# TROPICAL SCHEME THEORY

Size: px
Start display at page:

Transcription

1 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), however, quotient objects are instead specified by congruences. Definition 5.1. A congruence on a semiring is an equivalence relation that preserves the operations. That is, if r r and s s then and r + s r + s rs r s. Example 5.2. The trivial congruence, in which r r if and only if r = r. Example 5.3 (Bourne relations the congruence associated to an ideal). Let R be a semiring and let I R be an ideal. Then define a congruence I by setting r I r if there are elements a and a in I such that r + a = r + a. Note that if r I r and s I s then r + s I r + s and rs I r s. Lemma 5.4. If R is additively idempotent and I R is an ideal then r I r if and only if there is some b I such that r + b = r + b. Proof. ( ) This is immediate from the definition. ( ) Say r I r so there exist a, a I with r + a = r + a. Then we have r + a + a = r + a + a = r + a = r + a = r + a + a = r + a + a. So setting b = a + a we have r + b = r + b. Remark 5.5. Let R be an additively cancellative semiring. Define a new semiring S = (R R, +, ), with + being the obvious addition and multiplication given by (a, b) (c, d) = (ac + bd, ad + bc). The set = {(a, a) a R} is an ideal of S and S/ is a ring. Comment 5.6 (Sam). This construction seems to be introducing subtractionm i.e. we might think of the equivalence class of (a, b) in S/ as corresponding to a b. As an example, if we start with R = (N, +, ) then the quotient S/ is canonically isomorphic to Z. Remark 5.7 (Relation to the ring case). If R is a ring and I R is an ideal then R/I is canonically isomorphic to R/ I. Date: September 21/26, 2017, Speaker: Kalina Mincheva, Scribe: Netanel Friedenberg. 1

2 2 TROPICAL SCHEME THEORY Remark 5.8. If is a congruence on R then J = {x R x 0} is an ideal of R. Relation between ideals and congruences in additively idempotent semirings If R is a ring then there is a natural bijection between ideals in R and congruences on R, given by Remarks 5.7 and 5.8. We now explore the relationship between ideals and congruences in additively idempotent semirings. Let R be an additively idempotent semiring and let J be an ideal of R. Proposition 5.9. The congruence J is generated by {(x, 0) x J}. Proof. Let be the equivalence relation generated by {(x, 0) x J}. Suppose x J. Since R is additively idempotent, 0+x = x+x, and hence x J 0. This shows that is contained in J. We now prove that J is contained in. Suppose x J y. By definition, this means that there is some z J such that x + z = y + z. Then z 0 and hence x x + z. Since x + z = y + z and y + z y, it follows that x y, as required. Definition The saturation of the ideal J R is Equivalently, J = {x R x J 0}. J = {x R x + z = z for some z J}. Proposition Saturation is a closure operation on ideals in R. In other words, the saturation J is an ideal, J J, J = J, and if I J then Ī J. Proof. That J is an ideal is immediate from Remark 5.8. The containment J J is an immediate consequence of additive idempotence. If I J then the containment Ī J is also obvious. It remains to show that J = J. Suppose x J. Then there is some z J such that x + z = z. And since z J there is some z J such that z + z = z. Then and hence x J. x + z = x + z + z = z + z = z, Example The saturation of an ideal J in the polynomial ring T[x 1,..., x n ] is the smallest monomial ideal that contains J. In other words, if J is generated by {f 1,..., f m } then J is generated by the monomials x a1 1 xan n that appear with nonzero coefficient in one of the generators f i. In particular, if J contains a polynomial with nonzero constant term then J = T[x 1,..., x n ]. The analogous statement holds in B[x 1,..., x n ]. In particular, the ideal (x, x + 1) in B[x] considered in Lecture 2 is not saturated, and its saturation is B[x]. Definition A congruence is cancellative if ab ac implies that either b c or a 0. Theorem If J is saturated and cancellative, then J is prime. Proof. Suppose J is saturated and cancellative, and ab J. Then a b J a 0, so either a J 0 or b J 0. Since J is saturated, this implies a J or b J.

