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A module $_RU$ is projective module its annihilator in case $U$ is a projective $R/l_R(U)$ module.
Prove that these are equivalent:
(a) $_RU$ is projective module its annihilator;
(b) $U$ is a projective $R/I$ module for some ideal $I\subseteq{l_R(U)}$;
(c) $U$ is $U^A$-projective for every set $A$.
I have proved that $(a)\Rightarrow(b)$ and $(b)\Rightarrow(c)$, but I can't prove that $(c)\Rightarrow(a)$. Could any body give me some help? Thanks a lot!

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  • $\begingroup$ It is a question in GTM13 page 13,the 15th question $\endgroup$ – mazicai Feb 19 '14 at 1:21
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Proof (1) To prove $(a)\Rightarrow(b)$, we just need to take $I=l_R(U)$.
$(b)\Rightarrow(c):$ $U$ is a projective $R/I$ module, so $U^A$ is a $R/I$ module. And by the definition of projective module,for any epimorphism $g: M\rightarrow N$ and each homomorphism $f:U\rightarrow N$, there is a homomorphism $h:M\rightarrow U$ such that $f=gh$.Since $g$ is epic, then $U^A/ker {\ g}$ is isomophic to $N$. Thus $N$ is an $R/I$ module.Therefor by $(b)$, we can get that for any epimorphism $g: {U^A}\rightarrow N$ and each homomorphism $f:U\rightarrow N$, there is a homomorphism $h:U\rightarrow U^A$ such that $f=gh$.Therefor $U$ is $U^A$-projective for any set $A$.
$(c)\Rightarrow(a):$ Since for any set $A$, $_RU$ is $U^A$-projective,then $_{R/l_R(U)}U$ is $U^A$-projective. Because $_{R/l_R(U)}U$ is the generator of $_{R/l_R(U)}M$ and $_{R/l_R(U)}U$ is $U^A$-projective for any set $A$. Therefore $U$ is projective $_{R/l_R(U)}U$ module. Thus $_RU$ is projective modulo its annihilator.
(2) To prove these statements, we just need to reverse the arrows in (1).

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