Let $A$ be a ring (we don't assume that $A$ is commutative), and $\frak a$ a left ideal of $A$. Let $(E_\lambda)$ be a system of left $A$-modules. There is a natural homomorphism of modules $$ \phi :\mathfrak{a} \prod E_\lambda \to \prod \mathfrak{a} E_\lambda $$ and $\phi$ is injective. I would like to prove that if $\frak a$ is finitely generated, then $\phi$ is onto. Under the additional assumption that $A\mathfrak{a}=\mathfrak{a}A$ this is easy but otherwise I did not succeed.
Can you help? Thanks.
In fact, the result doesn't hold without additional assumption. If $M$ is the free semigroup over some set $X$ and $A=\mathbf{Z}[M]$ then it is easy to construct a counter-example. If we ask that $\frak a$ be generated by some finite central system, then the result holds with easy proof.