3 TROPICAL SCHEME THEORY 3 Remark The converse to Theorem 5.14 is false. For examples of prime ideals J that are not saturated and for which J is not cancellative, see [Les11, Remark 3.14]. Kernels of homomorphisms Definition Let ϕ : R S be a morphism of semirings. The kernel of ϕ is ker ϕ = {(a, b) R R ϕ(a) = ϕ(b)} Similarly, if C S S is any congruence on S then we can consider the preimage congruence ϕ 1 (C) = {(a, b) R R (ϕ(a), ϕ(b)) C}. Then ker(ϕ) = ϕ 1 ( ) where S S is the trivial congruence. Equivalently, φ 1 (C) is the kernel of the induced map R S/C. Remark 5.17 ( Ideal-theoretic kernels versus congruence-theoretic kernels ). The ideal-theoretic kernel of a morphism ϕ : R S is ϕ 1 (0). Consider surjective morphisms T[x] T. The only such morphism that has non-trivial ideal-theoretic kernel is the one which sends x 0 T =. On the other hand there are many congruences that can be realized as ker ϕ. For example, for each t T we have the evaluation morphism ϕ t : T[x] T f(x) f(t). If we let be the intersection of the kernels of all such ϕ t, then T[x]/ is CPL Z (R). Proposition The only additively idempotent semiring with no nontrivial proper congruences is the Boolean semiring B. Proof. Let R be an additively idempotent semiring. Suppose R has zero-divisors, i.e. there are x, y R\{0} such that xy = 0. Consider C = {(a, b) xa = xb}. Then C is a congruence on R. It is nontrivial because (y, 0) C and proper because (1, 0) / C. Now suppose R has no zero-divisors. Then we can define a morphism ϕ : R B by { 0 B if x = 0 R ϕ(x) =. 1 B if x 0 R This is a morphism because, in an additively idempotent semiring, if a + b = 0 then both a and b are 0. Then ker ϕ is a proper subset of R R, and so either ker ϕ is nontrivial or ϕ is an isomorphism onto B. (Since B has no nontrivial automorphisms, this isomorphism is unique.) Remark Even more is true: if R is an arbitrary commutative semiring (not necessarily additively idempotent) with no nontrivial proper congruences then either R = B or R is a field. See [Go99, Proposition 8.11]. Remark The Boolean semifield B does not have any nontrivial proper ideals either. More generally, semifields do not have any nontrivial proper ideals. On the other hand, by Proposition 5.18 we know that semifields other than B will have nontrivial proper congruences.

4 4 TROPICAL SCHEME THEORY Example The tropical semiring T has a unique nontrivial proper congruence and the quotient of T by this congruence is B. To see this, suppose is a nontrivial congruence on T. Then a b for two distinct elements a and b in T, with a 0 T. Multiplying the relation a b by a 1, we may assume a = 1 T. If b = 0 T, then is not proper (i.e. it identifies everything with 0 T ). It remains to consider the case where b is nonzero and there is a nontrivial relation 1 T b. Taking inverses on both sides, we also have 1 T = b 1. Since inverse in T is negation in R, we may therefore assume that b is positive in R. For any c in the interval (0, b) we then have 1 T = 1 T + c b + c = c. Repeating the argument with b replaced by b n in T (which is nb in R), for all positive integers n, we see that 1 T is identified with every positive element of R. Taking inverses then shows that 1 T is identified with every element of T 0 R, and hence T/ = B, as claimed. Goal: We want to show one of the pieces of information that congruences can capture is a suitable notion of Krull dimension (of a semiring/algebra). We will see later that this can also be related to the dimension of geometric objects. In order to do this, we will define prime congruences and semifields of fractions and show that we can extract the dimension from the semiring of fractions. Semifield of fractions: Every (multiplicatively) cancellative semiring R embeds into its semifield of fractions, denoted Frac(R). : If R is a ring then R is cancellative if and only if R has no zero-divisors. This fails for semirings. While we still have that a cancellative semiring has no zero-divisors, the converse is false. For example, T[x] has no zero divisors but (x 2 + 1)(x 2 + x + 1) = (x 2 + x + 1) 2. The elements of Frac(R) are equivalence classes in R (R\{0}) under the equivalence relation (r 1, s 1 ) (r 2, s 2 ) r 1 s 2 = r 2 s 1. The operations on Frac(R) are given by (r 1, s 1 ) + (r 2, s 2 ) = (r 1 s 2 + r 2 s 1, s 1 s 2 ) and (r 1, s 1 ) (r 2, s 2 ) = (r 1 r 2, s 1 s 2 ). We can also localize by certain smaller subsets of R, that is, construct S 1 R for S R. See [Go99, Chapter 11]. Prime congruences Let R be a semiring, and let C R R be a congruence. Definition 5.22 (Twisted product). The twisted product of α = (α 1, α 2 ) and β = (β 1, β 2 ) in R R is α β = (α 1 β 1 + α 2 β 2, α 1 β 2 + α 2 β 1 ).

