Is a submodule of a free module also free ? For me it looks natural that yes, but in my course it's written that it's only true for a module over a PID and I don't really understand why. Any example ?
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No: Take a projective module that isn't free, for example. – anomaly Aug 17 '18 at 13:31
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I don't know what it is. If we consider $\mathbb R[X_1,X_2,....]$ aver itself, does $(X_1,X_2;...)$ work ?@anomaly – user352653 Aug 17 '18 at 13:34
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@user352653: https://en.wikipedia.org/wiki/Projective_module – anomaly Aug 17 '18 at 13:35
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It's not only true for a module over a PID. You have just been told it is true for a module over a PID. It is also true for some other rings, like hereditary local rings which aren't PIDs. – rschwieb Aug 17 '18 at 15:22
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A counter-example
If $K$ is a field, consider the ring of polynomials in two indeterminates $R=K[X,Y]$. The ideal $I=(X,Y)$ is a submodule of the rank $1$ free $R$-module $R$. It is not free, since if it were, it would have a single generator. Can you find a polynomial $F(X,Y)$ which is a divisor of both $X$ and $Y$?
Bernard
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I just wanted to add for those wondering: here it is used that a submodule $M\subset R^n$ that is free, has rank $k\leq n$. Robin Chapman gives an elementary proof of this fact here:
https://mathoverflow.net/questions/30860/ranks-of-free-submodules-of-free-modules
+1 For the nice counter-example, because it generalises well: Any ring is free over itself of rank $1$, so any ideal that has more than one generator, works.
– Sha Vuklia Jun 22 '23 at 14:36
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Small (smallest?) counterexample: $(2)\subseteq \Bbb Z_4$ as $\Bbb Z_4$-modules.
Arthur
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but $(2)$ is generated by $2$, no ? So it's a module of 1 dimension ? – user352653 Aug 17 '18 at 13:37
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@user352653 It has only one generator, but ${2}$ isn't a basis, since you don't have unique linear combinations to make elements. For instance, $2 = 1\cdot 2 = 3\cdot 2$. Thus the module isn't free. – Arthur Aug 17 '18 at 13:45