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Let $R$ be a commutative ring with identity and $M$ a $R$-module. I'm interested in under what condition we can find an element $m\in M$ satisfing that $i:M\to M\otimes_R M, x\mapsto x\otimes m$ is an epimorphism.

I've found out an easy sufficient condition: there exists $m\in M$ such that, for every $n\in M$, we can find $m'\in M, r_1,r_2\in R$ that satisfy $r_1m'=n$ and $r_2m'=m$ (roughly speaking, every $n\in M$ can be expressed as $\frac{r_1}{r_2}m$, it's also equal to say that every finitely generated submodule of $M$ is contained in a cyclic submodule). But I'm not sure if this condition is necessary, too.

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  • $\begingroup$ But is this condition ever satisfied? $\endgroup$ – Bernard Jun 1 '15 at 11:30
  • $\begingroup$ @Bernard Yes, I found a sufficient condition to guarantee it as I posted above. $\endgroup$ – Censi LI Jun 1 '15 at 11:35
  • $\begingroup$ @Bernard A simple case is $M = R$ and $m = 1$. $\endgroup$ – Najib Idrissi Jun 1 '15 at 11:35
  • $\begingroup$ Tensoring by $R$ is trivial. I meant satisfied in non-trivial cases. $\endgroup$ – Bernard Jun 1 '15 at 11:48
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    $\begingroup$ @Censi LI: I don't know. I was looking for a finitely generated, locally cyclic $R$-module. $\endgroup$ – Bernard Jun 1 '15 at 12:15
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Some thoughts about this situation:

If $M$ is finitely generated, no homomorphism from $M$ to $M \otimes_R M$ can be an epimorphism, unless $M$ is locally cyclic.

Indeed, if there were such an epimorphism, consider a prime ideal $\mathfrak p$ and its residual field $k=R_{\mathfrak p}/\mathfrak p R_{\mathfrak p}$. Tensoring with $k$ over $R$, there would result an epimorphism: $$M\otimes_R k\to (M\otimes_R M)\otimes_R k\simeq (M\otimes_R k)\otimes_k(M\otimes_R k)$$ As $\dim_k\bigl((M\otimes_R k)\otimes_k(M\otimes_R k)\bigr)=\bigl(\dim_k(M\otimes_R k)\bigr)^2$, this implies $\dim_k(M\otimes_R k)=0$ or $1$.

Now $M\otimes_R k\simeq M_{\mathfrak p}/\mathfrak pM_{\mathfrak p}$, Nakayama's lemma implies $ M_{\mathfrak p}$ is a cyclic $R_{\mathfrak p}$-module.

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  • $\begingroup$ It seems that locally cyclic is also sufficient if $M$ is finitely generated, so when $R$ is a Dedekind domain, all ideals of $R$ satisfy the property. What do you think about the infinitely generated case when $R=\mathbb Z$? $\endgroup$ – Censi LI Jun 1 '15 at 20:54
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The sufficient condition you described is not necessary. in fact, since $\mathbb Z_{p^{\infty}}\otimes\mathbb Z_{p^{\infty}}=0$, for any index set $I$, $\mathbb Z_{p^{\infty}}^{(I)}$ meets your requirement.

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