Susan Marie Cooper

Exercise B14 - Tony’s Notes, Exercise 9B, part 2, page 6: Let M_{i} (i ∈ I) be a family of R-modules, and let M = ⊕M_{i}. Then

M is R-flat ⇔ each M_{i} is R-flat.

Proof. ⟸. Suppose that each M_{i} is R-flat. Let A ⟶^{f} B ⟶^{g} C be an exact sequence of R-modules and R-module homomorpims. Then, by definition of a flat R-module,

A ⊗_{R} M_{i} ⟶^{f⊗id_{i}} B ⊗_{R} M_{i} ⟶^{g⊗id_{i}} C ⊗_{R} M_{i}  (1)

is again exact for all i ∈ I, where id_{i} : M_{i} ⟶ M_{i} is the identity map on M_{i}.

Now recall the isomorphisms from the Proposition on page 1 (part 2) of Tony’s notes and Exercise C1. With these we have

A ⊗_{R} M = A ⊗_{R} (⊕M_{i}) ≅ (⊕M_{i}) ⊗_{R} A ≅ ⊕(M_{i} ⊗_{R} A) ≅ ⊕(A ⊗_{R} M_{i}).

Similarly, we have that B ⊗_{R} M ≅ ⊕(B ⊗_{R} M_{i}) and C ⊗_{R} M ≅ ⊕(C ⊗_{R} M_{i}). Define φ_{1} : ⊕(A ⊗_{R} M_{i}) → ⊕(B ⊗_{R} M_{i}), φ_{2} : ⊕(B ⊗_{R} M_{i}) → ⊕(C ⊗_{R} M_{i}) by

φ_{1}((a_{1}, a_{2}, …)) = ((f ⊗ id_{1})(a_{1}), (f ⊗ id_{2})(a_{2}), …), and similarly φ_{2}((b_{1}, b_{2}, …)) = ((g ⊗ id_{1})(b_{1}), (g ⊗ id_{2})(b_{2}), …) (where a_{i} ∈ A ⊗_{R} M_{i} and b_{i} ∈ B ⊗_{R} M_{i}).

These maps are clearly well-defined. Morever, they give rise to the following commutative diagram.

A ⊗_{R} M ⟶^{f⊗id} B ⊗_{R} M ⟶^{g⊗id} C ⊗_{R} M
α_{A} ↑≅↓ β_{A} α_{B} ↑≅↓ β_{B} α_{C} ↑≅↓ β_{C}
⊕(A ⊗_{R} M_{i}) ⟶^{φ_{1}} ⊕(B ⊗_{R} M_{i}) ⟶^{φ_{2}} ⊕(C ⊗_{R} M_{i})

where id : M ⟶ M is the identity map on M.

Claim 1: The bottom row of this diagram is exact.

To see this take b = (b_{1}, b_{2}, …) ∈ Im(φ_{1}). Then there exists a = (a_{1}, a_{2}, …) ∈ ⊕(A ⊗_{R} M_{i}) such that φ_{1}(a) = b, i.e. b_{i} = (f ⊗ id_{i})(a_{i}) for all i. So we have

φ_{2}(b) = ((g ⊗ id_{1})(b_{1}), (g ⊗ id_{2})(b_{2}), …)
= ((g ⊗ id_{1})((f ⊗ id_{1})(a_{1})), (g ⊗ id_{2})((f ⊗ id_{2})(a_{2})), …)
= (0, 0, …)

(by exactness of (1)). Thus, b ∈ Ker(φ_{2}), and hence Im(φ_{1}) ⊆ Ker(φ_{2}).

Now take c = (c_{1}, c_{2}, …) ∈ Ker(φ_{2}) ⟹ φ_{2}(c) = ((g⊗id_{1})(c_{1}), (g⊗id_{2})(c_{2}), …) = (0, 0, …). But, by exactness of (1), we have that there exists s_{i} ∈ A ⊗_{R} M_{i} such that c_{i} = (f ⊗ id_{i})(s_{i}) (where s_{i} = 0 if c_{i} = 0). Thus, s = (s_{1}, s_{2}, …) ∈ ⊕(B ⊗_{R} M_{i}) is such that φ_{1}(s) = c. Thus, c ∈ Im(φ_{1}), and hence Ker(φ_{2}) ⊆ Im(φ_{1}). Therefore, Ker(φ_{2}) = Im(φ_{1}) and Claim 1 is proved.

Claim 2: The top row of our commutative diagram is exact.

To see this note that a sequence isomorphic to an exact sequence is again an exact sequence.

It follows from the above claim that M is R-flat.

⟹. Suppose M is R-flat. Let A ⟶^{f} B ⟶^{g} C be an exact sequence of R-modules and R-module homomorphisms. Then, by definition of a flat R-module,

A ⊗_{R} M ⟶^{f⊗id} B ⊗_{R} M ⟶^{g⊗id} C ⊗_{R} M  (2)

is again exact, where id : M ⟶ M is the identity map on M.

