Let $A$ be an algebra over some field $\mathbb F$ and $E$ a (left) $A$-module. Write $$ A\cdot E:=\{a\cdot x:a\in A\wedge x\in E\} $$ Say $E$ is non-trivial iff $A\cdot E\neq 0$. Say $E$ is simple if it is non-trivial and the only submodules of $E$ are $0$ and $E$.
Now, let $M$ be a maximal (hence proper) left ideal in $A$. I'd like to show that $A/M$ is a simple left $A$-module.
I can prove the second half of the definition, but it's not obvious to me that $A/M$ is non-trivial. This is the same as finding $x,y\in A$ such that $xy\not\in M$, so we'd be done if we can show that $M$ is prime (or rather that $xy\in M\to x\in M\vee y\in M$), but we don't know that $A$ is commutative, which is usually relevant for that proof. We'd also be done if $M$ is modular (and the case I'm interested in is when $A$ is commutative and $M$ is modular), but I'd like to know if things still work in this generality.
For reference, this is a remark made by Dales in his Banach Algebras and Automatic Continuity book (page 61).
(I'm assuming that you don't require $A$ to be unital since that case is obvious)
This is not true. Consider the $\Bbb F$-algebra given by the ideal $A:=(x)$ inside $\Bbb F[x]$ and let $M=A^2=(x^2)$. Then $M$ is maximal in $A$, but $A \cdot (A/M) = 0$.