= Solution
{c}
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> gives
$$
yx^2=x^2y+2x,\qquad y^2x^2=x^2y^2+4xy+2.
$$
Both $y^2x^2$ and $x^2y^2$ lie in $J=Rx^2+Ry^2$, so their difference
$$
h=y^2x^2-x^2y^2=4xy+2
$$
belongs to $J$. Multiplying on the left and subtracting another known element of the <left ideal> gives
$$
yh-4xy^2=4(yx-xy)y+2y=6y\in J.
$$
Because 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 conclude
$$
\boxed{R/J=0,\qquad\dim_{\mathbb C}(R/J)=0.}
$$
Thus the answer is yes for every $n\geq1$. 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 $J=R$.
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