modularity of elliptic curves
... II. Reducing an elliptic curve modulo p and computing its conductor In this paper, we are concerned with elliptic curves whose coefficients are rational numbers. By multiplying the entire equation by the coefficents’ greatest common denominator, we come up with a cubic whose coefficients are integer ...
... II. Reducing an elliptic curve modulo p and computing its conductor In this paper, we are concerned with elliptic curves whose coefficients are rational numbers. By multiplying the entire equation by the coefficents’ greatest common denominator, we come up with a cubic whose coefficients are integer ...
2. Groups I - Math User Home Pages
... the other hand, suppose that the kernel is trivial. We will suppose that f (x) = f (y), and show that x = y. Left multiply f (x) = f (y) by f (x)−1 to obtain eH = f (x)−1 · f (x) = f (x)−1 · f (y) By the homomorphism property, eH = f (x)−1 · f (y) = f (x−1 y) Thus, x−1 y is in the kernel of f , so ( ...
... the other hand, suppose that the kernel is trivial. We will suppose that f (x) = f (y), and show that x = y. Left multiply f (x) = f (y) by f (x)−1 to obtain eH = f (x)−1 · f (x) = f (x)−1 · f (y) By the homomorphism property, eH = f (x)−1 · f (y) = f (x−1 y) Thus, x−1 y is in the kernel of f , so ( ...
Dualizing DG modules and Gorenstein DG algebras
... [8]. Moreover, in what follows, a few well known results in this subject are used; for these we quote from [4] whenever possible, in the interest of uniformity. A few words about notation: Most graded objects that appear here are assumed to be graded homologically; for instance, a graded set X is a ...
... [8]. Moreover, in what follows, a few well known results in this subject are used; for these we quote from [4] whenever possible, in the interest of uniformity. A few words about notation: Most graded objects that appear here are assumed to be graded homologically; for instance, a graded set X is a ...
Haar Measures for Groupoids
... Now, we can also say that P ROPOSITION 2.21. ( f0 : f )−1 ≤ Iϕ ( f ) ≤ ( f : f0 ). Proof. By 2.20 (4), ( f0 : f )−1 ( f0 : ϕ) ≤ ( f : ϕ) ≤ ( f : f0 )( f0 : ϕ). For fixed ϕ, then, the functional Iϕ is left-invariant (by Lemma 2.20 (1)). We now show that, in a certain sense, Iϕ is approximately additi ...
... Now, we can also say that P ROPOSITION 2.21. ( f0 : f )−1 ≤ Iϕ ( f ) ≤ ( f : f0 ). Proof. By 2.20 (4), ( f0 : f )−1 ( f0 : ϕ) ≤ ( f : ϕ) ≤ ( f : f0 )( f0 : ϕ). For fixed ϕ, then, the functional Iϕ is left-invariant (by Lemma 2.20 (1)). We now show that, in a certain sense, Iϕ is approximately additi ...
Equivalence Relations and Partial Orders ()
... people like “A is a brother of B” or “A is B’s aunt” or “A and B are neighbors”. In mathematics, we have relations on sets of numbers like “≤”, “>”, and “sum to a rational number”. Another familiar relation is that of “⊆” when dealing with sets. It is quite useful to abstract the concept of equality ...
... people like “A is a brother of B” or “A is B’s aunt” or “A and B are neighbors”. In mathematics, we have relations on sets of numbers like “≤”, “>”, and “sum to a rational number”. Another familiar relation is that of “⊆” when dealing with sets. It is quite useful to abstract the concept of equality ...
9 Radical extensions
... Proof. By proposition 119 it is enough to prove that A5 is not solvable. Suppose that it is: A5 = B0 ⊃ B1 ⊃ · · · ⊃ Br = 1 with Bi +1 normal in Bi and Bi / Bi +1 abelian. Let f : B0 → B0 / B1 denote the natural homomorphism. As B0 / B1 is abelian we have for all a, b ∈ B0 1 = f (a) f (b) f ( a)−1 f ...
... Proof. By proposition 119 it is enough to prove that A5 is not solvable. Suppose that it is: A5 = B0 ⊃ B1 ⊃ · · · ⊃ Br = 1 with Bi +1 normal in Bi and Bi / Bi +1 abelian. Let f : B0 → B0 / B1 denote the natural homomorphism. As B0 / B1 is abelian we have for all a, b ∈ B0 1 = f (a) f (b) f ( a)−1 f ...
HYPERELLIPTIC JACOBIANS AND SIMPLE GROUPS U3 1
... By Remark 3.2, the double transitivity implies that the F2 [U3 (q)]-module QB is absolutely simple. Since SU3 (Fq ) → U3 (q) is surjective, the corresponding F2 [SU3 (Fq )]-module QB is also absolutely simple. Recall that dimF2 (QB ) = #(B) − 1 = q 3 = 23m . By Theorem 4.3, there are no absolutely s ...
... By Remark 3.2, the double transitivity implies that the F2 [U3 (q)]-module QB is absolutely simple. Since SU3 (Fq ) → U3 (q) is surjective, the corresponding F2 [SU3 (Fq )]-module QB is also absolutely simple. Recall that dimF2 (QB ) = #(B) − 1 = q 3 = 23m . By Theorem 4.3, there are no absolutely s ...
Mathematics 220 Homework for Week 6 Due February
... then we claim that x R 0. This is because 3x − 7 × 0 = 6k = 2(3k) is even. On the other hand if x = 2k + 1 is odd, we claim that x R 1. This is because 3x − 7 × 1 = 3(2k + 1) − 7 = 6k − 4 = 2(3k − 2) is even. Therefore every even number is equivalent to 0 and every odd number is equivalent to 1. On ...
