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

48 3. EXACT SEQUENCES OF MODULES
3.3. Exactness of Hom
3.3.1. Proposition. Given an exact sequence
0
f
M
g
N
L
and an R-module K. Then the following sequence is exact
0
HomR(K, M)
HomR(K, N)
HomR(K, L)
Proof. Given h : K → M such that f ◦ h = 0 then h = 0 since f is injective.
Given k : K → N such that g ◦ k = 0 then by 3.1.4 there is h : K → M such that
f ◦ h = k. So the sequence is exact.
3.3.2. Proposition. Given a sequence
0
f
M
g
N
L
Such that for any module K, the following sequence is exact
0
HomR(K, M)
Then the original sequence is exact.
HomR(K, N)
HomR(K, L)
Proof. For K = M the identity 1M is mapped to g ◦ f ◦ 1M = 0 so it is a 0sequence.
Assume f(x) = 0. Take K = R then 1x is mapped to f ◦ 1x = 1 f(x) =
0. Therefore 1x = 0 and so x = 0. That insures that f is injective. Assume
g(y) = 0. Take K = R then 1y is mapped to g ◦ 1y = 1 g(y) = 0. There exists a
homomorphism h : R → M such that f ◦ h = 1y. By 2.5.12 h = 1x and therefore
f(x) = y. The original sequence is now proven exact.
3.3.3. Proposition. Given an exact sequence
M f
g
N
L
and a module K. Then the following sequence is exact
0
HomR(L, K)
HomR(N, K)
0
HomR(M, K)
Proof. Given h : L → K such that h ◦ g = 0 then h = 0 since g is surjective.
Given k : N → K such that k ◦ f = 0 then by 3.1.5 there is h : L → K such that
h ◦ g = k. So the sequence is exact.
3.3.4. Proposition. Given a sequence
M f
g
N
L
such that for any module K, the following sequence is exact
0
HomR(L, K)
Then the original sequence is exact.
HomR(N, K)
0
HomR(M, K)
Proof. For K = L the identity 1L is mapped to 1L ◦g ◦f = 0 so it is a 0-sequence.
Take K = Cok g then pg : L → Cok g has pg ◦ g = 0, but by exactness 0 is the
unique homomorphism satisfying this, so pg = 0. Therefore Cok g = 0 and g is
surjective. Take K = Cok f, p : N → Cok f the projection. p ◦ f = 0 so by
exactness there exists a unique q : L → Cok f such that q ◦ g = p. It follows that
Ker g ⊂ Ker p = Im f. All together the original sequence is exact.