Exact Sequences I - Illustrative Examples

#module_theory

Before diving into the definition of an exact sequence (and morphisms of exact sequences), we examine a few illustrative examples.

Submodules and quotient modules

Suppose $N$ is a submodule of an $R$ -module $M$ , where $R$ is a ring (with unity). We then have a pair of $R$ -module morphisms connecting the submodule $N$ to the quotient module $M / N$ , namely the (injective) inclusion morphism $i : N \to M$ and the (surjective) projection morphism $π : M \to M / N$ , and these morphisms are related by the fact that $\ker (π) = im (i) :$

N \overset{i}{\to} M \overset{π}{\to} M / N .

Direct sums of modules

Suppose $M_{1}$ and $M_{2}$ are two $R$ -modules. We have seen that their direct sum $M_{1} \oplus M_{2}$ and direct product $M_{1} \times M_{2}$ are isomorphic as $R$ -modules. (Later we'll see this as an instance of something called a biproduct. Also note that if we use our general construction for direct sums and products, in which we replace the notions of formal sums and ordered tuples by certain special set maps, then these two modules aren't just isomorphic, they're literally the same.) The former comes with injective $R$ -module morphisms $j_{i} : M_{i} \to M_{1} \oplus M_{2}$ ; the latter comes with surjective $R$ -module morphisms $p_{i} : M_{1} \times M_{2} \to M_{i}$ . If we let $f : M_{1} \oplus M_{2} \tilde{\to} M_{1} \times M_{2}$ be the $R$ -module isomorphism that sends $m_{1} + m_{2} \mapsto (m_{1}, m_{2})$ , then pre-composing each of the projections $p_{i}$ with $f$ gives surjective $R$ -module morphisms $π_{i} : M_{1} \oplus M_{2} \to M_{i}$ .

What does all of this have in common with the previous example? There is now a very similar connection between the module $M_{1}$ and the module $M_{2}$ by way of the module $M_{1} \oplus M_{2}$ , namely a sequence of $R$ -module morphisms

M_{1} \overset{j_{1}}{\to} M_{1} \oplus M_{2} \overset{π_{2}}{\to} M_{2}

where $j_{1}$ is injective, $π_{2}$ is surjective, and $\ker (π_{2}) = im (j_{1})$ .

Generators and relations for a module

Suppose $M$ is an $R$ -module and $X_{1}$ is any subset of the elements of $M$ . The submodule of $M$ generated by $X_{1}$ is exactly the image of the $R$ -module morphism

F (X_{1}) \overset{π_{1}}{\to} M

where $F (X_{1})$ is the free $R$ -module on $X_{1}$ and $π$ is the $R$ -module morphism corresponding to the inclusion $i : X_{1} \to U (M)$ . In particular, the set $X_{1}$ generates $M$ as an $R$ -module exactly when $π_{1}$ is surjective.

The kernel of $π_{1}$ is the submodule consisting of all formal sums $\sum_{x \in X_{1}} r_{x} \cdot x$ that simplify to $0_{M}$ in the module $M$ . In other words, it consists of all $R$ -linear relations among the elements in $X_{1}$ . If we let $i_{1} : \ker (π_{1}) \to F (X_{1})$ be the inclusion morphism, then we have a sequence of morphisms

\ker (π_{1}) \overset{i_{1}}{\to} F (X_{1}) \overset{π_{1}}{\to} M

where $i_{1}$ is injective, $π_{1}$ is surjective, and $\ker (π_{1}) = im (i_{1})$ . This information amounts to the classic "generators and relations" description of a module.

However, as the suspicious subscripts might indicate, there is a new feature available in this example. Unlike in the previous examples, here it is clear that we can continue this process. That's because $\ker (π_{1})$ is an $R$ -module, and so a set of generators for $\ker (π_{1})$ corresponds to another surjection

F (X_{2}) \overset{π_{2}^{'}}{\to} \ker (π_{1}) .

Composing with $i_{1}$ then gives a sequence of morphisms

F (X_{2}) \overset{π_{2}}{\to} F (X_{1}) \overset{π_{1}}{\to} M .

While it's true $π_{2}$ is no longer surjective, we still do have $\ker (π_{1}) = im (π_{2})$ . And we can once again continue the process, finding the kernel of $π_{2}$ , then a set of generators for that kernel, hence a surjection from another free module onto that kernel, and so on. In doing so, we are slowly building a free resolution of the module $M$ :

\dots \overset{π_{n + 1}}{\to} F (X_{n}) \overset{π_{n}}{\to} F (X_{n - 1}) \overset{π_{n - 1}}{\to} \dots \overset{π_{2}}{\to} F (X_{1}) \overset{π_{1}}{\to} M .

Moreover, at every spot in this sequence we have $\ker (π_{n}) = im (π_{n + 1})$ .

One could reasonably hope that most (maybe even all?) properties of $M$ are encoded in this sequence of morphisms. There's a catch, though. The choice of generators (at every step!) is not unique. So for a given module $M$ there could be (and almost always are) many other such sequences (of free modules with these kernel-image relationships). How could we compare one sequence to another? We would probably want a notion of "morphism" between such sequences...

Suggested next note

Exact Sequences II - Exact Sequences