redupticslaz

2022-04-27

I have to prove that if P is a R-module , P is projective right there is a family $\left\{{x}_{i}\right\}$ in P and morphisms ${f}_{i}:P\Rightarrow R$ such that for all $x\in P$

$x=\sum _{i\in I}{f}_{i}\left(x\right){x}_{i}$

where for each $x\in P,\text{}{f}_{i}\left(x\right)=0$ for almost all $i\in I$.

August Moore

Beginner2022-04-28Added 17 answers

Step 1

Put the$f}_{i$ together to form one giant f from P to $R}^{\left(I\right)$ , the direct sum of I copies of the ring R. The condition that

$x\sum {f}_{i}\left(x\right){x}_{i}$

just means that there is some$g:{R}^{\left(I\right)}\Rightarrow P$ such that $g\left(f\left(x\right)\right)=x$ , namely $g\left(\begin{array}{ccc}{r}_{1}& \text{}{r}_{2}& \cdots \end{array}\right)={r}_{1}{x}_{1}+{r}_{2}{x}_{2}+\cdots$ . In other words, P is a direct summand of the free module $R}^{\left(I\right)$

Put the

just means that there is some

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