Math 306, Spring 2012 Homework 1 Solutions
... (5) (5 pts) Prove that an integral domain R with a finite number of elements is a field. (Hint: For each nonzero a ∈ R, consider the map λa : R → R given by λa (r) = ar for all r ∈ R. Prove that λa is injective and use the fact that any injective function on a finite set is surjective.) Solution: Le ...
... (5) (5 pts) Prove that an integral domain R with a finite number of elements is a field. (Hint: For each nonzero a ∈ R, consider the map λa : R → R given by λa (r) = ar for all r ∈ R. Prove that λa is injective and use the fact that any injective function on a finite set is surjective.) Solution: Le ...
13. Dedekind Domains
... maximal ideals in A(X) are exactly the ideals of points, and so our smoothness assumption is the same as saying that all localizations at maximal ideals are regular. In fact, irreducible smooth curves over algebraically closed fields are the main geometric examples for Dedekind domains. However, the ...
... maximal ideals in A(X) are exactly the ideals of points, and so our smoothness assumption is the same as saying that all localizations at maximal ideals are regular. In fact, irreducible smooth curves over algebraically closed fields are the main geometric examples for Dedekind domains. However, the ...
EUCLIDEAN RINGS 1. Introduction The topic of this lecture is
... And our rings Z, Z[i], Z[ω], and k[X] are all Euclidean, so they are UFDs. This all may seem pointlessly Byzantine, but the issues here are already live in utterly elementary contexts. For example: • Math 112 exercises have to dance around the question For what positive integers n is Z/nZ a field? E ...
... And our rings Z, Z[i], Z[ω], and k[X] are all Euclidean, so they are UFDs. This all may seem pointlessly Byzantine, but the issues here are already live in utterly elementary contexts. For example: • Math 112 exercises have to dance around the question For what positive integers n is Z/nZ a field? E ...
2008-09
... Let I, J be ideals of a ring R such that I + J = R. Prove that I J = IJ and if IJ = 0, then R ; R R . ...
... Let I, J be ideals of a ring R such that I + J = R. Prove that I J = IJ and if IJ = 0, then R ; R R . ...
Solutions - Dartmouth Math Home
... We know that Z[i] is a subring of C. Because C is a field, C is necessarily an integral domain. If αβ = 0 for some α, β ∈ Z[i], we can also consider this equality in C. Since C is an integral domain, either α = 0 or β = 0. This shows that Z[i] is an integral domain. 8. Recall that any fraction can b ...
... We know that Z[i] is a subring of C. Because C is a field, C is necessarily an integral domain. If αβ = 0 for some α, β ∈ Z[i], we can also consider this equality in C. Since C is an integral domain, either α = 0 or β = 0. This shows that Z[i] is an integral domain. 8. Recall that any fraction can b ...
Math 396. Modules and derivations 1. Preliminaries Let R be a
... We now apply the notion of quotient M/JM of a R-module M modulo an ideal J of R in order to show how linear algebra is a helpful device in the study of modules over general commutative rings. A nonzero module M over R is finite free of rank n > 0 over R if there exists a finite R-basisP of size n; t ...
... We now apply the notion of quotient M/JM of a R-module M modulo an ideal J of R in order to show how linear algebra is a helpful device in the study of modules over general commutative rings. A nonzero module M over R is finite free of rank n > 0 over R if there exists a finite R-basisP of size n; t ...
Garrett 11-04-2011 1 Recap: A better version of localization...
... O integral over o and prime ideal p of o, there is at least one prime ideal P of O such that P ∩ o = p. P is maximal if and only if p is maximal. p · O 6= O. [Here use Nakayama, localization.] Now o is a domain, integrally closed in its field of fractions k. For K/k finite Galois, the Galois group G ...
... O integral over o and prime ideal p of o, there is at least one prime ideal P of O such that P ∩ o = p. P is maximal if and only if p is maximal. p · O 6= O. [Here use Nakayama, localization.] Now o is a domain, integrally closed in its field of fractions k. For K/k finite Galois, the Galois group G ...
Valuations and discrete valuation rings, PID`s
... Proof: We first have to show that each nonzero r ∈ R has a factorization into a finite product of irreducible elements. If r can’t be written as a finite product of irreducibles, then there has to be a factorization r = st in which neither s or t are units. Furthermore one of s or t can’t be written ...
... Proof: We first have to show that each nonzero r ∈ R has a factorization into a finite product of irreducible elements. If r can’t be written as a finite product of irreducibles, then there has to be a factorization r = st in which neither s or t are units. Furthermore one of s or t can’t be written ...
MTE-06-2008
... Let I, J be ideals of a ring R such that I + J = R. Prove that I J = IJ and if IJ = 0, then R ; R R . ...
... Let I, J be ideals of a ring R such that I + J = R. Prove that I J = IJ and if IJ = 0, then R ; R R . ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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
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 ...