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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.
- Projective equivalence is an equivalence relation.
- Any projective module is projectively equivalent to the zero module.
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(Schanuel's Lemma). Given two short exact sequences:
![$ \xymatrix{0\ar[r]&B_1\ar[r]&P\ar[r]&A_1\ar[r]&0}$](http://images.planetmath.org/cache/objects/6505/js/img1.png)
with $A_1\sim A_2$ , then $B_1\sim B_2$ .![$ \xymatrix{0\ar[r]&B_2\ar[r]&Q\ar[r]&A_2\ar[r]&0}$](http://images.planetmath.org/cache/objects/6505/js/img2.png)
- Schanuel's Lemma can be generalized. Given two projective resolutions:
![$ \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}$](http://images.planetmath.org/cache/objects/6505/js/img3.png)
with $A_1\sim A_2$ , then $\operatorname{Ker}(p_n)\sim\operatorname{Ker}(q_n)$ for all $n\geq0$![$ \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}$](http://images.planetmath.org/cache/objects/6505/js/img4.png)
- The concept of projective equivalence between two modules can be generalized to any abelian categories having enough projectives.
projective equivalence is owned by Chi Woo.
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