Just a quick question, as the title says let $M,N$ be $R$-modules, if $M\oplus N$ is free, is then $M\oplus 0$ also free? My initial feeling is "no" because it would be isomorphic to $M$ then which need not be free.
Asked
Active
Viewed 106 times
0
-
It's very unclear what you are asking.... – Sep 27 '15 at 09:42
-
1It is because I typoed, i meant for it to be $M\oplus 0$ and not $M\oplus N$ again, my bad! – Zelos Malum Sep 27 '15 at 09:45
-
What are $M$ and $N$? Are they modules over some ring $R$? – Arthur Sep 27 '15 at 09:51
-
Correct, I forgot to add that vital information, it has been added, my apologise. – Zelos Malum Sep 27 '15 at 09:52
-
1This should help you answer your question. – RayX Sep 27 '15 at 09:58
1 Answers
2
RayX has linked an answer to your question in a comment, but I'll write it down here as well: Over the ring $\Bbb Z_6$, you have the two modules $\Bbb Z_2$ and $\Bbb Z_3$. Their direct sum $\Bbb Z_2\oplus\Bbb Z_3 \cong \Bbb Z_6$ is clearly free, but neither $\Bbb Z_2$ nor $\Bbb Z_3$ are free.
In general, modules $M$ for which there exist a module $N$ such that $M\oplus N$ is free are called projective. If there is a free module $N$ such that $M\oplus N$ is free, then $M$ is called stably free. There are projective modules that are not stably free (the example above is one such), and there are stably free modules that are not free (examples of that are usually a bit more complicated).
Arthur
- 199,419
-
so it was my gut feeling then as it was isomorphic, I pretty much knew this stuff about projective modules but I was wondering and almost hoping there was some hidden thing that could render it free as a submodule, ah well that's life! Thank you though. – Zelos Malum Sep 27 '15 at 10:11