Free presentation

{{Short description|In algebra, a module over a ring}}

{{About|describing a module over a ring|specifying generators and relations of a group|presentation of a group}}

{{one source |date=May 2024}}

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

:$\backslash bigoplus\_\{i\; \backslash in\; I\}\; R\; \backslash \; \backslash overset\{f\}\; \backslash to\backslash \; \backslash bigoplus\_\{j\; \backslash in\; J\}\; R\; \backslash \; \backslash overset\{g\}\backslash to\backslash \; M\; \backslash to\; 0.$

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: $F\; \backslash \; \backslash overset\{g\}\backslash to\backslash \; M\; \backslash to\; 0$, but then the kernel of g is again a quotient of a free module: $F\text{'}\; \backslash \; \backslash overset\{f\}\; \backslash to\backslash \; \backslash ker\; g\; \backslash to\; 0$. 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:

: $\backslash bigoplus\_\{i\; \backslash in\; I\}\; N\; \backslash \; \backslash overset\{f\; \backslash otimes\; 1\}\; \backslash to\backslash \; \backslash bigoplus\_\{j\; \backslash in\; J\}\; N\; \backslash to\; M\; \backslash otimes\_R\; N\; \backslash to\; 0.$

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

For left-exact functors, there is for example

{{math_theorem|name=Proposition|Let F, G be left-exact contravariant functors from the category of modules over a commutative ring R to abelian groups and θ a natural transformation from F to G. If $\backslash theta:\; F(R^\{\backslash oplus\; n\})\; \backslash to\; G(R^\{\backslash oplus\; n\})$ is an isomorphism for each natural number n, then $\backslash theta:\; F(M)\; \backslash to\; G(M)$ is an isomorphism for any finitely-presented module M.}}

Proof: Applying F to a finite presentation $R^\{\backslash oplus\; n\}\; \backslash to\; R^\{\backslash oplus\; m\}\; \backslash to\; M\; \backslash to\; 0$ results in

:$0\; \backslash to\; F(M)\; \backslash to\; F(R^\{\backslash oplus\; m\})\; \backslash to\; F(R^\{\backslash oplus\; n\}).$

This can be trivially extended to

:$0\; \backslash to\; 0\; \backslash to\; F(M)\; \backslash to\; F(R^\{\backslash oplus\; m\})\; \backslash to\; F(R^\{\backslash oplus\; n\}).$

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