MTE-6-AST-2004
... For any three subsets A, B, C of a set U, A C if and only if A Bc C. The set of all mappings from {1, 2, , n} to itself form a group with respect to composition of maps. For any two elements a, b of a group G, o(ab) = o(ba). The set of elements of GL2 (R) whose orders divide a fixed numbe ...
... For any three subsets A, B, C of a set U, A C if and only if A Bc C. The set of all mappings from {1, 2, , n} to itself form a group with respect to composition of maps. For any two elements a, b of a group G, o(ab) = o(ba). The set of elements of GL2 (R) whose orders divide a fixed numbe ...
Dmytro Taranovsky
... An inner product between two vectors u and v, a real number, is defined such that = , = + , u≠0⇒∃w ≠0. Metric tensor, g≡EiEj. Conjugate
metric tensor, g, is such that gijgjk= ki . Metric is a smooth (that is infinitely differentiable) tenso ...
... An inner product between two vectors u and v, a real number
F-SINGULARITIES AND FROBENIUS SPLITTING
... for any a1 , . . . , an ∈ R such that a1 xp1 + · · · + an xpn = 0, then ai ∈ m[p ] . We now begin step 2. For this, we begin with a Lemma. Lemma 1.3. [Lec64, Lemma 3] If f1 , . . . , fn are Lech-independent elements and f1 ∈ gR for some g ∈ R, then g, f2 , . . . , fn is also Lech-independent. Furthe ...
... for any a1 , . . . , an ∈ R such that a1 xp1 + · · · + an xpn = 0, then ai ∈ m[p ] . We now begin step 2. For this, we begin with a Lemma. Lemma 1.3. [Lec64, Lemma 3] If f1 , . . . , fn are Lech-independent elements and f1 ∈ gR for some g ∈ R, then g, f2 , . . . , fn is also Lech-independent. Furthe ...
Cohomology as the derived functor of derivations.
... and a natural 1-1 correspondence between Der^T.M) and {a:T->M x R\eb0oi = eb}. Let r€0 be the category of sets in Case G, of unitary /¿-modules in Cases A and L, and of "supplemented" unitary X-modules (in the obvious sense) in Case S. Then there is a natural inclusion functor of <&into ^0. Moreover ...
... and a natural 1-1 correspondence between Der^T.M) and {a:T->M x R\eb0oi = eb}. Let r€0 be the category of sets in Case G, of unitary /¿-modules in Cases A and L, and of "supplemented" unitary X-modules (in the obvious sense) in Case S. Then there is a natural inclusion functor of <&into ^0. Moreover ...
Solutions to Homework 9 46. (Dummit
... which says that there is a non-trivial linear relation between any two basis elements of Q. Hence Q is a free Z-module of rank at most 1. But a/2b is not in the Z-span of a/b for any a/b ∈ Q, which says that Q cannot have rank 1 either. Thus Q is not a free Z-module. (b) Suppose that I were a free R ...
... which says that there is a non-trivial linear relation between any two basis elements of Q. Hence Q is a free Z-module of rank at most 1. But a/2b is not in the Z-span of a/b for any a/b ∈ Q, which says that Q cannot have rank 1 either. Thus Q is not a free Z-module. (b) Suppose that I were a free R ...
Counterexamples in Algebra
... Commutative Ring with Identity that is Not an Integral Domain. Z × Z, Z/6Z. Commutative Ring without Identity. 2Z, {0, 2} in Z/4Z. Noncommutative Ring without Identity. M2 (2Z). Noncommutative Division Ring with Identity. The real quarternion H. Ring with Cyclic Multiplicative Group. R = Z/nZ with n ...
... Commutative Ring with Identity that is Not an Integral Domain. Z × Z, Z/6Z. Commutative Ring without Identity. 2Z, {0, 2} in Z/4Z. Noncommutative Ring without Identity. M2 (2Z). Noncommutative Division Ring with Identity. The real quarternion H. Ring with Cyclic Multiplicative Group. R = Z/nZ with n ...
