Past exam of the non-commutative rings course of the University of Oxford / 2023 / 3 (Filtered Weyl modules)

Use the Weyl algebra convention $[y_i,x_j]={\delta }_{ij}$, with $[x_i,x_j]=[y_i,y_j]=0$, where the brackets are commutators. Its order filtration of a Weyl algebra isThus $x_i$ has degree zero and $y_i$ degree one. By the ordered monomial basis of a Weyl algebra, the associated graded ring iswhere ${\xi }_i$ is the symbol of $y_i$.

The Rees ring of a filtered algebra is the graded ringwhere $t$ is a central indeterminate. Here it is generated by $x_i$, $Y_i=y_{i}t$, and $t$, with degrees $0,1,1$, respectively, and relationsIt interpolates between the associated graded ring and the original filtered ring:

A good filtration for the order filtration on $M$ is an increasing exhaustive family of complex vector subspaces $(M_j{)}_{j\in \mathbb{Z}}$, zero for sufficiently negative $j$, with $F_{i}R{M}_j\subseteq M_{i+j}$, such that the associated graded moduleis a finitely generated module over $A$. Equivalently, there are finitely many $m_1,\dots ,m_s$ and integer shifts $d_1,\dots ,d_s$ such that $M_j={\sum }_{\ell =1}^s(F_{j-d_{\ell }}R)m_{\ell }$. In particular, choose any finite generating family $m_1,\dots ,m_s$ of $M$ and putThis is a good filtration for the order filtration.

For $M\ne 0$, its dimension of a Weyl algebra module for the order filtration isHere support of a module, annihilator of a module, and Krull dimension are taken over the commutative polynomial ring $A$. The corresponding algebraic set is the order characteristic variety of a Weyl algebra module. This dimension is independent of the chosen good filtration for the order filtration; one may set $d(0)=-\infty$.

The pieces $F_{j}R$ generally have infinite complex dimension, since they contain all polynomials in the $x_i$. Thus a polynomial-growth definition based directly on ${\dim }_{\mathbb{C}}{M}_j$ would be inappropriate for this order filtration of a Weyl algebra.

For the submodule $N\subseteq M$, define its subspace filtration and the quotient filtration on the quotient module $Q=M/N$ byThey are compatible filtrations of a module, since multiplication by $F_{i}R$ sends $N_j$ into $N_{i+j}$ and $Q_j$ into $Q_{i+j}$.

In each degree, inclusion and the quotient map induceThe first map is injective: if $n\in N_j$ belongs to $M_{j-1}$, then $n\in N\cap M_{j-1}=N_{j-1}$. The second map is surjective by the definition of $Q_j$.

For exactness in the middle, a class $m+M_{j-1}$ lies in $\ker {\pi }_j$ precisely when $m\in M_j$ and $m+N\in Q_{j-1}$. This means $m=m^{\prime }+n$ for $m^{\prime }\in M_{j-1}$ and $n\in N$. Since $m,m^{\prime }\in M_j$, one also has $n\in N_j$, andConversely, every class from $N_j$ maps to zero in $Q_j/Q_{j-1}$. Taking the direct sum over all degrees gives the short exact sequence of graded modulesThe maps respect the $\mathrm{gr}R$-action because they come from module homomorphisms compatible with the filtrations of a module. Consequentlyas $\mathrm{gr}R$-modules. This proves exactness of associated graded modules for induced filtrations.

Furthermore $A=\mathrm{gr}R$ is a Noetherian ring, and $\mathrm{gr}M$ is a finitely generated module. Its submodule $\mathrm{gr}N$ and its quotient module $\mathrm{gr}(M/N)$ are therefore finitely generated modules. Hence the induced subspace filtration and quotient filtration are both good filtrations for the order filtration.

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 $J=R$.