# Download Algebra and number theory: proceedings of a conference held by M. hammed Boulagouaz, Jean-Pierre Tignol, Mohammed PDF

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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Extra resources for Algebra and number theory: proceedings of a conference held in Fez, Morocco

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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?