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Let $R$ be a ring and $M$ be a $R$-module. Is it true that $ \bigcap \limits _{f \in \hom(M,R)} \ker f = 0$ ? If not, are there (necessary and) sufficient conditions to be put on $R$ and $M$ guaranteeing an affirmative answer?

Intuitively, but not rigorously, I would consider $0 \ne x \in \bigcap \limits _{f \in \hom(M,R)} \ker f$ and write $M$ as an internal direct sum of $Rx$ and an algebraic complement. Then I would consider the "projection onto $Rx$" $f(rx + y) = r$ and notice that $f(x)=1$ but, on the other side, $f(x)=0$, a contradiction which would imply $x=0$. The main problem here is the existence of that algebraic complement ($M$ could be an indecomposable module, or even a simple one). If $R$ is a commutative field, this proof seems to hold, but this is a very restricted setting.

Of course, one may come up with a different proof approach, one that does not stumble where mine does.

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    You cannot write $M$ as an internal direct sum of two submodules. In general, this is only true when $R$ is semi-simple (e.g. a field). Your property doesn't even hold in the commutative case. Try $M:= \mathbb Z/2\mathbb Z$ and $R := \mathbb Z$. – 2017-02-15
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    @user218931: I know that, read the end of my question. I am looking for at least clear hypotheses under which that can be done. In addition, notice that my approach to the proof should not be confused with the problem itself: there may be other proof approaches working in situations in which my approach wouldn't. – 2017-02-15
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    Yes, sorry that I didn't read carefully. But my example remains, since you asked, if it holds if $R$ is an integral domain. Also, the condition on $R$ being semi-simple is a partial answer. Note that your property holds for all free $R$-modules and that it is closed under taking submodules. – 2017-02-15
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    [Torsionless modules.](https://en.wikipedia.org/wiki/Torsionless_module) – 2017-02-15

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No, this is certainly not true in general. Let $R$ be a ring, and $M$ a (left) $R$-module. The dual module of $M$ is defined by the right $R$-module $M^* = \mathrm{hom}_R(M,R)$. By repeating this procedure one obtains the double dual module $M^{**} = (M^{*})^{*}$ (which is a left $R$-module). One has the canonical homomorphism $c_M\colon M\to M^{**}$ sending $x$ to the functional $f\mapsto f(x)$ on $M^*$.

Your condition is equivalent with the statement that $c_M$ is injective. This is not true in general! One can come up with examples where $M\neq 0$ and $M^* = 0$ (e.g., take $R = \mathbf Z$ and $M = \mathbf Q$). Since dualizing is an additive notion, one can also easily construct examples where $M^*\neq 0$ and $c_M$ not injective. Modules for which $c_M$ is injective are also known as torsionless or semi-reflexive modules.

If you search for it you'll find some other equivalent formulations of this notion, and also sufficient conditions implying it.