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7.5. LOCAL ARTINIAN RING 91
(1) M ⊗R S is a finite length S-module.
(2) In general
ℓS(M ⊗R S) ≤ ℓS(S/P S) · ℓR(M)
(3) If R → S is flat then
ℓS(M ⊗R S) = ℓS(S/P S) · ℓR(M)
Proof. The case M = k(P ) is clear. Conclude by induction.
7.4.9. Exercise. (1) Let K ⊂ L be a finite field extension and W a finite vector space
over L. Show that
rankK(W ) = rankK(L) · rankL(W )
(2) Let K ⊂ L be a finite field extension and V a finite vector space over K. Show that
rankL(V ⊗K L) = rankK(V )
7.5. Local artinian ring
7.5.1. Lemma. Let (R, P ) be a local artinian ring and M a finite module. The
following are equivalent.
(1) M = 0.
(2) HomR(k(P ), M) = 0.
(3) HomR(M, k(P )) = 0.
Proof. Clear by a filtration.
7.5.2. Proposition. Let (R, P ) be a local artinian ring and k(P ) ⊂ E an injective
envelope.
(1) The are isomorphisms
k(P ) HomR(k(P ), k(P )) HomR(k(P ), E)
(2) For a finite module M, the module HomR(M, E) has finite length and
ℓR(HomR(M, E)) = ℓR(M)
Proof. (1) A nonzero homomorphism f : k(P ) → E has Im f = k(P ) since the
extension is essential. (2) This follows from (1) by use of a filtration.
7.5.3. Corollary. Let (R, P ) be a local artinian ring and k(P ) ⊂ E an injective
envelope. Then E has finite length
ℓR(E) = ℓR(R)
7.5.4. Proposition. Let (R, P ) be a local artinian ring and k(P ) ⊂ E an injective
envelope. There is a natural isomorphism for any finite module M
x ↦→ evx : M → HomR(HomR(M, E), E)
Proof. The case M = k(P ) is clear by 7.5.2. Let 0 → N → M → L → 0 be a
short exact sequence. Then the following diagram has exact rows.
0
0
N
HomR(HomR(N, E), E)
M
HomR(HomR(M, E), E)
Conclusion by the five lemma 3.2.8 and induction on the length.
L
HomR(HomR(L, E), E)
0
0