5 TROPICAL SCHEME THEORY 5 Congruences are always closed under twisted product. Even more is true, if C contains α = (α 1, α 2 ) and if β = (β 1, β 2 ) is an arbitrary element of R R, then α β is in C. Remark Suppose C is the congruence induced by an ideal I in a ring. Then I is prime if and only if α β C implies that α C or β C. To see this, note that α is in C if and only if α 1 α 2 is in I. So we are simply saying that I is prime if and only if (α 1 α 2 )(β 1 β 2 ) I implies either (α 1 α 2 ) I or (β 1 β 2 ) I. Definition The congruence C is prime if it is a proper subset of R R and α β C implies α C or β C. Definition The congruence C is irreducible if it has no nontrivial expression as an intersection of congruences. In other words, C is irreducible if whenever C is an intersection of two congruences C 1 C 2 then C = C 1 or C = C 2. The following results on prime ideals appear in [JM17]. See the original paper for the proofs that are omitted. Theorem 5.26 ([JM17] Theorem 2.12). If R is additively idempotent then a congruence C R R is prime if and only if it is cancellative and irreducible. Recall from Definition 5.13 that if C is a cancellative congruence then R/C is a cancellative semiring, in other words, (a, b) C implies (a, c) C or (b, 0) C. Remark While the definition of prime congruence makes sense in any semiring, the conclusion of Theorem 5.26 fails for semirings that are not additively idempotent. An additively idempotent semiring carries a natural partial order, in which a b if and only if a + b = b. Semirings which are totally ordered play a special role in the theory. Proposition If C is a prime congruence on an additively idempotent semiring R then R/C is totally ordered. Proof. Assume without loss of generality that C = and this is prime. Note that the twisted product (a, a + b) (b, a + b) = (a 2 + ab + b 2, a 2 + ab + b 2 ), which is in. Therefore (a, a + b) or (b, b + a). This shows that a b or b a, so R is totally ordered. Remark Note that if R/C is totally ordered it does not mean that C is prime. However, if R/C is also cancellative then C is prime. In other words, a congruence C is prime if and only if the quotient R/C is cancellative and totally ordered. See [JM17, Proposition 2.10]. Example By the preceding remark, it is sensible to try to classify prime congruences on a semiring but considering the possible total orderings on the quotient. We now carry this through for the Laurent polynomial semiring over the Boolean semifield B[x ± ]. Let C be a prime congruence on B[x ± ]. We consider the various possibilities for the order relation between [x] and 1 in B[x ± ]/C.

