Topics in algebra Chapter IV: Commutative rings and modules I - 1

6 Hom and tensor product
The first part of this section is to study the functors HomR(−, M) and
HomR(M, −), where M is an R-module. Recall that the set of all Rhomomorphism
from M to N, denoted by HomR(M, N), is an R-module
with the action of R given by rf(x) = f(rx) for r ∈ R and x ∈ M.
Lemma 6.1 Let 0 −→ N1
φ
−→ N2
ψ
−→ N3 be an exact sequence; then
HomR(M, −) induces the exact sequence
0 −→ HomR(M, N1)
˜φ
˜ψ
−→ HomR(M, N2) −→ HomR(M, N3),
where ˜ φ(f) = φ ◦ f for f ∈ HomR(M, N1), and ˜ ψ(g) = ψ ◦ g for g ∈
HomR(M, N2).
Proof. ˜ φ is 1-1:
Let f ∈ HomR(M, N1) with ˜ φ(f) = 0; then φ(f(x)) = 0 for all x ∈ M, so
that f(x) = 0 for all x ∈ M as φ is 1-1, it follows that f = 0.
˜φ
˜ψ
HomR(M, N1) −→ HomR(M, N2) −→ HomR(M, N3) is exact:
It is easy to see that ˜ ψ ◦ ˜ φ = 0. Therefore Im ˜ φ ⊆ ker ˜ ψ. To show ker ˜ ψ ⊆
Im ˜ φ, let g ∈ ker ˜ ψ; then ψ(g(x)) = 0 for every x ∈ M. Let x ∈ M; then
there is a unique element y ∈ N1 such that φ(y) = g(x). Now, define a
map f : M −→ N1 by f(x) = y; then it is easy to check that f is an
R-homomorphism and ˜ φ(f) = g.
We say that HomR(M, −) is exact if it preserve every short exact se-
quences,i.e., if 0 −→ N1
HomR(M, N1)
φ
−→ N2
˜φ
−→ HomR(M, N2)
ψ
−→ N3 −→ 0 is a s.e.s., then 0 −→
˜ψ
−→ HomR(M, N3) −→ 0 stays exact.
Lemma 6.2 Let N1
φ
−→ N2
ψ
−→ N3 −→ 0 be an exact sequence; then
HomR(−, M) induces the exact sequence
0 −→ HomR(N3, M)
˜ψ
˜φ
−→ HomR(N2, M) −→ HomR(N1, M),
where ˜ ψ(f) = f ◦ ψ for f ∈ HomR(N3, M), and ˜ φ(g) = g ◦ φ for g ∈
HomR(N2, M).
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