Projective cover

In the branch of abstract mathematics called category theory, a projective cover of an object X is in a sense the best approximation of X by a projective object P. Projective covers are the dual of injective envelopes.

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Definition

Let \mathcal{C} be a category and X an object in \mathcal{C}. A projective cover is a pair (P,p), with P a projective object in \mathcal{C} and p a superfluous epimorphism in Hom(P, X).

If R is a ring, then in the category of R-modules, a superfluous epimorphism is then an epimorphism p�: P \to X such that the kernel of p is a superfluous submodule of P.

Properties

Projective covers and their superfluous epimorphisms, when they exist, are unique up to isomorphism.

The main effect of p having a superfluous kernel is the following: if N is any proper submodule of P, then p(N) \ne M[1]. Informally speaking, the superfluous kernel causes P to cover M optimally, in some sense. This does not depend upon the projectivity of P, it is true of all superfluous epimorphisms.

If (P,p) is a projective cover of M, and P' is another projective module with an epimorphism p':P'\rightarrow M, then there is a split epimorphism α from P' to P such that p\alpha=p'

Unlike injective envelopes, which exist for every left (right) R-module regardless of the ring R, left (right) R-modules do not in general have projective covers. A ring R is called left (right) perfect if every left (right) R-module has a projective cover in R-Mod (Mod-R).

A ring is called semiperfect if every finitely generated left (right) R-module has a projective cover in R-Mod (Mod-R). "Semiperfect" is a left-right symmetric property.

A ring is called lift/rad if idempotents lift from R/J to R, where J is the Jacobson radical of R. The property of being lift/rad can be characterized in terms of projective covers: R is lift/rad if and only if direct summands of the R module R/J (as a right or left module) have projective covers. (Anderson, Fuller 1992, p. 302)

Examples

In the category of R modules:

See also

References

  1. ^ Proof: Let N be proper in P and suppose p(N)=M. Since ker(p) is superfluous, ker(p)+NP. Chose x in P outside of ker(p)+N. By the surjectivity of p, there exists x' in N such that p(x' )=p(x ),, whence xx' is in ker(p). But then x is in ker(p)+N, a contradiction.