Lang's Algebra page 613 equivalent definitions of Flat Module. F3 $\implies$ F1. How?

Question

We are working in a category of modules and their morphisms.

F3. For every injection $0 \to E' \to E$ the sequence $0 \to F\otimes E' \to F \otimes E$ is exact.

F1. For every exact sqequence $E' \to E \to E''$ the sequence $F\otimes E' \to F \otimes E \to F\otimes E''$ is exact.

Question. How does one prove that F3 $\implies$ F1?

The hint in the book says:

[...] by writing down the kernel and image of the map $E' \to E$ and applying F3.

---

Attempt. Given $E' \xrightarrow{f} E \xrightarrow{g} E''$ we have that if $K = \ker f$ and $I = \text{im f}$, then $0 \to K \to E' \to I \to 0$ is exact. Applying F3 and the fact that $\otimes$ is right exact gives us that:

$$ 0 \to F\otimes K \to F\otimes E' \to F\otimes I \to 0 $$

is exact. But we can do the same thing for $g$ so that:

$$ 0 \to F\otimes K' \to F\otimes E\to F\otimes I' \to 0 $$

But $K' = \ker g = \text{im } f = I$. We can't simply glue at $F\otimes I$ and compose the two maps - doesn't work.

Answer 1

This answer is really a comment that won't fit as a comment.

You've probably solved this problem already, but I want to show you a way to draw diagrams so as to make it super easy to visualise how the various morphisms fit together. Personally, I find these diagrams extremely helpful when constructing these kinds of proofs - without them, I feel a bit lost.

I'll explain the approach in the context of the $F \otimes .$ functor (flat modules), but idea is also applicable to the $\text{Hom}(P,.)$ functor (projective modules) and to the $\text{Hom}(.,I)$ functor (injective modules).

Original diagram

We start with the diagram below.

This is a commutative diagram. Both triangles commute.

All three of the diagonal arrow sequences (in black) are exact sequences.

The horizontal arrow sequence (in blue) is also an exact sequence.

Tensoring with $F$ (when $F$ is not necessarily flat)

Suppose $F$ is some module, not necessarily flat. If we apply the $F \otimes .$ functor to our diagram, then we get a new diagram, as shown below.

This too is a commutative diagram - both triangles commute.

The three diagonal arrow sequences (in black) are exact, but notice that a couple of $0 \to .$ arrows from the the original diagram have disappeared. This is because the $F \otimes \text{Im}(f) \to F \otimes E$ and $F \otimes \text{Im}(g) \to F \otimes E''$ morphisms are not necessarily injective.

The horizontal arrow sequence (in blue) is not necessarily exact. The most we can say about the horizontal arrow sequence is that $$ \text{Im}(1 \otimes f) \subset \text{Ker}(1 \otimes g).$$ Please verify this claim using the diagram. It's a simple exercise in "diagram-chasing".

Tensoring with a flat $F$

So how are things different when $F$ is flat? When $F$ is flat, then the diagram looks like this.

The $0 \to .$ arrows have been restored, because the $F \otimes \text{Im}(f) \to F \otimes E$ and $F \otimes \text{Im}(g) \to F \otimes E''$ morphisms are injective, since $F$ is flat.

What's especially important is that the $F \otimes \text{Im}(g) \to F \otimes E''$ morphism is injective. This means that any element in the kernel of the $F \otimes E \to F \otimes E''$ morphism is also in the kernel of the $F \otimes E \to F \otimes \text{Im}(g)$ morphism. If you finish off this "diagram-chasing" argument, you should be able to deduce that $$ \text{Ker}(1 \otimes g) \subset \text{Im}(1 \otimes f).$$

Thus, the horizontal arrow sequence (in blue) is an exact sequence.

Summary

A nice way to summarise our findings is:

• In general, tensoring an exact sequence by some module $F$ does not necessarily produce an exact sequence (though it does produce a chain complex).
• But if $F$ is flat, then we do produce an exact sequence.