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Let $M$ be any $R$-module.

We can construct a free resolution as follows:

Let $M_1$ be the free module with basis in one to one correspondence with generators of $M$. Then the map $\phi:M \to M_1$ is necessarily surjective. Let $M_2 = \operatorname{ker}(\phi)$.

I don't see how the sequence : $0 \to M_2 \to M_1 \to M \to 0$ fails to be a finite resolution of $M$?

I am claiming that the map $M_2 \to M_1$ is the inclusion map and thus has kernel $0$.

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    $\begingroup$ Why do you expect the kernel to be free? $\endgroup$ – Moishe Kohan Sep 15 '16 at 1:03
  • $\begingroup$ For free groups any subgroup is free thus the kernel. I thought a similar result would hold for free submodules. $\endgroup$ – user7090 Sep 15 '16 at 1:05
  • $\begingroup$ this is true for some rings but not for others. $\endgroup$ – Moishe Kohan Sep 15 '16 at 1:06
  • $\begingroup$ Ok. Interesting. Eisenbud gives an example of constructing a free resolution via the kernel although he did this in the case that $R$ is graded. $\endgroup$ – user7090 Sep 15 '16 at 1:08
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If $I$ is a nonprincipal ideal of a ring, then $0\to I\to R\to R/I\to 0$ is a counterexample.

If $I$ were free and generated by more than one basis elements $u,v$, then $uv -vu=0$ contradicts the basis property.

In fact, the property of having all right ideals free in a commutative ring characterizes PID's. That is why it works for Abelian groups (since they are just $\mathbb Z$ modules.)

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