6 6 TROPICAL SCHEME THEORY If [x] = 1, then C contains the congruence generated by (x, 1). Note that B[x ± ]/ (x, 1) = B, which has no nontrivial prime congruences. It follows that C = (x, 1). If [x] > 1, then [x i ] > [x j ] for i > j. Therefore, every Laurent polynomial will be congruent to its monomial with highest exponent. For example, x 2 +x+1 x 2. So we see that B[x ± ]/C = (Z { }, max, +). If [x] < 1 then B[x ± ]/C = (Z { }, min, +), by a similar argument. A computation similar to that in the preceding example shows that there are no infinite chains of prime congruences on B[x]. Note, however, that there are infinite chains of prime ideals in B[x] (as we saw in Lecture 2, Example 2.7), and B[x, y] has infinite chains of (non-prime) cancellative congruences. See [JM15, Proposition 2.12]. Definition Let R be an additively idempotent semiring. The Krull dimension dim R is the number of strict inclusions in the longest chain of prime congruences. Note: There may be maximal chains of prime congruences of different lengths, but there is a unique longest chain. Theorem 5.32 ([JM15] Theorem 3.16). Let A be any (additively idempotent) semiring. Then dim A[x 1,..., x n ] = dim A + n. In particular: dim B[x] = 1 dim B[x 1,..., x n ] = n dim T[x 1,..., x n ] = n + 1 : This behavior is better than that of Krull dimension for arbitrary commutative rings R. If R is a ring then we only have dim R + 1 dim R[x] 2 dim R + 1. For an example where dim R > 0 and dim R[x] = 2 dim R + 1, see [Se54]. Definition The ideal-theoretic kernel of a congruence C on a semiring R is {x R (x, 0) C}. In other words, ker C is the ideal-theoretic kernel of the quotient map R R/C. Theorem 5.34 ([JM15] Proposition 3.15). Let R be (multiplicatively) cancellative and totally ordered. Then dim R = dim Frac(R). Moreover, the prime congruences C of R with ker C = 0 form a chain of maximum length. To show the above statement we need to relate the congruences of R and Frac(R). If C is a congruence on R we denote by C Frac(R) the congruence generated by C in Frac(R). On the other hand, if C is a congruence on Frac(R) we can restrict C to R and consider the congruence C R = (a, b) C a, b R. Proposition 5.35 ([JM15] Proposition 3.8 (ii)). Let R be a cancellative semiring and let C be a congruence of R such that R/C is cancellative and C has trivial kernel. Then C Frac(R) R = C. Conversely, if C is a congruence in Frac(R) then C R Frac(R) = C.

7 TROPICAL SCHEME THEORY 7 This proposition gives us the dimension equality once we know that the claimed prime congruences form a chain of maximum length. When R is not cancellative, let P 0 P 1 P k be a chain of maximum length. Then consider R/P 0, and look at P 1 P k. Question 5.36 (Daping). Is is true that dim A = dim S 1 A, for some S A? References [Go99] J. Golan, Semirings and Their Applications, Kluwer, Dordrecht, 1999 [JM17] D. Joó and K. Mincheva, Prime congruences of idempotent semirings and a Nullstellensatz for tropical polynomials, Selecta Mathematica (2017), doi: /s x [JM15] D. Joó and K. Mincheva, On the dimension of the polynomial and the Laurent polynomial semiring, arxiv: [Les11] P. Lescot Absolute Algebra II-Ideals and Spectra, Journal of Pure and Applied Algebra 215(7), 2011, [Se54] A. Seidenberg, On the dimension theory of rings II, Pacific J. Math. 4 (1954)

### 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

### 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

### TROPICAL SCHEME THEORY. Idempotent semirings

TROPICAL SCHEME THEORY Idempotent semirings Definition 0.1. A semiring is (R, +,, 0, 1) such that (R, +, 0) is a commutative monoid (so + is a commutative associative binary operation on R 0 is an additive

### Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman

Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman October 17, 2006 TALK SLOWLY AND WRITE NEATLY!! 1 0.1 Integral Domains and Fraction Fields 0.1.1 Theorems Now what we are going

### 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

### 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

### RINGS: SUMMARY OF MATERIAL

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

### 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

### 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,

### 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)

### φ(xy) = (xy) n = x n y n = φ(x)φ(y)

Groups 1. (Algebra Comp S03) Let A, B and C be normal subgroups of a group G with A B. If A C = B C and AC = BC then prove that A = B. Let b B. Since b = b1 BC = AC, there are a A and c C such that b =

### 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

### HOMEWORK 3 LOUIS-PHILIPPE THIBAULT

HOMEWORK 3 LOUIS-PHILIPPE THIBAULT Problem 1 Let G be a group of order 56. We have that 56 = 2 3 7. Then, using Sylow s theorem, we have that the only possibilities for the number of Sylow-p subgroups

### Theorem 5.3. Let E/F, E = F (u), be a simple field extension. Then u is algebraic if and only if E/F is finite. In this case, [E : F ] = deg f u.

5. Fields 5.1. Field extensions. Let F E be a subfield of the field E. We also describe this situation by saying that E is an extension field of F, and we write E/F to express this fact. If E/F is a field

### 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)

### MATH RING ISOMORPHISM THEOREMS

MATH 371 - RING ISOMORPHISM THEOREMS DR. ZACHARY SCHERR 1. Theory In this note we prove all four isomorphism theorems for rings, and provide several examples on how they get used to describe quotient rings.

