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

90 7. MODULES OF FINITE LENGTH
(5) Show that a ring with finitely many ideals is artinian.
(6) Let R be a principal ideal domain and a = 0. Show that R/(a) is artinian.
7.4. Localization
7.4.1. Proposition. Any artinian ring is the product of finitely many artinian local
rings. Let R be artinian with maximal ideals P1, . . . , Pk. Then there is n1, . . . , nk
such that P n1
1 . . . P nk
k = 0 and
R R/P n1
nk
1 × · · · × R/Pk Each R/P ni
1 is a local artinian ring.
Proof. This follows from 7.3.11, 7.3.12 and Chinese remainders 1.4.2.
7.4.2. Example. A reduced artinian ring is a finite product of fields.
7.4.3. Corollary. Let R be an artinian ring and U ⊂ R a multiplicative subset.
The ring of fractions U −1 R is an artinian ring.
7.4.4. Example. Let P ⊂ R be a maximal ideal and M a module. Assume s /∈ P
and n ∈ N.
(1) Rs + P n = A.
(2) Scalar multiplication by s : M/P n M → M/P n M, x → sx is an isomorphism.
(3) The canonical map M/P n M → (M/P n M)P is an isomorphism.
(4) If M is finite then M/P n M has finite length.
7.4.5. Proposition. Let R be artinian with maximal ideals P1, . . . , Pk. A finite
module M has a decomposition
M
The length is
Proof. Let R R/P n1
ℓR(M) =
R/P nk
k and MPi M ⊗R R/P ni
i .
1
i
i
MPi
ℓRP i (MPi )
nk
× · · · × R/Pk . Then M M ⊗R R/P n1
1 × · · · × M ⊗R
7.4.6. Proposition. Let M be an R-module of finite length and Ann(M) ⊂ P1, . . . , Pk
the maximal ideals. Then MPi is a finite length RPi-module and
ℓR(M) =
ℓRP (MPi i )
Proof. The ring R/ Ann(M) R/P n1
1
i
× · · · × R/P nk
k
by 7.3.16 and 7.4.1.
7.4.7. Proposition. Let (R, P ) → (S, Q) be a local ring homomorphism and assume
that k(P ) → k(Q) is a finite extension. If N is a finite length S-module, then
N is a finite length R-module and
Proof. Reduce to N = k(Q).
ℓR(N) = rank k(P )(k(Q)) · ℓS(N)
7.4.8. Proposition. Let (R, P ) → (S, Q) be a local ring homomorphism and assume
that S/P S is a finite length S-module. Let M be a finite length R-module.