... then we claim that x R 0. This is because 3x − 7 × 0 = 6k = 2(3k) is even. On the other hand if x = 2k + 1 is odd, we claim that x R 1. This is because 3x − 7 × 1 = 3(2k + 1) − 7 = 6k − 4 = 2(3k − 2) is even. Therefore every even number is equivalent to 0 and every odd number is equivalent to 1. On ...
Interval-valued Fuzzy Vector Space
... Abstract. The interval-valued fuzzy vector space with respect to the interval-valued fuzzy algebra is defined and ingestigated a lot of interesting properties. Each result is illustrated by a suitable exmaple. Keywords: Interval-valued fuzzy sets, interval-valued fuzzy vector space, subspace. AMS Ma ...
... Abstract. The interval-valued fuzzy vector space with respect to the interval-valued fuzzy algebra is defined and ingestigated a lot of interesting properties. Each result is illustrated by a suitable exmaple. Keywords: Interval-valued fuzzy sets, interval-valued fuzzy vector space, subspace. AMS Ma ...
Galois Groups and Fundamental Groups
... C(w).) Now C \ {1, 5} is a twice-punctured plane and hence homotopy equivalent to a figure-eight, which has fundamental group isomorphic to the free group on two generators. In particular, the Galois group of this extension can be generated by two elements. Of course, all groups of order ≤ 6 can be ...
... C(w).) Now C \ {1, 5} is a twice-punctured plane and hence homotopy equivalent to a figure-eight, which has fundamental group isomorphic to the free group on two generators. In particular, the Galois group of this extension can be generated by two elements. Of course, all groups of order ≤ 6 can be ...
elements of finite order for finite monadic church-rosser
... idempotents are specific elements of finite order, this is a restriction of problem (*). In §2 this restricted problem is solved for finite, special, Church-Rosser Thue systems, and in §3 this solution is extended to finite, monadic, Church-Rosser Thue systems. Then this result is used in §4 to esta ...
... idempotents are specific elements of finite order, this is a restriction of problem (*). In §2 this restricted problem is solved for finite, special, Church-Rosser Thue systems, and in §3 this solution is extended to finite, monadic, Church-Rosser Thue systems. Then this result is used in §4 to esta ...
THE SYLOW THEOREMS AND THEIR APPLICATIONS Contents 1
... non-abelian groups with order less than 60. • Lemma 4.1 eliminates groups of prime order, for they are cyclic and thus abelian. • Any group of order pk m where m < p and k 6= 0 will have a single Sylow psubgroup, since np ≡p 1 and np | m is only satisfied by np = 1. Uniqueness of a Sylow p-subgroup ...
... non-abelian groups with order less than 60. • Lemma 4.1 eliminates groups of prime order, for they are cyclic and thus abelian. • Any group of order pk m where m < p and k 6= 0 will have a single Sylow psubgroup, since np ≡p 1 and np | m is only satisfied by np = 1. Uniqueness of a Sylow p-subgroup ...
Rings and modules
... In particular, if f : M → N is an A -module homomorphism, and K is a submodule of ker(f ), then f induces an A -module homomorphism g: M/K → N, m + K 7→ f (m). 2.4. For A -modules M, N the intersection M ∩ N is an A -module. So if M, N are contained in a larger module L, one can speak about the mini ...
... In particular, if f : M → N is an A -module homomorphism, and K is a submodule of ker(f ), then f induces an A -module homomorphism g: M/K → N, m + K 7→ f (m). 2.4. For A -modules M, N the intersection M ∩ N is an A -module. So if M, N are contained in a larger module L, one can speak about the mini ...
Representations of Locally Compact Groups
... Gelfand-Naimark theorem, which we shall prove, states that every commutative C*-algebra is of the form C0 (X) for some locally compact Hausdorff space X, and more generally every C*-algebra is a C*-subalgebra of L(H) for some Hilbert space H by the (noncommutative) Gelfand-Naimark theorem. The algeb ...
... Gelfand-Naimark theorem, which we shall prove, states that every commutative C*-algebra is of the form C0 (X) for some locally compact Hausdorff space X, and more generally every C*-algebra is a C*-subalgebra of L(H) for some Hilbert space H by the (noncommutative) Gelfand-Naimark theorem. The algeb ...
MATH 436 Notes: Finitely generated Abelian groups.
... Example 2.10. We have previously seen that any nontrivial subgroup H of Z1 = Z is of the form dZ for a nonnegative integer d ≥ 1 and is hence itself free Abelian of rank 1 with basis {d}. This led to the classification of all one-generated Abelian groups as the cyclic groups Z or Z/dZ for d ≥ 1. No ...
... Example 2.10. We have previously seen that any nontrivial subgroup H of Z1 = Z is of the form dZ for a nonnegative integer d ≥ 1 and is hence itself free Abelian of rank 1 with basis {d}. This led to the classification of all one-generated Abelian groups as the cyclic groups Z or Z/dZ for d ≥ 1. No ...
ALGEBRA HANDOUT 2: IDEALS AND
... 1. Ideals in Commutative Rings In this section all groups and rings will be commutative. 1.1. Basic definitions and examples. Let R be a (commutative!) ring. An ideal of R is a subset I of R satisfying: (IR1) I is a subgroup of the additive group of R. (IR2) For any r ∈ R and any i ∈ I, ri ∈ I. We o ...
... 1. Ideals in Commutative Rings In this section all groups and rings will be commutative. 1.1. Basic definitions and examples. Let R be a (commutative!) ring. An ideal of R is a subset I of R satisfying: (IR1) I is a subgroup of the additive group of R. (IR2) For any r ∈ R and any i ∈ I, ri ∈ I. We o ...
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.