MATH 254A: RINGS OF INTEGERS AND DEDEKIND DOMAINS 1

... all of algebraic number theory. 2. Dedekind domains We are now ready to prove the following theorem: Theorem 2.1. A ring of integers OK is a Dedekind domain: i.e., it satisfies (i) OK is Noetherian; (ii) Every non-zero prime ideal of OK is maximal; (iii) OK is integrally closed in its field of fract ...

... all of algebraic number theory. 2. Dedekind domains We are now ready to prove the following theorem: Theorem 2.1. A ring of integers OK is a Dedekind domain: i.e., it satisfies (i) OK is Noetherian; (ii) Every non-zero prime ideal of OK is maximal; (iii) OK is integrally closed in its field of fract ...

MATH 254A: DEDEKIND DOMAINS I 1. Properties of Dedekind

... any prime ideal containing x, because m is principal it must be equal to (x). This shows by (i) that any PID is a UFD. But the same argument, if x is a generator for any non-zero prime ideal, shows that (x) is maximal. We already showed that any UFD is integrally closed, so every PID is also a Dedek ...

... any prime ideal containing x, because m is principal it must be equal to (x). This shows by (i) that any PID is a UFD. But the same argument, if x is a generator for any non-zero prime ideal, shows that (x) is maximal. We already showed that any UFD is integrally closed, so every PID is also a Dedek ...

Principal Ideal Domains

... This section of notes roughly follows Sections 8.1-8.2 in Dummit and Foote. Throughout this whole section, we assume that R is a commutative ring. Definition 55. Let R b a commutative ring and let a, b ∈ R with b , 0. (1) a is said to be multiple of b if there exists an element x ∈ R with a = bx. In ...

... This section of notes roughly follows Sections 8.1-8.2 in Dummit and Foote. Throughout this whole section, we assume that R is a commutative ring. Definition 55. Let R b a commutative ring and let a, b ∈ R with b , 0. (1) a is said to be multiple of b if there exists an element x ∈ R with a = bx. In ...

1 Principal Ideal Domains

... b.) d = ax + by for some x, y ∈ R. c.) d is unique up to multiplication by a unit in R We’ve already proved all of these - the only thing we needed the Euclidean Domain property for was a way to find that GCD. In PIDs, we at least are guaranteed that the GCD will exist, although it may in general be ...

... b.) d = ax + by for some x, y ∈ R. c.) d is unique up to multiplication by a unit in R We’ve already proved all of these - the only thing we needed the Euclidean Domain property for was a way to find that GCD. In PIDs, we at least are guaranteed that the GCD will exist, although it may in general be ...

MAXIMAL AND NON-MAXIMAL ORDERS 1. Introduction Let K be a

... is integrally closed. Combining this with Proposition 2.7, we get that OK is a Dedekind domain. Proposition 3.2. Let R ( OK be a non-maximal order. Then R is not integrally closed, and therefore is not a Dedekind domain. Proof: Because R ( OK , there is some β ∈ OK , β ∈ / R. Note that β ∈ OK ⊆ F ...

... is integrally closed. Combining this with Proposition 2.7, we get that OK is a Dedekind domain. Proposition 3.2. Let R ( OK be a non-maximal order. Then R is not integrally closed, and therefore is not a Dedekind domain. Proof: Because R ( OK , there is some β ∈ OK , β ∈ / R. Note that β ∈ OK ⊆ F ...

Problem Set 1 - University of Oxford

... Solution: From the theory of cyclic groups, a homomorphism from the abelian group Z to any group is completely determined by the image of 1 ∈ Z: if f : Z → (R, +) is a homomorphism of abelian groups then f (n) = f (1)+f (1)+. . . f (1) (n terms on the righthand side) when n ≥ 0 and f (n) = −f (−n) ...

... Solution: From the theory of cyclic groups, a homomorphism from the abelian group Z to any group is completely determined by the image of 1 ∈ Z: if f : Z → (R, +) is a homomorphism of abelian groups then f (n) = f (1)+f (1)+. . . f (1) (n terms on the righthand side) when n ≥ 0 and f (n) = −f (−n) ...

Dedekind domains and rings of quotients

... Proof. For each a, chose nΛ such that Pl<* is principal, say = A aa. Let S be the multiplicatively closed set generated by all aa. By Theorem 1-4, As is not a principal ideal domain, hence A8 must have an infinite number of non-principal prime ideals by Corollary 1-6. These come from non-principal p ...

... Proof. For each a, chose nΛ such that Pl<* is principal, say = A aa. Let S be the multiplicatively closed set generated by all aa. By Theorem 1-4, As is not a principal ideal domain, hence A8 must have an infinite number of non-principal prime ideals by Corollary 1-6. These come from non-principal p ...

