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

Proof. By 3.1.13 there is a split exact sequence
so the statement follows from 6.5.3.
0
6.5. FINITE PRESENTED MODULES 81
v
L
u
N
M
6.5.5. Corollary. Let f : M → N be a homomorphism.
(1) If M is finite and Im f finite presented, then Ker f is finite.
(2) If Ker f, Im f are finite presented, then M is finite presented.
(3) If N is finite presented and Im f finite, then Cok f is finite presented.
(4) If Im f, Cok f are finite presented, then N is finite presented.
Proof. Use the sequences 3.1.8.
6.5.6. Proposition. Let R be a ring and M, N finite presented modules.
(1) M ⊕ N is finite presented.
(2) M ⊗R N is finite presented.
Proof. (1) This is clear from 6.5.4. (2) If M = R n then M ⊗R N is finite presented
by (1). In general chose u : R n → M → 0 exact with finite kernel. The sequence
Ker u ⊗R N → R n ⊗R N → M ⊗R N → 0 is exact. So Ker u ⊗ 1N is finite.
Conclusion by 6.5.5.
6.5.7. Proposition. Given submodules N, L ⊂ M. Then
(1) If M/N, M/L are finite and M/N + L is finite presented, then M/N ∩ L is
finite.
(2) If M/N, M/L are finite presented and M/N ∩ L is finite, then M/N + L is
finite presented.
Proof. Use the sequence 3.2.7.
6.5.8. Proposition. Let R be a ring and U a multiplicative subset.
(1) For a finite module M and any module N the natural homomorphism
0
U −1 HomR(M, N) → Hom U −1 R(U −1 M, U −1 N)
is injective.
(2) For a finite presented module M and any module N the homomorphism
U −1 HomR(M, N) → Hom U −1 R(U −1 M, U −1 N)
is a natural isomorphism.
Proof. (1) If M = R this is an isomorphism. Then this is also an isomorphism
for M = R n since both functors respect finite direct sums. In general choose
0 → K → R n → M → 0 exact. There is a diagram
0
0
U −1 HomR(M, N)
HomU −1R(U −1M, U −1N)
U −1 HomR(R n , N)
HomU −1R(U −1Rn , U −1N) giving injectivity. (2) In (1) K is finite, so the last vertical map is injective. Conclusion
by the five lemma 3.2.8.
U −1 HomR(K, N)
HomU −1R(U −1K, U −1N)