Free presentation

Short description: In algebra, a module over a ring

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

$\underset{i\in I}{⨁}R\stackrel{f}{\to }\underset{j\in J}{⨁}R\stackrel{g}{\to }M\to 0.$

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: $F\stackrel{g}{\to }M\to 0$, but then the kernel of g is again a quotient of a free module: $F^{\prime }\stackrel{f}{\to }\ker g\to 0$. 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:

$\underset{i\in I}{⨁}N\stackrel{f\otimes 1}{\to }\underset{j\in J}{⨁}N\to M{\otimes }_{R}N\to 0.$

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

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 $\theta :F(R^{\oplus n})\to G(R^{\oplus n})$ is an isomorphism for each natural number n, then $\theta :F(M)\to G(M)$ is an isomorphism for any finitely-presented module M.

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. $◻$

See also

• Coherent module
• Finitely related module
• Fitting ideal
• Quasi-coherent sheaf

References

• Eisenbud, David, Commutative Algebra with a View Toward Algebraic Geometry, Graduate Texts in Mathematics, 150, Springer-Verlag, 1995, ISBN:0-387-94268-8.