By M. hammed Boulagouaz, Jean-Pierre Tignol, Mohammed Boulagouaz
This examine demonstrates the most important manipulations surrounding Brauer teams, graded jewelry, staff representations, excellent periods of quantity fields, p-adic differential equations, and rationality difficulties of invariant fields - showing a command of the main complicated tools in algebra. It describes new advancements in noncommutative valuation thought and p-adic research.
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Over three hundred sequences and plenty of unsolved difficulties and conjectures with regards to them are awarded herein. The publication comprises definitions, unsolved difficulties, questions, theorems corollaries, formulae, conjectures, examples, mathematical standards, and so on. ( on integer sequences, numbers, quotients, residues, exponents, sieves, pseudo-primes/squares/cubes/factorials, nearly primes, cellular periodicals, services, tables, prime/square/factorial bases, generalized factorials, generalized palindromes, and so forth.
Extra resources for Algebra and number theory: proceedings of a conference held in Fez, Morocco
Since a&− and − &a preserve arbitrary sups for all a ∈ Q, they have right adjoints and we shall denote them by a −→r − and a −→l − respectively. 2. Let Q be a quantale, a, b, c ∈ Q. Then (1) a&(a −→r b) ≤ b; (2) a −→r (b −→r c) = b&a −→r c; Again, analogous results hold upon replacing −→r by −→l . 3. Let Q be a quantale, An element c of Q is called cyclic, if a −→r c = a −→l c for all a ∈ Q. d ∈ Q is called a dualizing element, if a = (a −→l d) −→r d = (a −→r d) −→l d for all a ∈ Q. 4. A quantale Q is called a Girard quantale if it has a cyclic dualizing element d.
If 0 is a cyclic dualizing element of Q, then Q is strictly two-sided. Proof. Assume 0 is a cyclic dualizing element in Q. Then 0⊥ = 0 −→ 0 = 1 is the unit of Q, hence ∀a ∈ Q, a&1 = 1&a = a, this finished the proof. 6. If Q is a two-sided Girard quantale, then the unique cyclic dualizing element is the least element 0. Proof. If Q is a two-sided Girard quantale, then we have a = a&e ≤ a&1 ≤ a for all a ∈ Q. Similarly, we have 1&a = a. Thus Q is strictly two-sided. Suppose d is a cyclic dual element in Q, ⊥ is the unary operation induced by d, then we have d = 1 −→ d = 1⊥ = 0.
Problem 2. Find relationships between each of Z(m + n) and Z(mn) with Z(m) and Z(n). Problem 3. Find all values of n such that (1) Z(n + 1) = Z(n), (2) Z(n) divides Z(n + 1), (3) Z(n + 1) divides Z(n). Problem 4. Are there solutions to (1) Z(n + 2) = Z(n + 1) + Z(n)? (2) Z(n) = Z(n + 1) + Z(n + 2)? (3) Z(n+2) =Z(n + 1)Z(n)? (4) Z(n) =Z(n+1)Z(n + 2)? (5) Z(n+2) + Z(n) =2Z(n + 1)? (6) Z(n+2)Z(n) =[Z(n+1)]2 ? Problem 5. For a given m, how many n are there such that Z(n) =m? Moreover, if Z(n) = m has only one solution, what are the conditions on m?