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 $M$.



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> Question: if $\xi \otimes R/I$ is exact for arbitrary left ideal $I$, is $\xi$ pure exact sequence?
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I found this question [here](https://math.stackexchange.com/questions/2452775/f-otimes-id-m-otimes-a-i-rightarrow-n-otimes-a-i-is-injective-for-every), 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!



$R/I\otimes$ pure exact sequence
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.