Math 400 Spring 2016 – Test 3 (Take
... Solution: It’s false. The smallest (lowest order) counterexample is G = D4 , H = {R0 , R180 , v, h}, K = {R0 , v}. (Prove each subgroup is normal in the one larger by noting index 2; prove K 6C G by noting dvd 6∈ K. 2. (8) Let G be the quaternion group (also known as Q8 ) that you worked with on Tes ...
... Solution: It’s false. The smallest (lowest order) counterexample is G = D4 , H = {R0 , R180 , v, h}, K = {R0 , v}. (Prove each subgroup is normal in the one larger by noting index 2; prove K 6C G by noting dvd 6∈ K. 2. (8) Let G be the quaternion group (also known as Q8 ) that you worked with on Tes ...
This is the syllabus for MA5b, as taught in Winter 2016. Syllabus for
... finitely generated then M torsion iff M is not faithfull (HW). The endomorphism ring EndR (M ) for M a R-module. Schur’s lemma. M is a R-module iff there is a ring homomorphism R to End(M ). If R → A is a ring homomorphism, then A is an R-module. hom(M, N ) is an R-module. If R is commutative: homR ...
... finitely generated then M torsion iff M is not faithfull (HW). The endomorphism ring EndR (M ) for M a R-module. Schur’s lemma. M is a R-module iff there is a ring homomorphism R to End(M ). If R → A is a ring homomorphism, then A is an R-module. hom(M, N ) is an R-module. If R is commutative: homR ...
Module (mathematics)
... over the ring of integers Z in a unique way. For n > 0, let nx = x + x + ... + x (n summands), 0x = 0, and (−n)x = −(nx). Such a module need not have a basis—groups containing torsion elements do not. (For example, in the group of integers modulo 3, one cannot find even one element which satisfies t ...
... over the ring of integers Z in a unique way. For n > 0, let nx = x + x + ... + x (n summands), 0x = 0, and (−n)x = −(nx). Such a module need not have a basis—groups containing torsion elements do not. (For example, in the group of integers modulo 3, one cannot find even one element which satisfies t ...
18.703 Modern Algebra, The Isomorphism Theorems
... any category, the product is unique, up to unique isomorphism. The proof proceeds exactly as in the proof of the uniqueness of a categorical quotient and is left as an exercise for the reader. Lemma 10.11. The product of groups is a categorical product. That is, given two groups G and H, the group G ...
... any category, the product is unique, up to unique isomorphism. The proof proceeds exactly as in the proof of the uniqueness of a categorical quotient and is left as an exercise for the reader. Lemma 10.11. The product of groups is a categorical product. That is, given two groups G and H, the group G ...
15. The functor of points and the Hilbert scheme Clearly a scheme
... (1) If F is any functor from C to the category of sets, the natural transformations between hX and F are in natural correspondence with the elements of F (X). (2) h is an equivalence of categories with a full subcategory of Hom(C ◦ , D). Proof. Given a natural transformation α : hX −→ F, we assign ...
... (1) If F is any functor from C to the category of sets, the natural transformations between hX and F are in natural correspondence with the elements of F (X). (2) h is an equivalence of categories with a full subcategory of Hom(C ◦ , D). Proof. Given a natural transformation α : hX −→ F, we assign ...
Physics 129B, Winter 2010 Problem Set 1 Solution
... This should be the multiplication table for all groups of order 2, because we made no assumption for the group except that it is of order 2. So all groups of order 2, one of which is C2 , are isomorphic to one another. Next, consider a group of order 3, which has three elements e, a, and b. We can s ...
... This should be the multiplication table for all groups of order 2, because we made no assumption for the group except that it is of order 2. So all groups of order 2, one of which is C2 , are isomorphic to one another. Next, consider a group of order 3, which has three elements e, a, and b. We can s ...
22. Quotient groups I 22.1. Definition of quotient groups. Let G be a
... Theorem 22.2. Let G be a group and H a normal subgroup of G. Then the quotient set G/H is a group with respect to the operation ∗ defined by (Q). Proof. (G0) G/H is closed under ∗ by definition of cosets. (G1) Associativity of ∗ follows from the associativity of the group operation on G: for any g, ...
