Free presentation

In algebra, a free presentation of a module M over a commutative ring R is an exact sequence of R-modules:


 * $$\bigoplus_{i \in I} R \ \overset{f} \to\ \bigoplus_{j \in J} R \ \overset{g}\to\ M \to 0.$$

Note the image under g of the standard basis generates M. In particular, if J is finite, then M is a finitely generated module. If I and J are finite sets, then the presentation is called a finite presentation; a module is called finitely presented if it admits a finite presentation.

Since f is a module homomorphism between free modules, it can be visualized as an (infinite) matrix with entries in R and M as its cokernel.

A free presentation always exists: any module is a quotient of a free module: $$F \ \overset{g}\to\ M \to 0$$, but then the kernel of g is again a quotient of a free module: $$F' \ \overset{f} \to\ \ker g \to 0$$. The combination of f and g is a free presentation of M. Now, one can obviously keep "resolving" the kernels in this fashion; the result is called a free resolution. Thus, a free presentation is the early part of the free resolution.

A presentation is useful for computation. For example, since tensoring is right-exact, tensoring the above presentation with a module, say N, gives:


 * $$\bigoplus_{i \in I} N \ \overset{f \otimes 1} \to\ \bigoplus_{j \in J} N \to M \otimes_R N \to 0.$$

This says that $$M \otimes_R N$$ is the cokernel of $$f \otimes 1$$. If N is also a ring (and hence an R-algebra), then this is the presentation of the N-module $$M \otimes_R N$$; that is, the presentation extends under base extension.

For left-exact functors, there is for example

Proof: Applying F to a finite presentation $$R^{\oplus n} \to R^{\oplus m} \to M \to 0$$ results in
 * $$0 \to F(M) \to F(R^{\oplus m}) \to F(R^{\oplus n}).$$

This can be trivially extended to
 * $$0 \to 0 \to F(M) \to F(R^{\oplus m}) \to F(R^{\oplus n}).$$

The same thing holds for $$G$$. Now apply the five lemma. $$\square$$