Let $(R,\mathfrak m)$ be a commutative Noetherian local ring. Let $A$ be a finitely generated associative $R$-algebra. Let $x\in \mathfrak m$ be a non-zero-divisor on $A$ such that $xA\neq A$. If $A/J(A)$ is a division ring, then is $\dfrac{A/xA}{J(A/xA)}$ also a division ring ?
Here, $J(-)$ denotes Jacobson radical.
Of course, I would be done if $x\in J(A)$ and $J(A/xA)=J(A)/xA$. Is that true?
The analogue of "A commutative ring $R$ is local iff $R/J(R)$ is a field." is "A ring $R$ is local (has a unique maximal right ideal) iff $R/J(R)$ is a division ring." Of course by correspondence, the quotient of a (possibly noncommutative) local ring is again local. So yes, $A/xA$ is local because $A$ is local.