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In the exercise:

For every $R$-module $A$, show that $pd(A)=n$ implies $Ext_R^n(A, R) \neq 0.$ It is true for every $R$-module $A$ ? I think that $A$ should be finitely generated.

Napo Scott
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  • Hm, I can only prove it in the finitely generated case thus far, and I really need the biproduct. – Kevin Carlson Nov 16 '14 at 06:28
  • In fact, there is a free $R$-module $F$ such that $Ext_R^n(A, F)\neq 0$ – Napo Scott Nov 16 '14 at 07:00
  • I don't understand whether you're claiming you've solved the problem, extending the statement, or what. Certainly if the original statement holds, then your statement does! – Kevin Carlson Nov 16 '14 at 07:06
  • I can do only when $ A $ is finitely generated. – Napo Scott Nov 16 '14 at 07:31
  • On a different topic: please do not crosspost. I saw the same question posted to MathOverflow (where it is now closed) -- the problem with crossposting is that it leads to duplication of effort, hence a waste of time for busy professionals. Thanks. – user43208 Nov 17 '14 at 20:04

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Napo, I agree: it looks as though a hypothesis is missing. You are certainly right that the claim is provable if we add the condition of finite generation.

There is a problem with the exercise even in a very classical context: if we take $R = \mathbb{Z}$ and $n=1$, then the condition $\text{pd}(A) = 1$ is equivalent to $A$'s being nonprojective or nonfree. So even in this special case the exercise asks us to prove that $A$ nonfree (as a $\mathbb{Z}$-module) implies $\text{Ext}^1(A, \mathbb{Z}) \neq 0$. This is equivalent to the http://en.wikipedia.org/wiki/Whitehead_problem which is famously undecidable in ZFC.

If this is an exercise in a course and there are indeed no missing hypotheses, then you should bring this to the attention of your instructor, and perhaps consider also contacting the author of the textbook.

user43208
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  • Well, you have an answer. If this answer was helpful to you, then please consider upvoting and accepting (there's a box you can check for accepting an answer). Thanks. – user43208 Nov 18 '14 at 12:44