Before proving (hi), we recall a familiar fact (see [AM, Proposition 1.11], for
(4.3.2) If a maximal ideal m of a noetherian ring 17 is a not a minimal prime, then
m is not contained in the union of the minimal primes.
(hi) In view of (ii), it suffices to prove that (^m)n(Q ) = (^m)Q(nm)- Thus, we
may assume that ft is local with maximal ideal m. Moreover, if m is also a minimal
prime, then ft = ftm is a field, whence the proof becomes trivial; so we may also
assume that m is not minimal.
It suffices to show that the same prime ideals survive in both localizations (since
any localization is naturally isomorphic to the localization at the complement of
the union of the prime ideals that survive). Since the rings we are working with
are reduced, the denominator set in each localization can be taken to be the set of
regular elements of the ring in the subscript. Since regular elements of O remain
regular in ftmi no minimal prime of flm is destroyed by either localization. On
the other hand, the maximal ideals of both local rings ft and ftm contain regular
elements. For fJ, this holds because its maximal ideal is not contained in the union
of the minimal primes [(4.3.2) and Lemma 4.1(h)]. For ftm, this holds because
regular elements of ft remain regular in ftm. Therefore, the maximal ideal of ftm
is destroyed by both localizations. Since ftm has Krull dimension 1, the statement
In view of statements (i)-(iii) and (4.2.1), all of the rings in (i)-(iii) are reduced
artinian rings, and hence are direct sums of fields. This proves (4.3.1).
COROLLARY 4.4. Let T = (B^Th, where each T^ is an integral domain, be the
normalization of a reduced noetherian ring ft. Then each ( T / J Q , where Q Q(fl),
is the field of quotients ofT^.
P R O O F . The total quotient ring of T is TQ = ©^ Q?h)Q- Therefore each (T^Q
is a field or zero. The latter possibility never occurs because since ft is reduced
the natural map V TQ is an injection. Since the localization (Th)Q °f ^h is. a
field containing T^, it follows that ( I \ ) Q is the field of quotients of T^.
L E M M A 4.5. Let ft be a noetherian ring of Krull dimension 1 such that no
maximal ideal is a minimal prime ideal, and M G fingen(Q). Then MQ = 0 if and
only if M has finite length.
P R O O F . First, suppose that M has finite length. To prove that MQ = 0,
we may assume that M is a simple module and therefore is annihilated by some
maximal ideal m. Since m is not a minimal prime, (4.3.2) shows that m contains
an element c that is not in any minimal prime ideal of O. Thus, c acts invertibly
on MQ but also annihilates it, so that MQ = 0.
Conversely, suppose that MQ = 0. Since M is finitely generated, there is some
element c G ft UQ such that cM = 0. Thus, since ft has Krull dimension 1, M
is a finitely generated module over the artinian ring ft/ftc, and hence M has finite
5. Ideal-adic Completion s
N O T A T I O N AN D Q U I C K R E V I E W 5.1 (Completion). Let ft be a noetherian
ring, m a maximal ideal of Q, and M G fingen(O). Then we can identify the m-adic
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