# Projective modules

I have to prove that if $P$ is a $R$-module , $P$ is projective $\Leftrightarrow$ there is a family $\{x_i\}$ in $P$ and morphisms $f_i\colon P\to R$ such that for all $x\in P$ $$x= \sum_{i\in I} f_i(x)x_i$$ where for each $x\in P$, $f_i(x)=0$ for almost all $i\in I$.

Any help?

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I think you misstated when you wrote $f_i=0$ for almost all $i$; you mean that for each $x$, $f_i(x)=0$ for almost all $i$ (that is, the ones that are nonzero depend on $x$; otherwise, the statement would be that there are finitely many functions such that...) –  Arturo Magidin Sep 23 '10 at 20:04
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## 1 Answer

Put the fi together to form one giant f from P to R(I), the direct sum of I copies of the ring R. The condition that x = Sum f_i(x) x_i just means that there is some g:R(I)→P such that g(f(x)) = x, namely g((r1,r2,...)) = r1*x1 + r2*x2 + .... In other words, P is a direct summand of the free module R(I).

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This is nice, but only works if I is finite. Otherwise you only get a map into the direct product; and the direct product of free modules may not be free... (take ocuntably many copies of Z, for example). –  Dylan Wilson Sep 23 '10 at 22:23
No, this works for all index sets, precisely because the f_i(x) vanish for all but finitely many i. –  Jack Schmidt Sep 23 '10 at 22:37
Whoops! My bad. –  Dylan Wilson Sep 23 '10 at 22:43
Thank you!, sometimes its hard to remember all those equivalences. I was trying to go from the diagram def but couldnt find any relation. –  Angel Sep 24 '10 at 3:19
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