I have to prove that if P is a R-module , P is projective right there is a family&nbsp;{xi}&nbsp;in P and morphisms&nbsp;fi:P⇒R&nbsp;such that for all&nbsp;x∈Px=∑i∈Ifi(x)xiwhere for each&nbsp;x∈P,&nbsp;fi(x)=0&nbsp;for almost all&nbsp;i∈I.

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I have to prove that if P is a R-module , P is projective right there is a family&amp;nbsp;{xi}&amp;nbsp;in P and morphisms&amp;nbsp;fi:P⇒R&amp;nbsp;such that for all&amp;nbsp;x∈Px=∑i∈Ifi(x)xiwhere for each&amp;nbsp;x∈P,&amp;nbsp;fi(x)=0&amp;nbsp;for almost all&amp;nbsp;i∈I.

August Moore · Accepted Answer

Step 1Put 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 thatx∑fi(x)xijust means that there is some g:R(I)⇒P such that g(f(x))=x, namely g(r1 r2⋯)=r1x1+r2x2+⋯. In other words, P is a direct summand of the free module R(I)

I have to prove that if P is a R-module

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