By James W. Brewer, Edgar A. Rutter

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The f i e l d guarantee the and 9 ( P R t) = ~(PRp). By T a k e m in M c o m a x i m a l w i t h t and let I be the i d e a l of S g e n e r a t e d by m and ~ ( P R g e n e r a t e MS t. of S. Then I = M because As for (3), n o t e n o m i a l r i n g S/PS shows s h o w t h a t it s u f f i c e s say t h a t the N u l l s t e l l e n s a t z that M ~ R [ X I ] ~ ~(PRp) tf-p, w h e r e Hence tr+tf*, where ated maximal (reR, pcP). it s u f f i c e s f* d e n o t e s REMARKS. of apply in the p r o o f of (I) to c o n c l u d e shows domain of image is p r i n c i p a l ; (2) we h a v e 2.

I = (Yi) ' we know by Theorem 3 that I(~2) contains an R-sequence of length two. the minors of order 0 >0 For suppose >R (Ai) >B n n-i ~ But of the matrix (aij) . >R n-I Letting rank(~2) = n-i I(~ ) = (Ai) where 2 Thus, the sequence and [Ai} are is exact since the composition is clearly zero and Theorem 3 applies. ,n Although Theorem 3 provides us with a nice proof of Theorem i, it is not clear how that gets us any further into the problem of determining relations among the maps of an arbitrary finite free resolution.

Xn] ) + n. X n]) An appealing feature of Theorem i is that its proof is not only brief but also elementary and therefore it seems to make available from scratch, more readily than ever before, information about the ranks of prime ideals of R[X 1 ..... Moreover, Xn]. not only can Theorem 1 be used to prove the classical dimen- sion theorems, but it also yields immediately that dim (K[XI, .... Xn]) = n, when K is a field. , X n] have the same rank, then R is a Hilbert ring. Perhaps the most interesting part of the paper is Section 2 which is given over to the construction of counterexamples which arise in Section i.