# What is the importance of functors being left exact, right exact, and exact, especially in the context of modules?

I have just studied section 10.5 of Dummit and Foote, which covers exact sequences and applies them to projective, injective, and flat modules. There is a common theme: we start with a short exact sequence, then apply some functor ($$\mathrm{Hom}_R(P,\star)$$, $$\mathrm{Hom}_R(\star,Q)$$, and $$D\otimes_{R}\star)$$. Unfortunately, the functor is in general only right or left exact, and we may have to flip the order if the functor is contravariant.

Projective, injective, and flat modules, therefore, are the natural definition to make: they are the modules for which short exact sequences are indeed mapped to short exact sequences by the functor.

My question: Why do we care about these properties? While it seems nice that short exactness is preserved, why did someone care enough to define these notions of projective/injective/flat modules, and why did it make into an introductory textbook? Where would these notions become important later in my study of math?

I can give some reasons why someone working with modules ought to care about these properties and abstract them into definitions.

For flat modules, the definition arises naturally when we intuit something from the notation, and then discover that this notation can sometimes be misleading.

For injective and projective modules, the definition comes from looking at how homomorphisms naturally arise from other homomorphisms in certain situations, and asking the question "Does every homomorphism arise in this natural way?"

Flat:

Let $$P, M, N$$ be modules over your commutative ring $$R$$. Suppose that $$M$$ is a submodule of $$N$$. The generators of $$N \otimes_R P$$ are written as $$n \otimes p$$ for $$n \in N$$ and $$p \in P$$ and likewise for the generators of $$M \otimes_R P$$.

There is a natural $$R$$-module homomorphism $$\varphi: M \otimes_R P \rightarrow N \otimes_R P$$ given on generators by

$$\varphi(m \otimes p) = m \otimes p$$

By the notation, it looks like $$\varphi$$ doesn't actually do anything. From the notation, and since $$M$$ is a subset of $$N$$, it looks like $$M \otimes_R P$$ should be a submodule of $$N \otimes_R P$$. But this isn't always true. You cannot guarantee this for all inclusions of submodules $$M \subset N$$ unless $$P$$ is flat. This is one possible definition you can take for flat modules: modules for which the above intuition always works.

Injective:

Easier to motivate this for abelian groups (that is, $$\mathbb Z$$-modules). Let $$A$$ be a subgroup of an abelian group $$B$$. Let $$\mathbb C^{\ast}$$ be the (multiplicative) group of complex numbers. If $$\chi: A \rightarrow \mathbb C^{\ast}$$ is a group homomorphism, one might wonder whether it is possible to extend $$\chi$$ (possibly nonuniquely) to a group homomorphism $$\overline{\chi}: B \rightarrow \mathbb C^{\ast}$$. It turns out that this is indeed possible, because $$\mathbb C^{\ast}$$ is an injective object in the category of abelian groups.

You can take this as the definition of injective: an abelian group $$C$$ is injective if whenever $$A \subset B$$ are abelian groups, every homomorphism of $$A$$ into $$C$$ can be extended (possibly nonuniquely) to a homomorphism of $$B$$ into $$C$$.

This is a pretty useful property for abelian groups to have. I ended up randomly needing the fact that $$\mathbb C^{\ast}$$ is an injective abelian group in my dissertation.

Projective:

Let $$R$$ be a commutative ring with identity, and let $$I \subset J$$ be ideals of $$R$$. There is a natural ring homomorphism

$$R/I \rightarrow R/J$$

$$r+I \mapsto r+J$$

Now let $$B$$ be another commutative ring with identity. If you're given a homomorphism of $$B$$ into $$R/I$$, you can compose with the natural homomorphism above and get a homomorphism of $$B$$ into $$R/J$$. It's natural to ask whether there are homomorphisms of $$B$$ into $$R/J$$ which don't arise in this fashion. This is the sort of idea that projective modules are based on.

Let $$P$$ be an $$R$$-module, and let $$N$$ be a submodule of an $$R$$-module $$M$$. You have a natural $$R$$-module homomorphism $$\pi: M \rightarrow M/N$$. If $$\varphi: P \rightarrow M/N$$ is an $$R$$-module homomorphism, you might ask the question of whether $$\varphi$$ always comes from a (possibly nonunique) $$R$$-module homomorphism of $$P$$ into $$M$$. The answer "Yes for all choices of $$N \subset M$$" is equivalent to $$P$$ being projective.