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

34 2. MODULES
(3) The formation of partial homomorphism are again homomorphisms.
and
N → HomR(M, M ⊗R N)), y ↦→ (x ↦→ x ⊗ y)
M → HomR(N, M ⊗R N)), x ↦→ (y ↦→ x ⊗ y)
2.6.3. Proposition. Given a map µ : M × N → L such that the partial maps x ↦→
µ(x, y) : M → L and y ↦→ µ(x, y) : N → L are homomorphisms. Then there
exists a unique homomorphism u : M ⊗R N → L such that u(x ⊗ y) = µ(x, y).
⊗
M × N
M ⊗R N
µ u
L
Proof. By 2.6.1 M ⊗R N = F/F ′ . The homomorphism 2.4.11 F → K, (x, y) ↦→
µ(x, y) has F ′ in the kernel. 2.3.5 gives the homomorphism u.
2.6.4. Remark. Two homomorphisms u, v : M ⊗RN → L are equal if u(x⊗y) =
v(x ⊗ y) for all x ∈ M, y ∈ N.
2.6.5. Proposition. Let R be a ring and f : M → M ′ , g : N → N ′ homomorphisms
of modules. Then there is induced a homomorphism
Proof. The map south-east
M ⊗R N → M ′ ⊗R N ′ , x ⊗ y ↦→ f(x) ⊗ g(y)
M × N
f×g
M ′ × N ′
⊗
M ⊗R N
⊗
M ′
⊗R N ′
satisfies the assumptions in 2.6.3 to induce the right vertical map x ⊗ y ↦→ f(x) ⊗
g(y).
2.6.6. Definition. f ⊗ g : M ⊗R N → M ′ ⊗R N ′ is the induced homomorphism
2.6.5.
2.6.7. Proposition. Let R be a ring. The constructions
(1)
(2)
are functors.
M ↦→ M ⊗R N, f ↦→ f ⊗ 1N
N ↦→ M ⊗R N, g ↦→ 1M ⊗ g
Proof. Given also homomorphisms f ′ : M ′ → M ′′ and g ′ : N ′ → N ′′ . Then by
2.6.4
f ′ ◦ f ⊗ g ′ ◦ g = f ′ ⊗ g ′ ◦ f ⊗ g
The rest follows directly from 2.6.5.