A right $R$-module short exact sequence $\xi:0\rightarrow A \rightarrow B \rightarrow C\rightarrow 0$ is called pure if $\xi \otimes M$ is also a short exact sequence for arbitrary left $R$-module $M$.

Question: if $\xi \otimes R/I$ is exact for arbitrary left ideal $I$, is $\xi$ pure exact sequence?

I found this question here, but the question hasn't been solved.

Maybe this need the property: $\xi $ is pure exact if and only if $\mathrm{Hom}_{\mathbb Z}(\xi,\mathbb {Q/Z})$ is split. I don't know how to do. Thank you in advance!


The answer is no: In general this is not enough information to conclude $\xi$ is pure exact. This question is discussed in detail in T.Y. Lam's book "Lectures on Modules and Rings", at the end of Section 4. I recommend reading the presentation there, but I can summarize the process of obtaining a counterexample.

Start with a nonzero commutative ring $R$ for which the family $\mathcal{F}$ of all nonzero ideals has trivial intersection. Then set $A = R$ and $B=\prod_{I \in \mathcal{F}} R/I$, with $i : A \rightarrow B$ the natural map. Then $i$ is injective, and one shows that it remains injective after tensoring with any $R/I$ (4.95 in LMR). So the sequence $\xi: 0\rightarrow A \rightarrow B \rightarrow B/A \rightarrow 0$ satisfies your assumptions.

Next, one can show that $i \otimes_R N$ is not injective when $N$ is any finitely presented left $R$-module with $\bigcap_{I\in \mathcal{F}} IN \neq 0$ (4.96 in LMR).

For an explicit counterexample Lam applies the above to $R=k[x,y]/(x^2,xy,y^2)$ over a field $k$, where $N=k^3$ with $x$ and $y$ acting via $x(a,b,c) = (0,0,a)$ and $y(a,b,c) = (0,0,b)$. (Note $N$ is the indecomposable injective $R$-module.)

Finally, Lam notes that any $\xi : 0\rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ with $\xi \otimes_R R/I$ exact for all ideals $I$ and $B$ projective is pure exact. This is a result of Fieldhouse, which is also included as an exercise.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.