Past exam of the non-commutative rings course of the University of Oxford / 2023 / 3 / c / ii

In fact $J=R$, so the quotient module is zero. This requires an argument using left ideals: a left ideal is not generally closed under multiplication on the right.

Write $x=x_1$, $y=y_1$, so that $yx-xy=1$. The commutator product rule givesBoth $y^2{x}^2$ and $x^2{y}^2$ lie in $J=R{x}^2+R{y}^2$, so their differencebelongs to $J$. Multiplying on the left and subtracting another known element of the left ideal givesBecause the scalar $6$ is invertible in $\mathbb{C}$, it follows that $y\in J$, and hence $xy\in J$. Therefore $h-4xy=2\in J$, giving $1\in J$. We concludeThus the answer is yes for every $n\ge 1$. All ideal operations in the proof use left multiplication. In particular, simply replacing $J$ by the commutative ideal $(x_1^2,{\xi }_1^2)$ of the associated graded ring would miss the lower-order consequences that force $J=R$.