# 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. 1.

Projective equivalence is an equivalence relation.

2. 2.

Any projective module is projectively equivalent to the zero module.

3. 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. 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. 5.

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

Title projective equivalence ProjectiveEquivalence 2013-03-22 14:50:13 2013-03-22 14:50:13 CWoo (3771) CWoo (3771) 5 CWoo (3771) Definition msc 16E10 msc 18G20 msc 18G10 projectively equivalent Schanuel’s Lemma