... Theorem 22.2. Let G be a group and H a normal subgroup of G. Then the quotient set G/H is a group with respect to the operation ∗ defined by (Q). Proof. (G0) G/H is closed under ∗ by definition of cosets. (G1) Associativity of ∗ follows from the associativity of the group operation on G: for any g, ...
Introduction to tensor, tensor factorization and its applications
... 2. Tensor multiplication 3. Tensor rank Tensor factorization 1. CANDECOMP/PARAFAC factorization 2. Tucker factorization Applications of tensor factorization ...
... 2. Tensor multiplication 3. Tensor rank Tensor factorization 1. CANDECOMP/PARAFAC factorization 2. Tucker factorization Applications of tensor factorization ...
aa5.pdf
... (ii) Prove that if AeA = A then ψ is an isomorphism. (iii) Prove the following generalization of (ii): if AeA = A then, for any right A-module M and ...
... (ii) Prove that if AeA = A then ψ is an isomorphism. (iii) Prove the following generalization of (ii): if AeA = A then, for any right A-module M and ...
TENSOR PRODUCTS OF LOCALLY CONVEX ALGEBRAS 124
... of MiA3) is A3. Let/ be an element of Ai. If A3 is semisimple there is, for every e>0 and every finite set {x,} of elements of A3, a finite combination of elements of MiA3) approximating/within e on the set {x,}. By Corollary 2, each element of MiA3) is in the extension to A3 of Ai ® Ai and this set ...
... of MiA3) is A3. Let/ be an element of Ai. If A3 is semisimple there is, for every e>0 and every finite set {x,} of elements of A3, a finite combination of elements of MiA3) approximating/within e on the set {x,}. By Corollary 2, each element of MiA3) is in the extension to A3 of Ai ® Ai and this set ...
Algebra 411 Homework 5: hints and solutions
... (What is the condition if we have the product of three or more cyclic groups? There are several ways to answer this.) (a) hcf(12, 9) = 3 so not cyclic. (b) hcf(10, 85) = 5 so not cyclic. (c) hcf(4, 6) = 2 so not cyclic. (d) hcf(22, 21) = hcf(22, 65) = hcf(21, 65) = 1 so is cyclic. 6.6 We show that G ...
... (What is the condition if we have the product of three or more cyclic groups? There are several ways to answer this.) (a) hcf(12, 9) = 3 so not cyclic. (b) hcf(10, 85) = 5 so not cyclic. (c) hcf(4, 6) = 2 so not cyclic. (d) hcf(22, 21) = hcf(22, 65) = hcf(21, 65) = 1 so is cyclic. 6.6 We show that G ...
Solution
... basis is the same, except that column j turns into Cj − λ−1 Ci . Thus we can do column reductions on the matrix (wij ) by replacing the basis {ui }m i=1 . Similarly, we can do row reductions by replacing the basis {vj }nj=1 . A matrix has rank r if and only if it can be reduced to a matrix with r 1’ ...
... basis is the same, except that column j turns into Cj − λ−1 Ci . Thus we can do column reductions on the matrix (wij ) by replacing the basis {ui }m i=1 . Similarly, we can do row reductions by replacing the basis {vj }nj=1 . A matrix has rank r if and only if it can be reduced to a matrix with r 1’ ...
Groups with exponents I. Fundamentals of the theory and tensor
... In applications # will mostly be an embedding of rings. However, even in this case, the homomorphism A : G ---* G B'" is not always an embedding. Since in the abelian case the group G B results from by tensoring the A-module G by the ring B, appropriate examples can be found in many articles on comm ...
... In applications # will mostly be an embedding of rings. However, even in this case, the homomorphism A : G ---* G B'" is not always an embedding. Since in the abelian case the group G B results from by tensoring the A-module G by the ring B, appropriate examples can be found in many articles on comm ...
