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  • Cotorsion modules

A module M is called cotorsion if for all flat modules X, $Ext_R^1(X,M)=0$ .

  • Strongly cotorsion modules

M is called strongly cotorsion if for all modules X of finite flat dimension, $Ext_R^1(X,M)=0$ .

  • Questions

By definitions it is easy to observe that strongly cotorsion modules are cotorsion modules. But I fail to find some examples to show the two classes are different. Furthermore, what happens when the Ext functor is replace by the Tor functor?

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    $\begingroup$ If you replace $Ext$ by $Tor$ then any module will be cotorsion, and any flat module will be strongly co-torsion (and it will be if and only if over regular rings), but many other modules will not be strongly co-torsion. $\endgroup$
    – the L
    Commented Jul 29, 2012 at 9:23
  • $\begingroup$ @Liran: nevermind. Lemma 4.1.10 (page 94) in Weibel's Introduction to Homological Algebra says that the flat dimension of $X$ is $\leq d$ iff $Tor_R^{d+1}(X,M)=0$ for all $M$. So $X$ is flat iff $fd(X)=0$ iff $Tor_R^1(X,M)=0$ for all $M$, i.e. your comment is correct. My comment above shows that whenever $M$ is flat, $Tor_R^1(X,M)=0$. However, I suspect the class of modules for which $Tor(finite fd,-)=0$ is more than just flat modules $\endgroup$
    – David White
    Commented Jul 29, 2012 at 13:22
  • $\begingroup$ @David, that's very basic, a right $R$-module $M$ is flat iff $M\otimes_R-$ is exact iff $\operatorname{Tor}^1(M,-)=0$ iff $\operatorname{Tor}^n(M,-)=0$ for all $n>0$. $\endgroup$
    – Fernando Muro
    Commented Jul 29, 2012 at 13:31
  • $\begingroup$ @Liran, David, Fernando, thanks for your comments! $\endgroup$
    – TmobiusX
    Commented Jul 30, 2012 at 2:18
  • $\begingroup$ @Fernando: I was more interested in Liran's claim about strongly cotorsion modules. I wasn't really aware of this result about Tor into a flat module until I found that reference in Weibel. Knowing $X$ has finite flat dimension doesn't tell me $Tor^1(X,M)=0$. I'll bet one can get away with a lesser hypothesis on $M$ than flat in order to conclude this. $\endgroup$
    – David White
    Commented Jul 30, 2012 at 2:19

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Let me raise an example:

Let $R = k[[t]]$, $k$ be a field. $R$ is a regular ring as well as a Dedekind ring.

  1. $R$ is regular hence every $R$-module $X$ has finite flat dimension. Thus if $M$ is strong cotorsion, then $Ext^1_R(X,M) = 0$ for all $X$, so $M$ is injective.
  2. By (1) we have $k$ is not strong cotorsion. We shall show that $k$ is cotorsion.
  3. Since $R$ is Dedekind we have $X$ is flat iff $X$ is torsion-free (see http://en.wikipedia.org/wiki/Flat_module). So we have a short exact sequence $$0 \to X \to X \to X/tX \to 0.$$ The above exact sequence induces the following exact sequence $$Ext^1_R(X,k) \overset{t}{\to} Ext^1_R(X,k) \to Ext^2_R(X/tX,k) \cong 0$$ the isomorphism $Ext^2_R(X/tX,k) \cong 0$ follows from $R$ is a regular ring of dimension one. But $Ext^1_R(X,k) \overset{t}{\to} Ext^1_R(X,k)$ is a zero homomorphism. So $Ext^1_R(X,k) \cong 0$. The proof is complete
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