher Department of Mathematics, Michigan State University, East Lansing MI 48840 meier@math.msu.edu Mathematisches Seminar, Christian-Albrechts- Universität,

### INJECTIVE MODULES AND THE INJECTIVE HULL OF A MODULE, November 27, 2009

INJECTIVE ODULES AND THE INJECTIVE HULL OF A ODULE, November 27, 2009 ICHIEL KOSTERS Abstract. In the first section we will define injective modules and we will prove some theorems. In the second section,

### REPRESENTATION 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

### 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

### Part IX. Factorization

IX.45. Unique Factorization Domains 1 Part IX. Factorization Section IX.45. Unique Factorization Domains Note. In this section we return to integral domains and concern ourselves with factoring (with respect

### 38 Irreducibility criteria in rings of polynomials

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

### Projective and Injective Modules

Projective and Injective Modules Push-outs and Pull-backs. Proposition. Let P be an R-module. The following conditions are equivalent: (1) P is projective. (2) Hom R (P, ) is an exact functor. (3) Every

### 11. Finitely-generated modules

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

### NOTES FOR COMMUTATIVE ALGEBRA M5P55

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

### Topics in Module Theory

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

### Homological Methods in Commutative Algebra

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

### COHEN-MACAULAY RINGS SELECTED EXERCISES. 1. Problem 1.1.9

COHEN-MACAULAY RINGS SELECTED EXERCISES KELLER VANDEBOGERT 1. Problem 1.1.9 Proceed by induction, and suppose x R is a U and N-regular element for the base case. Suppose now that xm = 0 for some m M. We

### 4.3 Composition Series

4.3 Composition Series Let M be an A-module. A series for M is a strictly decreasing sequence of submodules M = M 0 M 1... M n = {0} beginning with M and finishing with {0 }. The length of this series

### 11 Annihilators. Suppose that R, S, and T are rings, that R P S, S Q T, and R U T are bimodules, and finally, that

11 Annihilators. In this Section we take a brief look at the important notion of annihilators. Although we shall use these in only very limited contexts, we will give a fairly general initial treatment,

### Structure of rings. Chapter Algebras

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

### 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.

### 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

### 43 Projective modules

43 Projective modules 43.1 Note. If F is a free R-module and P F is a submodule then P need not be free even if P is a direct summand of F. Take e.g. R = Z/6Z. Notice that Z/2Z and Z/3Z are Z/6Z-modules

### Tensor Product of modules. MA499 Project II

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

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

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

### A Do It Yourself Guide to Linear Algebra

A Do It Yourself Guide to Linear Algebra Lecture Notes based on REUs, 2001-2010 Instructor: László Babai Notes compiled by Howard Liu 6-30-2010 1 Vector Spaces 1.1 Basics Definition 1.1.1. A vector space

### 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

### DUAL MODULES KEITH CONRAD

DUAL MODULES KEITH CONRAD 1. Introduction Let R be a commutative ring. For two (left) R-modules M and N, the set Hom R (M, N) of all R-linear maps from M to N is an R-module under natural addition and

### 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

### 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

### Modules Over Principal Ideal Domains

Modules Over Principal Ideal Domains Brian Whetter April 24, 2014 This work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. To view a copy of this

### Proceedings of the Twelfth Hudson Symposium, Lecture Notes in Math. No. 951, Springer-Verlag (1982), 4l 46.

Proceedings of the Twelfth Hudson Symposium, Lecture Notes in Math. No. 951, Springer-Verlag (1982), 4l 46. MAXIMAL TORSION RADICALS OVER RINGS WITH FINITE REDUCED RANK John A. Beachy Northern Illinois

### A 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

### Chapter 5. Linear Algebra

Chapter 5 Linear Algebra The exalted position held by linear algebra is based upon the subject s ubiquitous utility and ease of application. The basic theory is developed here in full generality, i.e.,

### Homework #05, due 2/17/10 = , , , , , Additional problems recommended for study: , , 10.2.

Homework #05, due 2/17/10 = 10.3.1, 10.3.3, 10.3.4, 10.3.5, 10.3.7, 10.3.15 Additional problems recommended for study: 10.2.1, 10.2.2, 10.2.3, 10.2.5, 10.2.6, 10.2.10, 10.2.11, 10.3.2, 10.3.9, 10.3.12,

### Written Homework # 5 Solution

Math 516 Fall 2006 Radford Written Homework # 5 Solution 12/12/06 Throughout R is a ring with unity. Comment: It will become apparent that the module properties 0 m = 0, (r m) = ( r) m, and (r r ) m =

### STABLY FREE MODULES KEITH CONRAD

STABLY FREE MODULES KEITH CONRAD 1. Introduction Let R be a commutative ring. When an R-module has a particular module-theoretic property after direct summing it with a finite free module, it is said to

### Solution. That ϕ W is a linear map W W follows from the definition of subspace. The map ϕ is ϕ(v + W ) = ϕ(v) + W, which is well-defined since

MAS 5312 Section 2779 Introduction to Algebra 2 Solutions to Selected Problems, Chapters 11 13 11.2.9 Given a linear ϕ : V V such that ϕ(w ) W, show ϕ induces linear ϕ W : W W and ϕ : V/W V/W : Solution.

