0
$\begingroup$

Define $A:=F\{x,y\}/\langle xy-yx-1\rangle$ where $F\{x,y\}$ denotes the free algebra generated by $x,y$ and $F$ is a field. Let $I$ be a maximal left ideal of $A$, then $E=A/I$ is a simple $A$-module, could you explain me why?

1 Answers 1

3

By the Lattice Isomorphism Theorem, the submodules of $A/I$ are in one-to-one correspondence with the left ideals of $A$ that contain $I$.

Since $I$ is a maximal left ideal of $A$, what are the left ideals of $A$ that contain $I$?

This is completely independent of what $A$ is. If $A$ is any ring, and $I$ is any maximal left ideal of $A$, then the quotient module $A/I$ is a simple left $A$-module.