Over the next few posts, I'll talk more about axiomatizability of algebraic structures in first-order logic. Before I do that, we need to know about purity of exact sequences. So let's fix a ring $R$. An exact sequence

$$0\to A\to B\to C\to 0$$ in the category of left $R$ modules is called **pure** if for every right $R$-module $C'$, the sequence

$$0\to A\otimes_R C'\to B\otimes_R C'\to C\otimes_R C'\to 0$$ is exact. Notice how this is like a dual concept to flatness: a right $R$-module is flat if its associated tensor functor preserves every exact sequence in the category of left $R$-modules. Whereas, a sequence is pure if its preserved by every tensor product functor.

However, it turns out we can also characterize flatness in terms of purity.

**Theorem.**A left $R$-module $C$ is flat if and only if every exact sequence of the form $0\to A\to B\to C\to 0$ is pure.

This is quite interesting. In other words, this theorem says that flat modules are precisely those modules that appear as quotients exclusively in pure exact sequences. Let's see why this is true.

*Proof.*($\Rightarrow$) First, assume that $C$ is flat, and that $0\to A\to B\to C\to 0$ is an arbitrary exact sequence. We need to show that this sequence is pure. If $X$ is a right $R$-module, then applying $-\otimes X$ gives a long exact sequence of the form

$$\cdots\to {\rm Tor}_1(C,X)\to A\otimes X\to B\otimes X\to C\otimes X\to 0.$$ Since $C$ is flat, the group ${\rm Tor}_1(C,X)$ is zero. Therefore,

$$0\to A\otimes X\to B\otimes X\to C\otimes X\to 0$$ is an exact sequence. Thus the original sequence $0\to A\to B\to C\to 0$ is pure.

($\Leftarrow$) Now, assume that $C$ is a left $R$-module such that any sequence $0\to A\to B\to C\to 0$ is pure. We will prove that $C$ is flat.

Suppose that $0\to A'\to B'\to C'\to 0$ is an arbitrary exact sequence of right $R$-modules. We will prove that that image of this sequence under the functor $C\otimes-$ is exact. To do so, choose a sequence $0\to K\to F\to C\to 0$ where $F$ is a flat module. Such a sequence exists, since every module is the quotient of a free module, and free modules are flat. Form the following commutative diagram:

The exactness of the rows and columns is as presented: that is, all columns are exact (by purity), the middle row is exact (since $F$ is flat), and the top and bottom rows are right exact as the tensor product is a right-exact functor. To finish the proof, we need to show that the map $C\otimes A'\to C\otimes B'$ is injective, which is easily accomplished by the following diagram chase:

- Let $y\in C\otimes A'$ map to zero under $C\otimes A'\to C\otimes B'$. We need to show that $y = 0$.
- Since $F\otimes A'\to C\otimes A'$ is surjective, it maps some $x\in F\otimes A'$ to $y$.
- Let $z$ be the image of $x$ under $F\otimes A'\to F\otimes B'$.
- Commutativity of the diagram shows that the image of $z$ under $F\otimes B'\to C\otimes B'$ is zero.
- By exactness of the middle column, $z$ is the image of some $t$ under the map $K\otimes B'\to F\otimes B'$.
- The image of $t$ under $K\otimes B'\to K\otimes C'\to F\otimes C'$ is zero.
- Since $K\otimes C'\to F\otimes C'$ is injective, the image of $t$ under $K\otimes B'\to K\otimes C'$ is zero.
- Since the top row is exact at $K\otimes B'$, $t$ is the image of some $u\in K\otimes A'$ under the map $K\otimes A'\to K\otimes B'$.
- By construction, the image of $u$ under $0 = K\otimes A'\to F\otimes A'\to C\otimes A'$ is $y$, and therefore $y = 0$.

In our next post, we will see how purity can help with some axiomatizability results.