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) ...
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 ...
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 ...
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 ...
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 ...
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 ...