### 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.

### Chapter 3. Rings. The basic commutative rings in mathematics are the integers Z, the. Examples

Chapter 3 Rings Rings are additive abelian groups with a second operation called multiplication. The connection between the two operations is provided by the distributive law. Assuming the results of Chapter

### GEOMETRIC CONSTRUCTIONS AND ALGEBRAIC FIELD EXTENSIONS

GEOMETRIC CONSTRUCTIONS AND ALGEBRAIC FIELD EXTENSIONS JENNY WANG Abstract. In this paper, we study field extensions obtained by polynomial rings and maximal ideals in order to determine whether solutions

### φ(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)

### Formal power series rings, inverse limits, and I-adic completions of rings

Formal power series rings, inverse limits, and I-adic 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

### School of Mathematics and Statistics. MT5836 Galois Theory. Handout 0: Course Information

MRQ 2017 School of Mathematics and Statistics MT5836 Galois Theory Handout 0: Course Information Lecturer: Martyn Quick, Room 326. Prerequisite: MT3505 (or MT4517) Rings & Fields Lectures: Tutorials: Mon

### 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

### Computations/Applications

Computations/Applications 1. Find the inverse of x + 1 in the ring F 5 [x]/(x 3 1). Solution: We use the Euclidean Algorithm: x 3 1 (x + 1)(x + 4x + 1) + 3 (x + 1) 3(x + ) + 0. Thus 3 (x 3 1) + (x + 1)(4x

### MATH 8253 ALGEBRAIC GEOMETRY WEEK 12

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

### 1 Rings 1 RINGS 1. Theorem 1.1 (Substitution Principle). Let ϕ : R R be a ring homomorphism

1 RINGS 1 1 Rings Theorem 1.1 (Substitution Principle). Let ϕ : R R be a ring homomorphism (a) Given an element α R there is a unique homomorphism Φ : R[x] R which agrees with the map ϕ on constant polynomials

### Boolean Algebras, Boolean Rings and Stone s Representation Theorem

Boolean Algebras, Boolean Rings and Stone s Representation Theorem Hongtaek Jung December 27, 2017 Abstract This is a part of a supplementary note for a Logic and Set Theory course. The main goal is to

### 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

### Handout - Algebra Review

Algebraic Geometry Instructor: Mohamed Omar Handout - Algebra Review Sept 9 Math 176 Today will be a thorough review of the algebra prerequisites we will need throughout this course. Get through as much

### (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

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

### 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

### 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.

### Solutions to Some Review Problems for Exam 3. by properties of determinants and exponents. Therefore, ϕ is a group homomorphism.

Solutions to Some Review Problems for Exam 3 Recall that R, the set of nonzero real numbers, is a group under multiplication, as is the set R + of all positive real numbers. 1. Prove that the set N of

### Algebra SEP Solutions

Algebra SEP Solutions 17 July 2017 1. (January 2017 problem 1) For example: (a) G = Z/4Z, N = Z/2Z. More generally, G = Z/p n Z, N = Z/pZ, p any prime number, n 2. Also G = Z, N = nz for any n 2, since

### 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

### 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

### 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

### 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

### Rings. Chapter 1. Definition 1.2. A commutative ring R is a ring in which multiplication is commutative. That is, ab = ba for all a, b R.

Chapter 1 Rings We have spent the term studying groups. A group is a set with a binary operation that satisfies certain properties. But many algebraic structures such as R, Z, and Z n come with two binary

### Lecture 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:

### 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

### 0.1 Spec of a monoid

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

### 2a 2 4ac), provided there is an element r in our

MTH 310002 Test II Review Spring 2012 Absractions versus examples The purpose of abstraction is to reduce ideas to their essentials, uncluttered by the details of a specific situation Our lectures built

### Homework 10 M 373K by Mark Lindberg (mal4549)

Homework 10 M 373K by Mark Lindberg (mal4549) 1. Artin, Chapter 11, Exercise 1.1. Prove that 7 + 3 2 and 3 + 5 are algebraic numbers. To do this, we must provide a polynomial with integer coefficients

### Algebra Exam Fall Alexander J. Wertheim Last Updated: October 26, Groups Problem Problem Problem 3...

Algebra Exam Fall 2006 Alexander J. Wertheim Last Updated: October 26, 2017 Contents 1 Groups 2 1.1 Problem 1..................................... 2 1.2 Problem 2..................................... 2

### D-MATH Algebraic Geometry FS 2018 Prof. Emmanuel Kowalski. Solutions Sheet 1. Classical Varieties

D-MATH Algebraic Geometry FS 2018 Prof. Emmanuel Kowalski Solutions Sheet 1 Classical Varieties Let K be an algebraically closed field. All algebraic sets below are defined over K, unless specified otherwise.

