The injective hull of the ring of all eventually constant sequences of elements of $\mathbb{Z}_2$

Question

Let $R$ be the ring of all eventually constant sequences $(x_n)_{n\in \mathbb{N}}$ of elements of $\mathbb{Z}_2$. It's known that the injective hull $E(R_R)$ of $R_R$ is $S:=\prod_{n\in \mathbb{N}} \mathbb{Z}_2$. I need to check this myself. I know that $S_S$ is injective. But how can I prove that $S_R$ (i.e., $S$ as a right $R$-module) is injective ?!.

Answer 1

It suffices to show each $\mathbb Z_2$ is an injective $R$ module, since arbitrary products of injective modules are injective.

$R$ is commutative von Neumann regular (actually it is even a boolean ring) so its simple modules are injective. (Rings whose simple modules are injective are called V rings and their equivalence in commutative rings with VNR rings is not hard.)

Finally then, each factor of $\mathbb Z_2$ is given an $R$ module structure with the given module action (which is suggested to be coordinatewise). Concretely, each one looks like $e_iS$ where $e_i\in R$ is the element with $1$ on the $i$th index and zero elsewhere. These simple modules are mutually nonisomorphic, since they have distinct annihilators in $R$. But all of them only have two elements, so they can only be simple.