By Stephen McAdam (auth.)
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Example text
Rees ring of R[Y] with respect to Then and Also let Using the above claim, maps induces an isomorphism from Furthermore, $ I, and let (R,M). x e M for the Rees rings of Y be an indeterminate with respect to ~ it is also an Let with ~ = R [ t - l , lt] be x not in any minimal ~(l,x) =~(I) +I. ~(l,x) ker ~ c rad[(t'l,x)~(l,Y)]. 6]. be an ideal in a local ring induces an obvious homomorphism Therefore, sequence over I modulo I. the Rees ring of Proof: sequences over This was done by Katz in [Kz2].
23. 20, we can find I c J c IVf]R, case follows by P r o p o s i t i o n for such that if n > 0 pnv c_ IV. P e A*(J) ((i)~>(iv)) prime. 26. Let I ~ P be ideals of R with P prime. Assume dim R > 0. The following are equivalent. (i) P e A (1). (ii) P e A (IJ) for any ideal (iii) P e A (Ic) for any element (iv) There exists Proof: prove (ii) ~ (i) ~ an element Since (ii) and R (IJV : IV) = JV ~ shows that P e A (Ic). IV (IJ). equals In C h a p t e r 4, we w i l l P~A (J). and let prime, R a domain.
Asymptotic sequence a maximal a s y m p t o t i c s e q u e n c e from asymptotic sequence. If of Proposition P e A (I). that by Proposition Let We have of the result. be a prime ideal. 13. class. there is a since analytic spread does not exceed the minimal number of generators. 6, asymptotic is in the principal asymptotic sequence coming from with Q, RQ Therefore z(P) ~ h e i g h t P, gr (I) ~ height I. I0. gr(I) ~ gr (I) is obvious. and so by P r o p o s i t i o n 5 . 14. c qn be a saturated chain of primes in minimal prime of R.