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?

  • 1
    $\begingroup$ 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...) $\endgroup$ – Arturo Magidin Sep 23 '10 at 20:04

Put the $f_i$ 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 $g:R^{(I)}\to P$ such that $g(f(x)) = x$, namely $g((r_1,r_2,...)) = r_1x_1 + r_2x_2 +\cdots$. In other words, $P$ is a direct summand of the free module $R^{(I)}$.

  • $\begingroup$ 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). $\endgroup$ – Dylan Wilson Sep 23 '10 at 22:23
  • $\begingroup$ No, this works for all index sets, precisely because the f_i(x) vanish for all but finitely many i. $\endgroup$ – Jack Schmidt Sep 23 '10 at 22:37
  • $\begingroup$ Whoops! My bad. $\endgroup$ – Dylan Wilson Sep 23 '10 at 22:43

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