As in the first direction, A ⊗_{R} M ≅ ⊕(A ⊗_{R} M_{i}), B ⊗_{R} M ≅ ⊕(B ⊗_{R} M_{i}), and C ⊗_{R} M ≅ ⊕(C ⊗_{R} M_{i}). We also have the commutative diagram from the first direction.

Define the maps ρ_{A} : ⊕(A ⊗_{R} M_{i}) → A ⊗_{R} M_{i} by (a_{1}, a_{2}, …, a_{i}, …) = a_{i} for a_{i} ∈ A⊗_{R}M_{i}, ρ_{B} : ⊕(B⊗_{R}M_{i}) → B⊗_{R}M_{i} by (b_{1}, b_{2}, …, b_{i}, …) = b_{i} for b_{i} ∈ B⊗_{R}M_{i}, and ρ_{C} : ⊕(C ⊗_{R} M_{i}) → C ⊗_{R} M_{i} by (c_{1}, c_{2}, …, c_{i}, …) = c_{i} for c_{i} ∈ C ⊗_{R} M_{i}. Also define ι_{A} : A⊗_{R}M_{i} → ⊕(A⊗_{R}M_{i}) by ι(a) = (0, 0, …, 0, a, 0, …), where a ∈ A⊗_{R}M_{i} is in the i^{th} position, ι_{B} : B ⊗_{R} M_{i} → ⊕(B ⊗_{R} M_{i}) by ι(b) = (0, 0, …, 0, b, 0, …), where b ∈ B ⊗_{R} M_{i} is in the i^{th} position, and ι_{C} : C ⊗_{R} M_{i} → ⊕(C ⊗_{R} M_{i}) by ι(c) = (0, 0, …, 0, c, 0, …), where c ∈ C ⊗_{R} M_{i} is in the i^{th} position.

We clearly have the following commutative diagrams for all i ∈ I.

⊕(A ⊗_{R} M_{i}) ⟶^{φ_{1}} ⊕(B ⊗_{R} M_{i}) ⟶^{φ_{2}} ⊕(C ⊗_{R} M_{i})
ι_{A} ↑↓ ρ_{A}  ι_{B} ↑↓ ρ_{B}  ι_{C} ↑↓ ρ_{C}
A ⊗_{R} M_{i}  ⟶^{f⊗id_{i}} B ⊗_{R} M_{i}  ⟶^{g⊗id_{i}} C ⊗_{R} M_{i}

where ρ_{A} ∘ ι_{A} = id_{A⊗_{R}M_{i}}, ρ_{B} ∘ ι_{B} = id_{B⊗_{R}M_{i}}, ρ_{C} ∘ ι_{C} = id_{C⊗_{R}M_{i}}, and the maps id_{i} : M_{i} ⟶ M_{i} are the identity maps on M_{i}.

Since the top row of this commutative diagram is isomorphic to (2) (which is an exact sequence) we have that this row is an exact sequence.

Claim 3: The bottom row of this commutative diagram is exact for all i ∈ I. To prove this we let i ∈ I be arbitary. Take b ∈ B⊗_{R}M_{i} such that b ∈ Im((f⊗id_{i})).

Then there exists a ∈ A⊗_{R}M_{i} such that (f ⊗id_{i})(a) = b. Let ι_{A}(a) = a′ and ι(b) = b′. Then, by the commutativity of our diagram, φ_{1}(a′) = b′. Since our top row is exact we must have then that b′ ∈Ker(φ_{2}), i.e. φ_{2}(b′) = (0, 0, …). This implies that ρ_{C}(φ_{2}(b′)) = 0. But 0 = ρ_{C}(φ_{2}(b′)) = (g ⊗ id_{i})(ρ_{B}(b′)) = (g ⊗ id_{i})(b). That is, b ∈Ker((g ⊗ id_{i})). Thus, Im ((f ⊗ id_{i})) ⊆ Ker((g ⊗ id_{i})).

Finally, take b ∈ B⊗_{R}M_{i} such that b ∈ Ker((g⊗id_{i})). Then (g⊗id_{i})(b) = 0. Then ι_{C}((g ⊗ id_{i})(b)) = (0, 0, …). Let ι_{B}(b) = b′ ∈ ⊕(B ⊗_{R} M_{i}). Then, by commutativity of our diagram, φ_{2}(b′) = (0, 0, …) ⟹ b′ ∈Ker(φ_{2}). By exactness of the top row in our commutative diagram, we then have that b′ ∈Im(φ_{1}). So let a′ ∈ ⊕(A ⊗_{R} M_{i}) be such that φ_{1}(a′) = b′. Then ρ_{B}(φ_{1}(a′)) = ρ_{B}(b′) = b = (f ⊗ id_{i})(ρ_{A}(a′)) (by the commutativity of our diagram). This implies that b ∈Im((f ⊗ id_{i})). Therefore, Ker((g ⊗ id_{i})) ⊆ Im((f ⊗ id_{i})), and so Im((f ⊗ id_{i})) = Ker((g ⊗ id_{i})). Thus, the bottom row is exact. This completes the proof of Claim 3.

Since the above proof was done for arbitrary i ∈ I, it follows that each M_{i} is R-flat.