0
$\begingroup$

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 !?

$\endgroup$
5
  • $\begingroup$ Do you mean a long exact sequence? Should it begin with $0 \to A$ or just $A$? $\endgroup$ Jan 2, 2013 at 12:05
  • 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. $\endgroup$ Jan 2, 2013 at 14:35
  • 1
    $\begingroup$ Although with that interpretation, the sequence $A\rightarrow 0\rightarrow B$ provides an example. $\endgroup$ Jan 2, 2013 at 15:01
  • $\begingroup$ @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.) $\endgroup$ Jan 2, 2013 at 15:23
  • $\begingroup$ I give for granted that the question is about extensions, thus my answer. $\endgroup$ Jan 2, 2013 at 20:15

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.