0
$\begingroup$

Let us have a $P$ module. We have to prove the following statement:

$P$ is projective $\Leftrightarrow$ There is an $F$ free module, such that $F \cong F \oplus P$.

I have already seen the $\Rightarrow$ direction in a previously asked question, but can't really think of anything to prove the other way.

Any help appreciated!

$\endgroup$
  • 3
    $\begingroup$ If $F\cong F\oplus P$, then $P$ is a direct summand of a free module, which is one of the usual characterizations of a projective module. $\endgroup$ – Ben West Sep 13 '17 at 14:49
2
$\begingroup$

If $F\cong F\oplus P$, then $P$ is a direct summand of a free module, which is one of the usual characterizations of a projective module. – Ben West 16 mins ago

$\endgroup$
  • $\begingroup$ A stupid (!) question: if $F \cong F\oplus P$, how the ranks add up? Shouldn't it be "there exists a module $Q$ such that $Q\oplus P$ is a free module of finite rank"? $\endgroup$ – Krish Sep 13 '17 at 15:49
  • 1
    $\begingroup$ You should read to answer to the linked question in the OP. Also, there is no finite rank condition in the given characterization of projective. $\endgroup$ – David Hill Sep 13 '17 at 16:06
  • $\begingroup$ Ah!! I see now. Thanks for clearing the confusion. $\endgroup$ – Krish Sep 13 '17 at 16:10
  • $\begingroup$ @Xam I had already done that. $\endgroup$ – David Hill Sep 13 '17 at 20:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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