### Math 222A W03 D. Congruence relations

Math 222A W03 D. 1. The concept Congruence relations Let s start with a familiar case: congruence mod n on the ring Z of integers. Just to be specific, let s use n = 6. This congruence is an equivalence

### 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.

### 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

### 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

### 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.

### 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.

### Supplement. Dr. Bob s Modern Algebra Glossary Based on Fraleigh s A First Course on Abstract Algebra, 7th Edition, Sections 0 through IV.

Glossary 1 Supplement. Dr. Bob s Modern Algebra Glossary Based on Fraleigh s A First Course on Abstract Algebra, 7th Edition, Sections 0 through IV.23 Abelian Group. A group G, (or just G for short) is

### Solutions to odd-numbered exercises Peter J. Cameron, Introduction to Algebra, Chapter 2

Solutions to odd-numbered exercises Peter J Cameron, Introduction to Algebra, Chapter 1 The answers are a No; b No; c Yes; d Yes; e No; f Yes; g Yes; h No; i Yes; j No a No: The inverse law for addition

### Modern Computer Algebra

Modern Computer Algebra Exercises to Chapter 25: Fundamental concepts 11 May 1999 JOACHIM VON ZUR GATHEN and JÜRGEN GERHARD Universität Paderborn 25.1 Show that any subgroup of a group G contains the neutral

### A GENERAL THEORY OF ZERO-DIVISOR GRAPHS OVER A COMMUTATIVE RING. David F. Anderson and Elizabeth F. Lewis

International Electronic Journal of Algebra Volume 20 (2016) 111-135 A GENERAL HEORY OF ZERO-DIVISOR GRAPHS OVER A COMMUAIVE RING David F. Anderson and Elizabeth F. Lewis Received: 28 April 2016 Communicated

### Prime Rational Functions and Integral Polynomials. Jesse Larone, Bachelor of Science. Mathematics and Statistics

Prime Rational Functions and Integral Polynomials Jesse Larone, Bachelor of Science Mathematics and Statistics Submitted in partial fulfillment of the requirements for the degree of Master of Science Faculty

### Math 210B: Algebra, Homework 1

Math 210B: Algebra, Homework 1 Ian Coley January 15, 201 Problem 1. Show that over any field there exist infinitely many non-associate irreducible polynomials. Recall that by Homework 9, Exercise 8 of

### 1 Absolute values and discrete valuations

18.785 Number theory I Lecture #1 Fall 2015 09/10/2015 1 Absolute values and discrete valuations 1.1 Introduction At its core, number theory is the study of the ring Z and its fraction field Q. Many questions

### 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 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

### 14 Lecture 14: Basic generallities on adic spaces

14 Lecture 14: Basic generallities on adic spaces 14.1 Introduction The aim of this lecture and the next two is to address general adic spaces and their connection to rigid geometry. 14.2 Two open questions

### 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

### 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

### 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 structures I

MTH5100 Assignment 1-10 Algebraic structures I For handing in on various dates January March 2011 1 FUNCTIONS. Say which of the following rules successfully define functions, giving reasons. For each one

### (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

### Abstract Algebra, Second Edition, by John A. Beachy and William D. Blair. Corrections and clarifications

1 Abstract Algebra, Second Edition, by John A. Beachy and William D. Blair Corrections and clarifications Note: Some corrections were made after the first printing of the text. page 9, line 8 For of the

### CRing Project, Chapter 5

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

### ALGEBRA II: RINGS AND MODULES OVER LITTLE RINGS.

ALGEBRA II: RINGS AND MODULES OVER LITTLE RINGS. KEVIN MCGERTY. 1. RINGS The central characters of this course are algebraic objects known as rings. A ring is any mathematical structure where you can add

### Math 210B:Algebra, Homework 2

Math 210B:Algebra, Homework 2 Ian Coley January 21, 2014 Problem 1. Is f = 2X 5 6X + 6 irreducible in Z[X], (S 1 Z)[X], for S = {2 n, n 0}, Q[X], R[X], C[X]? To begin, note that 2 divides all coefficients

Graduate Preliminary Examination Algebra II 18.2.2005: 3 hours Problem 1. Prove or give a counter-example to the following statement: If M/L and L/K are algebraic extensions of fields, then M/K is algebraic.

