Projective modules over rings with many units

Let R be a commutative ring. Assume that every polynomial whose values generate the unit ideal actually takes on an invertible value. Then projective R-modules split into cyclic summands, and those of constant rank are free. A ring R (commutative with 1) satisfies the primitive condition if each f(x) = ao + + anXn that is primitive (Y:(aiR) = R) has some b in R with f(b) a unit. This condition, which guarantees the existence of many units in R, was introduced by van der Kallen [13]; he gave examples of rings satisfying the condition and established properties of K2(R) for such R. Subsequently it was shown [4], [5], [6] that the condition implies pleasant structural results about GL2(R) and Aut(GL2(R)). One step in this was a computational argument [6, 11.3] proving that if Q is a rank one direct summand of F where F is free of rank 2, then Q is free. We here will see that a much more general result is true. It actually applies to a slightly larger class of rings, and we begin by discussing them. I. Let f(X1, ... , Xn) be a polynomial over a ring R. We will say that f has local unit values if for each maximal ideal M of R there are b,, . . ., bn in RM with f(bl, ... , bn) invertible in RM. We can here replace the bi by elements of R congruent to them modulo M, so the condition says that not all values of f are in M; in other words, the values of f should generate the unit ideal of R. (This implies, of course, that the coefficients of f must generate the unit ideal.) We say that f has unit values if somef(b1, . . ., bn) is actually invertible in R. The rings we care about will be those in which every f with local unit values has unit values. Since elements are invertible iff they are so modulo the Jacobson radical J(R), it is evident that R has this property iff R/J(R) does. In particular, semilocal rings have the property. It is also easy to see that a product HIRi has the property iff all the factors do (consider maximal ideals of the form Mi x fljH i Rj). Further examples of rings with this property are given by the following propositions. PROPOSITION. Let R be a ring for which R/J(R) is von Neumann regular (= absolutely flat). Then polynomials with local unit values have unit values. PROOF. Replacing R by R/J(R), we may assume it is von Neumann regular. Suppose f has local unit values. For each maximal M pick b = (bl, ... , bn) with Received by the editors August 28, 1980. 1980 Mathematics Subject Classification Primary 13CO5. 'The work of both authors was supported in part by the National Science Foundation. ? 1981 American Mathematical Society 0002-9939/8 1/0000-0503/$02.00 455 This content downloaded from 157.55.39.231 on Thu, 06 Oct 2016 04:38:41 UTC All use subject to http://about.jstor.org/terms 456 B. I{. McDONALD AND W. C. WATEIRHOUSE f(b) a unit at M; then f(b) is still a unit on a neighborhood of M in Spec R. Since Spec R is a Boolean space, we can refine this covering to a finite covering by disjoint clopen sets U where we have f(bu) invertible on U. We may then choose b agreeing with bu on U, and f(b) will be invertible. E] This result could also be deduced from [2, Proposition 2]. The argument shows more generally that if we have a sheaf of rings over a Boolean space and the fibers have our property, so does the ring of global sections. We should also point out that by [1, 11.4, Exercise 16, p. 173] we have the following special case of the proposition: COROLLARY. Let R be zero-dimensional. Then polynomials with local unit values have unit values. LI PROPOSITION. Let S be an R-algebra which is a finitely generated free R-module. Suppose that over R, all polynomials with local unit values have unit values. Then the same is true over S. PROOF. Let s1, ... , sm be a basis of S over R. Given f(X1, ... , XJ) over S, take indeterminates Y 11..., Ymn, and define a polynomial g( Y) over R as the norm (from S to R) of f(> si Yi, . .. , E si Yi). Then setting Xj = E sirij makes f(X) invertible iff g(rij) is invertible, since units and only units have unit norms. Assume now that f has local unit values. If M is any maximal ideal of R, then SM is semilocal, so f has unit values in SM. Hence g has unit values in RM. By hypothesis then g has unit values in R. C] The main theorem will automatically allow us to replace "free" by "projective of constant rank" in this result. We now show exactly how our property is related to the primitive condition mentioned in the introduction. For this we need a pair of simple lemmas. LEMMA. Let R satisfy the primitive condition. Let fi(X) = E aXJ be a finite sequence of polynomials with Eij(aijR) = R. Then there is some b in R with E(fi(b)R) = R. PROOF. Choose an integer m greater than the degrees of all fi, and let g(X) = L f(X)Xm'. All aij occur as coefficients in g, so g is primitive. Hence some g(b) = E f(b)b'm is a unit, and in particular 2(fi(b))R = R. E] This allows us to deduce a multivariable extension of the condition: LEMMA. Let R satisfy the primitive condition. Let f(XI,... , XJ) = E a,Xa be a polynomial with E(a,R) = R. Then there are bl, . .. , b, in R with f(b1, .. ., bn) invertible. PROOF. Rewrite f as E, f,(X1)X8, where ,B = (i2, . .. , in). All the a, appear as the coefficients of the polynomials ff(XI). By the lemma there is some b, such that E(ffi(bj)R) = R. Then f(bl, X2,... Xn) again satisfies the hypothesis of the lemma, and the result follows by induction. El PROPOSITION. A ring R satisfies the primitive condition iff (1) every polynomial with local unit values has unit values and (2) every residue field R/ M is infinite. This content downloaded from 157.55.39.231 on Thu, 06 Oct 2016 04:38:41 UTC All use subject to http://about.jstor.org/terms PROJECTIVE MODULES OVER RINGS 457 PROOF. If R/M has finite cardinality q, then Xq X is a primitive polynomial with all values in M. Thus if R satisfies the primitive condition, (2) must hold. And any f(XI, .. ., X",) with local unit values has coefficients generating the unit ideal, so (1) holds by the last lemma. Conversely, if f is a primitive polynomial, then it is nontrivial modulo M, so by (2) it has a nonzero value modulo M. This means it has local unit values, so by (1) it has unit values. E1 This shows in particular that rings with our property can be very far from dimension zero. Indeed [3], if A is any ring and S is the set of primitive polynomials in A[x], then R = S-'A[x] satisfies the primitive condition and has maximal ideal space identical with that of A.