INTRODUCTION TO ALGEBRAIC GEOMETRY, CLASS 18 Contents
... corresponds to a point of Y mapping to zero.) Also, Bn ⊂ R. And also nBn ⊂ mR . Now Bn is also a DVR of K/k, so Bn = R by the Commutative Algebra Lemma. Hence if y is in mR , then y is in n. To say that y is in n, means that y, as a regular function on Y , vanishes at the corresponding point. As non ...
... corresponds to a point of Y mapping to zero.) Also, Bn ⊂ R. And also nBn ⊂ mR . Now Bn is also a DVR of K/k, so Bn = R by the Commutative Algebra Lemma. Hence if y is in mR , then y is in n. To say that y is in n, means that y, as a regular function on Y , vanishes at the corresponding point. As non ...
1 Principal Ideal Domains
... Theorem 1. Let R be a PID. Let a, b ∈ R be non-zero elements and let d be an element such that (d) = (a, b). Then: a.) d is a GCD of a and b. 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 Euclid ...
... Theorem 1. Let R be a PID. Let a, b ∈ R be non-zero elements and let d be an element such that (d) = (a, b). Then: a.) d is a GCD of a and b. 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 Euclid ...
Solutions - Dartmouth Math Home
... this equality, using the fact that N respects multiplication, and using N(1) = 1, we have N(α)N(β) = N(αβ) = N(1) = 1. But both N(α) and N(β) are integers, and their product is the multiplicative identity 1 of Z. Therefore N(α) ∈ Z× . Conversely, suppose N(α) ∈ Z× . By definition, there exists u ∈ Z ...
... this equality, using the fact that N respects multiplication, and using N(1) = 1, we have N(α)N(β) = N(αβ) = N(1) = 1. But both N(α) and N(β) are integers, and their product is the multiplicative identity 1 of Z. Therefore N(α) ∈ Z× . Conversely, suppose N(α) ∈ Z× . By definition, there exists u ∈ Z ...
Commutative Algebra Fall 2014/2015 Problem set III, for
... for every open set U associates (a set, an abelian group, a vector space, etc) S(U ). By default, we will assume that S(∅) is the terminal object in the category. Moreover for every pair of open sets U ⊆ V we have a morphism rV U : S(V ) → S(U ) such that for any triple U ⊆ V ⊆ W it holds rV U ◦ rW ...
... for every open set U associates (a set, an abelian group, a vector space, etc) S(U ). By default, we will assume that S(∅) is the terminal object in the category. Moreover for every pair of open sets U ⊆ V we have a morphism rV U : S(V ) → S(U ) such that for any triple U ⊆ V ⊆ W it holds rV U ◦ rW ...
Groups - CSE-IITK
... the notion of similarity now. Clearly if two sets are equal if and only if there is a bijection between them. But the bijection need not respect the composition. That means the composition properties of two groups might be completely different even if they have a bijection between them. Exercise 9. ...
... the notion of similarity now. Clearly if two sets are equal if and only if there is a bijection between them. But the bijection need not respect the composition. That means the composition properties of two groups might be completely different even if they have a bijection between them. Exercise 9. ...
Distribution of the Distributive Property
... 3(2 + x) to produce the equivalent expression 6 + 3x; apply the distributive property to the expression 24x + 18y to produce the equivalent expression 6(4x + 3y). ...
... 3(2 + x) to produce the equivalent expression 6 + 3x; apply the distributive property to the expression 24x + 18y to produce the equivalent expression 6(4x + 3y). ...
Algebra for Digital Communication
... (3) Let’s give explicit descriptions of these two homomorphisms, constructed, as usual, by sending [1] (in Z/4Z or Z/12Z) on 1R = [9]12 . Then using additivity, the only possibility is: f ([x]4 ) = f (x · [1]4 ) = x · f ([1]4 ) = x · [9]12 = [9x]12 , and g([x]12 ) = [9x]12 . We can then verify that ...
