This question might be trivial but I cann't see.
Let $A$ and $B$ be two modules. is it always possible to have an exact sequence which begins with $A$, ends with $B$ with all modules in the sequence (other than $A$ and $B$) projective !?
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$\begingroup$ Do you mean a long exact sequence? Should it begin with $0 \to A$ or just $A$? $\endgroup$– Martin BrandenburgCommented Jan 2, 2013 at 12:05
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1$\begingroup$ @Martin Brandenburg: It must mean just $A$. Otherwise, it would require embedding $A$ into a projective module (or directly into $B$), and any non-free $\mathbb Z$-module would provide a counterexample. $\endgroup$– Andreas BlassCommented Jan 2, 2013 at 14:35
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1$\begingroup$ Although with that interpretation, the sequence $A\rightarrow 0\rightarrow B$ provides an example. $\endgroup$– Jeremy RickardCommented Jan 2, 2013 at 15:01
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$\begingroup$ @Jeremy: Right, I missed that. So we don't seem to have any non-trivial interpretation of the question. What if we ask for just $A$ at the beginning but $B\to0$ at the end? Your idea would handle any $B$ of finite projective dimension, but there seems to be no such exact sequence if $A$ has finite projective dimension and $B$ doesn't. (Maybe I should stop trying to de-trivialize the question and let the OP tell us what (s)he actually wants.) $\endgroup$– Andreas BlassCommented Jan 2, 2013 at 15:23
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$\begingroup$ I give for granted that the question is about extensions, thus my answer. $\endgroup$– Fernando MuroCommented Jan 2, 2013 at 20:15
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Not in general. The keyword is stable module category, the quotient of the module category by the ideal of morphisms which factor through a projective. The leftmost term is functorial on the rightmost term in this category if all intermediate modules are projective. This imposes some restrictions. If you take a hereditary ring, eg the integers, you get easy counterexamples as any submodule of a projective module is projective.
answered Jan 2, 2013 at 11:13