Representation theory of finite groups
... (note that V is a k-vector space, so we already know how to scale by the ai ∈ k). The k[x]-submodules of V are precisely the T -invariant subspaces W of V used to develop the theory of canonical forms of matrices, that is, those W ≤ V such that T (W ) ⊆ W (check this!) In fact, finitely generated mo ...
... (note that V is a k-vector space, so we already know how to scale by the ai ∈ k). The k[x]-submodules of V are precisely the T -invariant subspaces W of V used to develop the theory of canonical forms of matrices, that is, those W ≤ V such that T (W ) ⊆ W (check this!) In fact, finitely generated mo ...
April 6: Groups Acting on Categories
... if their composition comes from C), H−equivariant object. Throughout the above discussion, we have mostly dealt with a weak/strong action of G on an abelian category C. However, this always induces a weak/strong action on the category of complexes C(C), and from there to a weak/strong action on the ...
... if their composition comes from C), H−equivariant object. Throughout the above discussion, we have mostly dealt with a weak/strong action of G on an abelian category C. However, this always induces a weak/strong action on the category of complexes C(C), and from there to a weak/strong action on the ...
FREE MODULES Throughout, let A be a commutative ring with 1. 1
... The proposition shows that in particular if the ring A is a field k then the free k-module on S is the k-vector space having basis {i(s) : s ∈ S}. We will use this fact in the next section. 4. Invariance of Rank Although we still don’t have the free A-module on a set S, we use the characteristic map ...
... The proposition shows that in particular if the ring A is a field k then the free k-module on S is the k-vector space having basis {i(s) : s ∈ S}. We will use this fact in the next section. 4. Invariance of Rank Although we still don’t have the free A-module on a set S, we use the characteristic map ...
3. Modules
... However, although many properties just carry over without change, others will turn out to be vastly different. Of course, proofs that are literally the same as for vector spaces will not be repeated here; instead we will just give references to the corresponding well-known linear algebra statements ...
... However, although many properties just carry over without change, others will turn out to be vastly different. Of course, proofs that are literally the same as for vector spaces will not be repeated here; instead we will just give references to the corresponding well-known linear algebra statements ...
THE TENSOR PRODUCT OF FUNCTION SEMIMODULES The
... and B are infinite, the map µ still remains injective. We recall the argument for injectivity in the proof of Proposition 3.3. The purpose of the present paper is to observe that, in marked contrast to the situation over fields, the canonical map µ ceases to be injective in general when one studies ...
... and B are infinite, the map µ still remains injective. We recall the argument for injectivity in the proof of Proposition 3.3. The purpose of the present paper is to observe that, in marked contrast to the situation over fields, the canonical map µ ceases to be injective in general when one studies ...
10. The isomorphism theorems We have already seen that given
... Proof. Consider the natural map G −→ G/H. The kernel, H, contains K. Thus, by the universal property of G/K, it follows that there is a homomorphism G/K −→ G/H. This map is clearly surjective. In fact, it sends the left coset gK to the left coset gH. Now suppose that gK is in the kernel. Then the le ...
... Proof. Consider the natural map G −→ G/H. The kernel, H, contains K. Thus, by the universal property of G/K, it follows that there is a homomorphism G/K −→ G/H. This map is clearly surjective. In fact, it sends the left coset gK to the left coset gH. Now suppose that gK is in the kernel. Then the le ...
Tensor product of modules
In mathematics, the tensor product of modules is a construction that allows arguments about bilinear maps (e.g. multiplication) to be carried out in terms of linear maps (module homomorphisms). The module construction is analogous to the construction of the tensor product of vector spaces, but can be carried out for a pair of modules over a commutative ring resulting in a third module, and also for a pair of a right-module and a left-module over any ring, with result an abelian group. Tensor products are important in areas of abstract algebra, homological algebra, algebraic topology and algebraic geometry. The universal property of the tensor product of vector spaces extends to more general situations in abstract algebra. It allows the study of bilinear or multilinear operations via linear operations. The tensor product of an algebra and a module can be used for extension of scalars. For a commutative ring, the tensor product of modules can be iterated to form the tensor algebra of a module, allowing one to define multiplication in the module in a universal way.