Commutative algebra - Department of Mathematical Sciences - old ...

8.6. PRIME FILTRATIONS OF MODULES 99
Proof. The set of submodules of M for which the theorem is true is nonempty by
8.2.6. Let N ⊂ M be maximal in this set. Suppose N = M. By 8.2.6 applied to
M/N there is a chain N ⊂ N ′ ⊂ M such that N ′ /N is isomorphic to an R-module
of the form R/P ′ where P ′ is a prime ideal. This contradicts the maximality of N.
So N = M.
8.6.4. Corollary. Let R be a nonzero noetherian ring. Then there exists a finite
filtration of ideals
0 = I0 ⊂ I1 ⊂ · · · ⊂ Ir−1 ⊂ Ir = R
such that Ii/Ii−1, i = 1, . . . , r is isomorphic to an R-module of the form R/Pi
where Pi is a prime ideal in R.
8.6.5. Example. In Z there is a filtration
0 =⊂ (p n ) ⊂ (p n−1 ) ⊂ · · · ⊂ (p) ⊂ Z
of any length with factors Z/(p) for any prime number p.
8.6.6. Proposition. A ring R is artinian if and only if it is noetherian and all prime
ideals are maximal.
Proof. By 7.3.13 an artinian ring has finite length and therefore noetherian. Primes
are maximal by 7.3.11. Conversely by 8.6.4 there is a finite composition series.
8.6.7. Corollary. Let R be a ring and M a module. The following are equivalent
(1) M has finite length.
(2) R/ Ann(M) is artinian and M is finite.
8.6.8. Proposition. If R ⊂ S be a finite extension. Then R is artinian if and only
if S is artinian.
Proof. Suppose S is artinian. By 8.3.9 R is noetherian. By 6.6.5 any prime ideal
in R is maximal. Conclusion by 8.6.6.
8.6.9. Proposition. Let R be a noetherian ring. The number of minimal prime
ideals is finite.
Proof. Choose a filtration 8.6.4, 0 = I0 ⊂ · · · ⊂ Ir = R with Ii/Ii−1 R/Pi,
where Pi is a prime ideal. Let P be a minimal prime ideal in R. Then (Ii/Ii−1)P
(R/Pi)P = 0 if and only if Pi ⊂ P . Thus P = Pi for some i since RP = 0.
8.6.10. Proposition. Let R be a noetherian ring such that the local rings RP are
domains for all maximal ideals P . Then R is a finite product of domains.
Proof. Let P1, . . . , Pn be the minimal primes 8.6.9. The intersection is 0 5.4.9 and
they are comaximal since a domain has a unique minimal prime. Conclusion by
Chinese remainders 1.4.2.
8.6.11. Exercise. (1) Compute a filtration 8.6.3 of the Z-module Z/(36).