### Math Introduction to Modern Algebra

Math 343 - Introduction to Modern Algebra Notes Field Theory Basics Let R be a ring. M is called a maximal ideal of R if M is a proper ideal of R and there is no proper ideal of R that properly contains

### Math 547, Exam 1 Information.

Math 547, Exam 1 Information. 2/10/10, LC 303B, 10:10-11:00. Exam 1 will be based on: Sections 5.1, 5.2, 5.3, 9.1; The corresponding assigned homework problems (see http://www.math.sc.edu/ boylan/sccourses/547sp10/547.html)

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

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

### 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

### Groups, Rings, and Finite Fields. Andreas Klappenecker. September 12, 2002

Background on Groups, Rings, and Finite Fields Andreas Klappenecker September 12, 2002 A thorough understanding of the Agrawal, Kayal, and Saxena primality test requires some tools from algebra and elementary

### 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

### 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

### (Rgs) Rings Math 683L (Summer 2003)

(Rgs) Rings Math 683L (Summer 2003) We will first summarise the general results that we will need from the theory of rings. A unital ring, R, is a set equipped with two binary operations + and such that

### DIHEDRAL GROUPS II KEITH CONRAD

DIHEDRAL GROUPS II KEITH CONRAD We will characterize dihedral groups in terms of generators and relations, and describe the subgroups of D n, including the normal subgroups. We will also introduce an infinite

### 5.1 Commutative rings; Integral Domains

5.1 J.A.Beachy 1 5.1 Commutative rings; Integral Domains from A Study Guide for Beginner s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair 23. Let R be a commutative ring. Prove the following

### 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

### A Harvard Sampler. Evan Chen. February 23, I crashed a few math classes at Harvard on February 21, Here are notes from the classes.

A Harvard Sampler Evan Chen February 23, 2014 I crashed a few math classes at Harvard on February 21, 2014. Here are notes from the classes. 1 MATH 123: Algebra II In this lecture we will make two assumptions.

### 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

### 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.

### 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

### 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

### Math 121 Homework 3 Solutions

Math 121 Homework 3 Solutions Problem 13.4 #6. Let K 1 and K 2 be finite extensions of F in the field K, and assume that both are splitting fields over F. (a) Prove that their composite K 1 K 2 is a splitting

### 1 First Theme: Sums of Squares

I will try to organize the work of this semester around several classical questions. The first is, When is a prime p the sum of two squares? The question was raised by Fermat who gave the correct answer

### 1.5 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 well-behaved or tractable. Two types considered here are the nil and

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

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

### MATH 1530 ABSTRACT ALGEBRA Selected solutions to problems. a + b = a + b,

MATH 1530 ABSTRACT ALGEBRA Selected solutions to problems Problem Set 2 2. Define a relation on R given by a b if a b Z. (a) Prove that is an equivalence relation. (b) Let R/Z denote the set of equivalence

### Math 4400, Spring 08, Sample problems Final Exam.

Math 4400, Spring 08, Sample problems Final Exam. 1. Groups (1) (a) Let a be an element of a group G. Define the notions of exponent of a and period of a. (b) Suppose a has a finite period. Prove that

### Semilattice Modes II: the amalgamation property

Semilattice Modes II: the amalgamation property Keith A. Kearnes Abstract Let V be a variety of semilattice modes with associated semiring R. We prove that if R is a bounded distributive lattice, then

### Universal Algebra for Logics

Universal Algebra for Logics Joanna GRYGIEL University of Czestochowa Poland j.grygiel@ajd.czest.pl 2005 These notes form Lecture Notes of a short course which I will give at 1st School on Universal Logic

### 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)