Sample Exam #1

... 1. (40) pts. Let a, b, d, p, n with b 0 and n > 1. Let and ~~ be rings.
Define or tell what is meant by the following:
(a) b divides a (b| a)
(b) d is the greatest common divisor of a and b (d = (a,b))
(c) p is prime
(d) a and b are relatively prime
(e) a is congruent to b modu ...
~~

... 1. (40) pts. Let a, b, d, p, n with b 0 and n > 1. Let

Chapter 2 Introduction to Finite Field

... Definition 2.1 (Finite field and Order of finite field). A finite field is a field F which has a finite number of elements, this number being called the order of the field, denoted by |F |. Theorem 2.1 (Subfield Isomorphic to Zp ). Every finite field has the order of a power of a prime number p and ...

... Definition 2.1 (Finite field and Order of finite field). A finite field is a field F which has a finite number of elements, this number being called the order of the field, denoted by |F |. Theorem 2.1 (Subfield Isomorphic to Zp ). Every finite field has the order of a power of a prime number p and ...

4. Lecture 4 Visualizing rings We describe several ways - b

... The second method of representing rings geometrically is to draw a point for each basis element. Here we assume that the ring is a free module over some nice ring such as a field. For example k[x, y] can be represented by a 2-dimensional quadrant of a 2-dimensional lattice. This is widely used in co ...

... The second method of representing rings geometrically is to draw a point for each basis element. Here we assume that the ring is a free module over some nice ring such as a field. For example k[x, y] can be represented by a 2-dimensional quadrant of a 2-dimensional lattice. This is widely used in co ...

Ma 5b Midterm Review Notes

... Throughout this section, let R denote a commutative ring. Recall. An ideal P ⊂ R is prime if it is proper and its complement is closed under multiplication, i.e. for any ab ∈ P, either a ∈ P or b ∈ P. An ideal is maximal if it is proper and is not properly contained in any other proper ideal. Let I ...

... Throughout this section, let R denote a commutative ring. Recall. An ideal P ⊂ R is prime if it is proper and its complement is closed under multiplication, i.e. for any ab ∈ P, either a ∈ P or b ∈ P. An ideal is maximal if it is proper and is not properly contained in any other proper ideal. Let I ...

Sol 2 - D-MATH

... 1. (a) Is there an integral domain that contains exactly 15 elements ? Solution : The only additive abelian group of order 15 is the cyclic group Z15 (you can see this with Sylow, and conclude that all groups of order 15 are isomorphic to the product Z3 × Z5 of cyclic groups – cf. your notes of last ...

... 1. (a) Is there an integral domain that contains exactly 15 elements ? Solution : The only additive abelian group of order 15 is the cyclic group Z15 (you can see this with Sylow, and conclude that all groups of order 15 are isomorphic to the product Z3 × Z5 of cyclic groups – cf. your notes of last ...

Notes - UCSD Math Department

... Claim. S is nonempty and bounded from above. Proof of claim. (1) By Theorem 1.20(a), since 1 > 0 and x 2 R, there exists a positive integer n such that n = n 1 > x. That is, n < x; which says that n 2 S. So S is nonempty. (2). By Theorem 1.20(a) again, since 1 > 0 and x 2 R, there exists a positive ...

... Claim. S is nonempty and bounded from above. Proof of claim. (1) By Theorem 1.20(a), since 1 > 0 and x 2 R, there exists a positive integer n such that n = n 1 > x. That is, n < x; which says that n 2 S. So S is nonempty. (2). By Theorem 1.20(a) again, since 1 > 0 and x 2 R, there exists a positive ...

Problem Score 1 2 3 4 or 5 Total - Mathematics

... Solution: Since I is a proper ideal, R/I has non-zero elements. Suppose R/I is a free R-module, then R/I has a non-empty basis, and in particular it contains at least one linearly independent element x + I, x ∈ R. Let i ∈ I. Then i(x + I) = ix + I = I since I is an ideal so ix ∈ I. I is the zero ele ...

... Solution: Since I is a proper ideal, R/I has non-zero elements. Suppose R/I is a free R-module, then R/I has a non-empty basis, and in particular it contains at least one linearly independent element x + I, x ∈ R. Let i ∈ I. Then i(x + I) = ix + I = I since I is an ideal so ix ∈ I. I is the zero ele ...

THE LOWER ALGEBRAIC K-GROUPS 1. Introduction

... so by the UMP of direct limits there’s a homomorphism det : GL(R) → R× given by sending the equivalence class [A] of A ∈ GLn (R) to detn (A). This induces a map det1 : K1 (R) → R× since E(R) ⊆ SL(R) ⊆ ker(det). We also have a canonical injection ι1 : R× = GL1 (R) GL(R) K1 (R), so we may consider ...

... so by the UMP of direct limits there’s a homomorphism det : GL(R) → R× given by sending the equivalence class [A] of A ∈ GLn (R) to detn (A). This induces a map det1 : K1 (R) → R× since E(R) ⊆ SL(R) ⊆ ker(det). We also have a canonical injection ι1 : R× = GL1 (R) GL(R) K1 (R), so we may consider ...