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Lectures on Orders

By Dr Daniel Chan


Question: How does one study a field? Herre are two answers: 1. Suppose K/Q is a finite field extension. Pick a subring R ⊂ K consisting of algebraic integers such that K(R) = K. Information about R gives information about K. There are many choices for R but the best choice is the one which is normal (i.e. integrally closed). 2. Let k be an algebraically closed field and K a field of finite transcendence degree over k. Pick an algebraic variety X with function field k(X) = K. Such an X is called a model for K. X gives information about K. There are many choice for X so we usually impose extra conditions for example: normal/smooth, projective, etc. The theory of orders studies “integral models ” for central simples algebras. In analogy with (1) this gives rise to non-commutative arithmetic (see [Rei03]) and in analogy with (2) to non-commutative algebraic geometry. Definition 1.1. Let R be a commutative, noetherian, normal domain. An R-order (or an order over R) is an R-algebra A such that: 1. A is finitely generated as an R-module (analogue of integral over R), 2. A is a torsion free R-module and K(R) ⊗R A is a central simple K(R)algebra. Note that torsion free implies A ֒ → K(R) ⊗R A so A is a subring of a central simple algebra. In fact, with the above notation and D: = K(R)⊗RA, we will say that A is an order in D. 1 Example 1.2. Let k be a field of characteristic ̸ = 2 and R = k[u,v]. R〈x,y〉 (x 2 − u,y 2 − v,xy + yx) = k〈x,y〉 (xy + yx). A = Proposition 1.3. Let R ⊂ S be an extension of commutative noetherian normal domains and A an R-order in D. Then A ⊗R S is an S-order in D ⊗K(R) K(S). Proof. Suppose {a1, · · ·,an} generates A as an R-module and let a ⊗ s ∈ A ⊗R S. Then for some r1, · · ·,rn ∈ R, a = a1r1 + · · · + anrn and so a ⊗ s = (a1r1 + · · · + anrn) ⊗ s = a1 ⊗ r1s + · · · + an ⊗ rns.

Publisher: 2011-06-01
Year: 2011
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