### Representing rank complete continuous rings

David Handelman
1976 Canadian Journal of Mathematics - Journal Canadien de Mathematiques
Given a suitable regular ring R, we construct a sheaf-like representation for R as a ring of continuous sections from a completely regular space to an appropriately toplogized disjoint union of factor rings corresponding to "extremal" pseudo-rank functions. Applied to rings which are complete with respect to a rank function this representation is an isomorphism, the completely regular space is extremally disconnected and compact, and the * 'stalks" are the simple factor rings. These factor
more » ... are discrete exactly if they are artinian, so the construction is not generally a sheaf. In particular, this yields an isomorphic representation for continuous geometries complete with respect to a (lattice) valuation, in terms of the simple homomorphic images. Let R denote a (von Neumann) regular ring. A pseudo-rank function if e, f are orthogonal idempotents. As a consequence, if aR < bR (for a, b G R, as right /^-modules) then N(a) ^ N(b); in general N(r + s) ^ N(r) + N(s), so N induces a uniform topology given by the pseudo-metric d N , d N (x, y) = N(x -y). If {TV} is a family of pseudo-rank functions on R, we shall abuse our terminology and refer to the uniform topology with gauge {d N }, as the topology induced by {N\. A pseudo-rank function N is called a rank function if d N is a metric, i.e. N(r) =0 implies r = 0. We denote by P(i£), the collection of pseudo-rank functions on R. P(R) is a subset of [0, l] R . Endowed with the relative product topology (equivalently, the point-open topology) P(R) becomes a compact, convex subset of [0, 1] R ([6, Lemma 7]). We denote by E(R) or E, the collection of extremal points of P(i^). If P(R) is non-empty, the Krein-Milman theorem tells us P(i?) is the closure of the convex hull of E(R)> so in particular E(R) is non-empty. If N e P(R), then ker N = {r G R\N(r) = 0} is a two-sided ideal (since N(r + s) ^ N(r) + N(s)\ N(rs) S N(r), N(s)). We define R N to be R / kevN . Then N induces a rank function on R N ([6, Lemma 5]), which we shall also call N.