Serre's Problem on Projective Modules by T.Y. Lam

By T.Y. Lam

A useful precis of study paintings performed within the interval from 1978 to the current

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0) · Q−1 = d · (x1 , . . , xn ). 4. 6) are more general than Kaplansky’s K-Hermite rings. We’ll also show that the property (6) above for a K-Hermite ring can be further strengthened. 26. Let R be K-Hermite. Then (A) R is both a Hermite ring and a Bézout ring. (B) Whenever b1 R + · · · + bn R = eR with n 2, there exist (y1 , . . , yn ) ∈ Umn (R) such that (b1 , . . , bn ) = e · (y1 , . . , yn ). Proof. 25) shows that R is Bézout. Given any β = (b1 , . . , bn ) ∈ Umn (R), take Q ∈ GLn (R) such that β Q = (d, 0, .

For n = 2, however, P (2) is not free over R2 . The best and quickest way to prove this is to use⎛a little bit⎞of topology. Indeed, suppose τ (2) = (x0 , x1 , x2 ) can x0 x1 x2 be completed into ⎝ f g h ⎠, with determinant equal to a unit e ∈ R2 . We think f g h of f, g, h, e, e−1 , . . as functions on S 2 : they are polynomial expressions in the “coordinate functions” x0 , x1 , x2 . Consider the continuous vector field on S 2 given by v ∈ S 2 → (f (v), g(v), h(v)) ∈ IR3 . Since e = f · (x1 h − x2 g) − g · (x0 h − x2 f ) + h · (x0 g − x1 f ) is clearly nowhere zero on S 2 , the vector (f (v), g(v), h(v)) is nowhere collinear with the vector v.

Xn form a free basis for each Pmi upon localization. If we define f : R n → P by f (ej ) = xj for all unit vectors ej , then fmi is an isomorphism for every i. 2), it follows that f itself is an isomorphism, so P ∼ = R n as desired. For later reference, we shall take this opportunity to make some other useful remarks about semilocal rings. 8. A commutative ring R is semilocal iff R/rad R is artinian. Proof. If R/rad R is artinian, it is a finite direct product of fields. Thus, Max R = Max (R/rad R) is finite.

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