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Suppose the projective dimension of a module $M$ is $n < \infty$. Does there exist a free $R$-module $F$ such that $\operatorname{Ext}^n(M, F) \not = 0$?

Can't we write the free module as a direct sum of $R$'s and then we just need to show $\operatorname{Ext}^n(M, R) \not = 0$? So does this work actually for any free module?

mathdragon
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  • If you show that $\operatorname{Ext}^n(M,R)\neq0$, then because there exists an $F'$ such that $F\cong R\oplus F'$ we have that $\operatorname{Ext}^n(M,F)\cong \operatorname{Ext}^n(M,R)\oplus \operatorname{Ext}^n(M,F')\neq0$. – Mariano Suárez-Álvarez May 15 '13 at 06:36
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    On the other hand, if $F\cong R^{(I)}$ is a direct sum of $|I|$ copies of $R$ for some set $I$, then it is not true in general that $\operatorname{Ext}^n(M,F)\cong\operatorname{Ext}^n(M,R)^{(I)}$, for $\operatorname{Ext}$ distributes only over finite direct sums in the second argument. – Mariano Suárez-Álvarez May 15 '13 at 06:39
  • I've proved somewhere on this site the following result: Let $R$ be a noetherian ring and $M\neq 0$ a finitely generated $R$-module with $\mathrm{pd}_R(M)=n$. Then $\mathrm{Ext}_R^n(M,R)\neq 0$. –  May 15 '13 at 08:27
  • @YACP, since in that case one can take f.g. projective modules in the resolution of $M$ and Ext is additive in the second argument, one can deduce that from my answer too. – Mariano Suárez-Álvarez May 15 '13 at 08:53
  • I couldn't find your answer when I looked for it before —if you recover it, please do add a link here! – Mariano Suárez-Álvarez May 15 '13 at 09:06
  • @MarianoSuárez-Alvarez It wasn't easy to find it, but finally I've got it: http://math.stackexchange.com/questions/210304/modules-with-projective-dimension-n-have-not-vanishing-mathrmextn –  May 15 '13 at 09:51
  • @YACP, searching is a pain way too often :-/ – Mariano Suárez-Álvarez May 15 '13 at 16:52

1 Answers1

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If you ask the question

if $M$ is an $R$-module of projective dimension $n$, is $\operatorname{Ext}^n(M,R)\neq0$?

then the answer is surpring: it depends. When $R=\mathbb Z$, this is essentially known as Whitehead's problem and Shelah proved that its answer depends on the specific set theory that you choose. Indeed, he showed that depending on the background set theory, there exist abelian groups $M$ which are not free (so that their projective dimension is $1$) such that $\operatorname{Ext}^1(M,\mathbb Z)=0$, or not.

$$\bullet\bullet\bullet$$

For the original question in which you look just for a free module:

Suppose the projective resolution of $M$ ends with $$\cdots\leftarrow P_{n-1}\xleftarrow{\hskip2ex d\hskip2ex }P_n\leftarrow 0$$ and suppose $P_n$ is a direct summand of the free module $F$. Applying the functor $\hom(\mathord-,F)$ we get a complex ending with $$\cdots\to\hom(P_{n-1},F)\xrightarrow{\hskip2ex d^*\hskip2ex }\hom(P_n,F)\to0$$ $\operatorname{Ext}^1(M,F)$ is the cokernel of the map $d$. To show it is not zero, it is enough to show that $d$ is not surjective.

Now let $i:P_n\to F$ be the inclusion and let $j:F\to P_n$ be a retraction of $i$, so that $ji=1_{P_n}$. The element $i\in\hom(P_n,F)$ is not in the image of $d$. Otherwise we'd have a map $r:P_{n-1}\to F$ such that $rd=i$, so composing with $j$ we'd have $jrd=ji=1_{P_n}$. It follows that $jr:P_{n-1}\to P_n$ is a retraction for the map $d$. This is impossible, because the projective dimension of $M$ is exactly $n$.

Mariano Suárez-Álvarez
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  • Interesting. So then trying to show $Ext^n(M, R)$ is not zero is probably the wrong way to go. I just want to show there is some free module $F$ where $Ext^n(M, F) \not = 0$. – mathdragon May 15 '13 at 06:53