suppose that $R$ be a domain, $M$ a torsion-free $R$-module and $V=E(M)$, the injective envelope of $M$. Is it true that if $M$ is torsion-free then $V$ is torsion-free?
I guess it is true because $V$ is divisible, (since it is injective), and so torsion-free. Is it necessary for $M$ to be torsion-free?
If $M$ has some torsion, then $V$ has some torsion. This is nothing surprising, since we have an embedding $M \subseteq V$.
On the other hand, suppose that $M$ is torsion-free and, for contradiction, that $x \in V$ is a torsion element, i.e. nonzero such that it is killed by some nonzero $r \in R$. Then $M \cap Rx =0$ (every nonzero element of $Rx$ is torsion), which contradicts the fact that $M$ is an essential submodule in $V$ (meets every non-trivial submodule non-trivially). Thus, no such element can exist, i.e. $V$ is torsion-free as well.