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

Use the quotient filtration induced from the order filtration of a Weyl algebra. The key step is to identify the entire associated graded module of the left ideal $I$, rather than merely take symbols of its displayed generators.

For every member $x^{\alpha }{y}^{\beta }$ of the ordered monomial basis of a Weyl algebra, right multiplication by the pure $y$-monomials givesNo commuting past an $x_i$ is necessary. It follows that $I$ is exactly the complex linear span of the basis monomials satisfying eitherIntersecting this linear span with each filtration piece and then taking symbols therefore yields the monomial idealBy b,

To determine its Krull dimension, compute the radical of an ideal:Indeed, $K$ lies in the displayed intersection, an intersection of prime ideals. Conversely, $({\xi }_1{\xi }_2{)}^3\in K$ and $({\xi }_1{\xi }_3{)}^3\in K$, so both generators of that intersection lie in $\sqrt{K}$. The equality of the intersection with $({\xi }_1{\xi }_2,{\xi }_1{\xi }_3)$ follows by factoring out ${\xi }_1$ and reducing modulo $({\xi }_2,{\xi }_3)$.

The order characteristic variety of a Weyl algebra module in this case has two irreducible components:They are coordinate affine spaces of dimensions $2n-1$ and $2n-2$, respectively, in the $2n$ coordinates $x_1,\dots ,x_n,{\xi }_1,\dots ,{\xi }_n$. The Krull dimension is the larger of these component dimensions. ThusIn particular, two displayed generators do not force a drop of two in dimension: their common factor creates the component ${\xi }_1=0$ of dimension $2n-1$.

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 $J=R$.