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Let $P$ be a nonzero projective left $R$-module. I want to show that $Hom(P,R)$ is nonzero.

Let $I$ be an index set and let $R^{I}$ denote $|I|$ copies of $R$. We have an epimorphism $f$ from $R^{I}$ to $P$ and since $P$ is projective there exists $g: P \rightarrow R^{I}$ such that $f \circ g = 1_{P}$. Now consider the projection $\pi: R^{I} \rightarrow R$ then the composition $\pi \circ g$ determines a morphism from $P$ to $R$. Is this nonzero? Why?

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    You say «consider the projection $\pi:R^I\to R$...» but there are *many* such projections (as many as $I$ has elements!) Which *one* do you want to consider?2012-03-20

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For each $i\in I$ let $p_i:R^I\to R$ be the $i$th projection. If all the compositions $p_i\circ g$ are zero, then the image of $g$ is contained in the intersection of all the kernels of all the $p_i$.

That intersection is the zero submodule.

Since $g$ is injective, this is only possible if $P$ itself is the zero module.