Consider the Jacobson algebra $R =k \langle x, y\rangle /(xy − 1)$, where $k$ is a field. Let $k^ω$ be the R-module whose elements are infinite sequences in $k$, with $y$ acting as the right shift operator, and $x$ the left shift operator. I want to show that the submodule $_{R}M ≤ k^ω$ of eventually zero sequences is a simple $R$-submodule.
To begin with, Let $N\le M$ be a nonzero left submodule and take $z\in N-\{0\}$. Then $z=(z_1,z_2,\dots)$ is an eventually zero sequence and $Rx$ is a left $R$-module (and left ideal). We only need to show $Rz=_{R}M$.
Can anyone show me how to continue from here? Thanks for help.
A nonzero module $M$ is simple if and only if, for every $z\in M$, $Rz=M$.
Let $z\ne0$, $z\in M$. Then we can assume the last nonzero term in $z$ is $1$. Then with left shifts we can say that the sequence $(1,0,0,\dotsc)\in Rz$. Hence, for every $a\in k$, $(a,0,0,\dotsc)\in Rz$. With right shifts we can move $a$ wherever we please.
An eventually zero sequence is a (finite) sum of sequences of this kind.