### 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 2 Linear Transformations

Chapter 2 Linear Transformations Linear Transformations Loosely speaking, a linear transformation is a function from one vector space to another that preserves the vector space operations. Let us be more

### RINGS ISOMORPHIC TO THEIR NONTRIVIAL SUBRINGS

RINGS ISOMORPHIC TO THEIR NONTRIVIAL SUBRINGS JACOB LOJEWSKI AND GREG OMAN Abstract. Let G be a nontrivial group, and assume that G = H for every nontrivial subgroup H of G. It is a simple matter to prove

### 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.

### 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

### 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

### 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

### Introduction to modules

Chapter 3 Introduction to modules 3.1 Modules, submodules and homomorphisms The problem of classifying all rings is much too general to ever hope for an answer. But one of the most important tools available

### 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

### Normed Vector Spaces and Double Duals

Normed Vector Spaces and Double Duals Mathematics 481/525 In this note we look at a number of infinite-dimensional R-vector spaces that arise in analysis, and we consider their dual and double dual spaces

### 1.8 Dual Spaces (non-examinable)

2 Theorem 1715 is just a restatement in terms of linear morphisms of a fact that you might have come across before: every m n matrix can be row-reduced to reduced echelon form using row operations Moreover,

### 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

### 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)

### Available online at J. Math. Comput. Sci. 4 (2014), No. 3, ISSN: ORDERINGS AND PREORDERINGS ON MODULES

Available online at http://scik.org J. Math. Comput. Sci. 4 (2014), No. 3, 574-586 ISSN: 1927-5307 ORDERINGS AND PREORDERINGS ON MODULES DONGMING HUANG Department of Applied Mathematics, Hainan University,

### Definition 3 A Hamel basis (often just called a basis) of a vector space X is a linearly independent set of vectors in X that spans X.

Economics 04 Summer/Fall 011 Lecture 8 Wednesday August 3, 011 Chapter 3. Linear Algebra Section 3.1. Bases Definition 1 Let X be a vector space over a field F. A linear combination of x 1,..., x n X is

### n-x -COHERENT RINGS Driss Bennis

International Electronic Journal of Algebra Volume 7 (2010) 128-139 n-x -COHERENT RINGS Driss Bennis Received: 24 September 2009; Revised: 31 December 2009 Communicated by A. Çiğdem Özcan Abstract. This

### BENJAMIN LEVINE. 2. Principal Ideal Domains We will first investigate the properties of principal ideal domains and unique factorization domains.

FINITELY GENERATED MODULES OVER A PRINCIPAL IDEAL DOMAIN BENJAMIN LEVINE Abstract. We will explore classification theory concerning the structure theorem for finitely generated modules over a principal

### MATH 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

### INTRO TO TENSOR PRODUCTS MATH 250B

INTRO TO TENSOR PRODUCTS MATH 250B ADAM TOPAZ 1. Definition of the Tensor Product Throughout this note, A will denote a commutative ring. Let M, N be two A-modules. For a third A-module Z, consider the

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

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

### 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

### Rings and Fields Theorems

Rings and Fields Theorems Rajesh Kumar PMATH 334 Intro to Rings and Fields Fall 2009 October 25, 2009 12 Rings and Fields 12.1 Definition Groups and Abelian Groups Let R be a non-empty set. Let + and (multiplication)

### 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.

### Infinite-Dimensional Triangularization

Infinite-Dimensional Triangularization Zachary Mesyan March 11, 2018 Abstract The goal of this paper is to generalize the theory of triangularizing matrices to linear transformations of an arbitrary vector

### 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

### 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

### 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

### Lecture 1. (i,j) N 2 kx i y j, and this makes k[x, y]

Lecture 1 1. Polynomial Rings, Gröbner Bases Definition 1.1. Let R be a ring, G an abelian semigroup, and R = i G R i a direct sum decomposition of abelian groups. R is graded (G-graded) if R i R j R i+j

### ALGEBRA QUALIFYING EXAM, FALL 2017: SOLUTIONS

ALGEBRA QUALIFYING EXAM, FALL 2017: SOLUTIONS Your Name: Conventions: all rings and algebras are assumed to be unital. Part I. True or false? If true provide a brief explanation, if false provide a counterexample

### SUMMARY ALGEBRA I LOUIS-PHILIPPE THIBAULT

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

### ON RIGHT S-NOETHERIAN RINGS AND S-NOETHERIAN MODULES

ON RIGHT S-NOETHERIAN RINGS AND S-NOETHERIAN MODULES ZEHRA BİLGİN, MANUEL L. REYES, AND ÜNSAL TEKİR Abstract. In this paper we study right S-Noetherian rings and modules, extending notions introduced by

### ALGEBRA QUALIFYING EXAM SPRING 2012

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

### CHARACTERS AS CENTRAL IDEMPOTENTS

CHARACTERS AS CENTRAL IDEMPOTENTS CİHAN BAHRAN I have recently noticed (while thinking about the skewed orthogonality business Theo has mentioned) that the irreducible characters of a finite group G are

### 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

### Total 100

Math 542 Midterm Exam, Spring 2016 Prof: Paul Terwilliger Your Name (please print) SOLUTIONS NO CALCULATORS/ELECTRONIC DEVICES ALLOWED. MAKE SURE YOUR CELL PHONE IS OFF. Problem Value 1 10 2 10 3 10 4

### 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...

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

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

### Chapter 4 Vector Spaces And Modules

Chapter 4 Vector Spaces And Modules Up to this point we have been introduced to groups and to rings; the former has its motivation in the set of one-to-one mappings of a set onto itself, the latter, in

### Presentation 1

18.704 Presentation 1 Jesse Selover March 5, 2015 We re going to try to cover a pretty strange result. It might seem unmotivated if I do a bad job, so I m going to try to do my best. The overarching theme