... (3) Let’s give explicit descriptions of these two homomorphisms, constructed, as usual, by sending [1] (in Z/4Z or Z/12Z) on 1R = [9]12 . Then using additivity, the only possibility is: f ([x]4 ) = f (x · [1]4 ) = x · f ([1]4 ) = x · [9]12 = [9x]12 , and g([x]12 ) = [9x]12 . We can then verify that ...
selected solutions to Homework 11
... / S. Thus, S is not closed under addition and is not a subring. (b) This statement is true. Proof. Let R be a ring with unity, 1. Let a, b ∈ S. This means that there exist c, d ∈ R such that ac = 1 and bd = 1. Then (ab)(dc) = a(bd)c = a(1)(c) = (ac) = 1 Thus, (ab) ∈ S and S is closed under multiplic ...
... / S. Thus, S is not closed under addition and is not a subring. (b) This statement is true. Proof. Let R be a ring with unity, 1. Let a, b ∈ S. This means that there exist c, d ∈ R such that ac = 1 and bd = 1. Then (ab)(dc) = a(bd)c = a(1)(c) = (ac) = 1 Thus, (ab) ∈ S and S is closed under multiplic ...
MATH 521A: Abstract Algebra Homework 7 Solutions 1. Consider
... different ones, such as f (x) = (x + 2)(x + x + 1), and 10 ways of picking the square of one, such as f (x) = (x2 + 2)2 . Hence there are 45 + 10 = 55 answers to this question. 6. Factor x7 − x as a product of irreducibles in Z7 [x]. By Fermat’s Little Theorem, x7 ≡ x (mod 7), for all integer x. Hen ...
... different ones, such as f (x) = (x + 2)(x + x + 1), and 10 ways of picking the square of one, such as f (x) = (x2 + 2)2 . Hence there are 45 + 10 = 55 answers to this question. 6. Factor x7 − x as a product of irreducibles in Z7 [x]. By Fermat’s Little Theorem, x7 ≡ x (mod 7), for all integer x. Hen ...
![Math 110B HW §5.3 – Solutions 3. Show that [−a, b] is the additive](http://s1.studyres.com/store/data/017359919_1-72a70245febeadd05992d7dba1b6dd48-300x300.png)
Math 110B HW §5.3 – Solutions 3. Show that [−a, b] is the additive
... Thus the right distributive property in Q holds. 8. Assume that the ring R is isomorphic to the ring R0 . Prove that if R is commutative, then R0 is commutative. Proof. Let φ : R → R0 be an isomorphism and assume R is commutative. Let a0 , b0 ∈ R0. Since φ is onto, there exists a, b ∈ R such that φ( ...
... Thus the right distributive property in Q holds. 8. Assume that the ring R is isomorphic to the ring R0 . Prove that if R is commutative, then R0 is commutative. Proof. Let φ : R → R0 be an isomorphism and assume R is commutative. Let a0 , b0 ∈ R0. Since φ is onto, there exists a, b ∈ R such that φ( ...
MAT 364 - Homework 9 Solutions
... Exercise 3.15– Show that if X is a topological space consisting of a finite number of points, with any topology, then X is compact. Proof: As X is finite, we can write X = {x1 , x2 , . . . , xn }. Let U = {Uα }α∈I be any open cover of X. Note that we do not assume anything about this cover - the ind ...
... Exercise 3.15– Show that if X is a topological space consisting of a finite number of points, with any topology, then X is compact. Proof: As X is finite, we can write X = {x1 , x2 , . . . , xn }. Let U = {Uα }α∈I be any open cover of X. Note that we do not assume anything about this cover - the ind ...
Birkhoff's representation theorem
This is about lattice theory. For other similarly named results, see Birkhoff's theorem (disambiguation).In mathematics, Birkhoff's representation theorem for distributive lattices states that the elements of any finite distributive lattice can be represented as finite sets, in such a way that the lattice operations correspond to unions and intersections of sets. The theorem can be interpreted as providing a one-to-one correspondence between distributive lattices and partial orders, between quasi-ordinal knowledge spaces and preorders, or between finite topological spaces and preorders. It is named after Garrett Birkhoff, who published a proof of it in 1937.The name “Birkhoff's representation theorem” has also been applied to two other results of Birkhoff, one from 1935 on the representation of Boolean algebras as families of sets closed under union, intersection, and complement (so-called fields of sets, closely related to the rings of sets used by Birkhoff to represent distributive lattices), and Birkhoff's HSP theorem representing algebras as products of irreducible algebras. Birkhoff's representation theorem has also been called the fundamental theorem for finite distributive lattices.