projective equivalence

Let $R$ be a ring with 1. Two $R$ -modules $A$ and $B$ are said to be projectively equivalent $A\sim B$ if there exist two projective $R$ -modules $P$ and $Q$ such that

A\oplus P\cong B\oplus Q.

Remarks.

1.

Projective equivalence is an equivalence relation.
2.

Any projective module is projectively equivalent to the zero module.
3.

(Schanuel’s Lemma). Given two short exact sequences:

$\xymatrix{0\ar[r]&B_{1}\ar[r]&P\ar[r]&A_{1}\ar[r]&0}$

$\xymatrix{0\ar[r]&B_{2}\ar[r]&Q\ar[r]&A_{2}\ar[r]&0}$

with $A_{1}\sim A_{2}$ , then $B_{1}\sim B_{2}$ .
4.

Schanuel’s Lemma can be generalized. Given two projective resolutions:

$\xymatrix{\ldots\ar[r]^{p_{3}}&P_{2}\ar[r]^{p_{2}}&P_{1}\ar[r]^{p_{1}}&P_{0}% \ar[r]^{p_{0}}&A_{1}\ar[r]&0}$

$\xymatrix{\ldots\ar[r]^{q_{3}}&Q_{2}\ar[r]^{q_{2}}&Q_{1}\ar[r]^{q_{1}}&Q_{0}% \ar[r]^{q_{0}}&A_{2}\ar[r]&0}$

with $A_{1}\sim A_{2}$ , then $\operatorname{Ker}(p_{n})\sim\operatorname{Ker}(q_{n})$ for all $n\geq 0$
5.

The concept of projective equivalence between two modules can be generalized to any abelian categories having enough projectives.

Title	projective equivalence
Canonical name	ProjectiveEquivalence
Date of creation	2013-03-22 14:50:13
Last modified on	2013-03-22 14:50:13
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	5
Author	CWoo (3771)
Entry type	Definition
Classification	msc 16E10
Classification	msc 18G20
Classification	msc 18G10
Defines	projectively equivalent
Defines	Schanuel’s Lemma