Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 2 / 6 / Solution

Fix a finite-dimensional nonnegative filtration of a ring giving $R$ its almost commutative algebra structure, and choose a good filtration of a module on $M$. When $\mathrm{gr}R$ is a standard graded algebra, the Hilbert-Serre theorem makes the cumulative Hilbert function $h_M(j)={\dim }_k{F}_{j}M$ eventually a polynomial. For $M\ne 0$, define the dimension of a filtered module and multiplicity of a filtered module byUse $d(0)=-\infty$ and $m(0)=0$. With positive weighted generators instead of a standard grading, the cumulative Hilbert function is eventually a quasipolynomial with a common positive leading coefficient across residue classes. Equivalently, $h_M(j)=c{j}^d+O(j^{d-1})$, and the definitions are $d(M)=d$ and $m(M)=d!c$. The common leading coefficient follows from monotonicity: interlacing consecutive residue classes forces their leading coefficients to agree. For a standard grading, multiplicity is a positive integer, since an integer-valued polynomial has an integral expansion in the binomial polynomials; weighted normalization can instead give a positive rational number.

Two good filtrations of a module for this fixed algebra filtration bound each other after fixed index shifts. The corresponding Hilbert functions therefore satisfy $h_F(j-a)\le h_G(j)\le h_F(j+b)$ for fixed $a,b$. Comparing their leading growth shows that the degree and leading coefficient agree. Thus $d(M)$ and $m(M)$ are independent of the module filtration; the fixed algebra filtration is part of the multiplicity convention.

For a submodule $N$, use its subspace filtration and the quotient filtration on $M/N$. These are good because $\mathrm{gr}R$ is Noetherian. The short exact sequence of filtered pieces givesNonzero leading coefficients are positive, soIf $d(N)=d(M/N)$, those leading coefficients add andThis is dimension and multiplicity in a filtered exact sequence.

For the Weyl algebra $A_n(k)$ in characteristic zero, Bernstein inequality for Weyl algebra modules asserts $d(M)\ge n$ for every nonzero finitely generated module $M$. Write its generators as $x_i,D_i$, with $[D_i,x_j]={\delta }_{ij}$ and all other generator commutators zero. The ordered monomials $x^{\alpha }{D}^{\beta }$ give the Bernstein filtrationTake a finite-dimensional generating subspace $V\ne 0$ of $M$ and put $M_j=A_{j}V$.

To prove faithful finite-step action of a Weyl algebra, consider $0\ne a\in A_j$ and choose a nonzero coefficient $c_{\alpha \beta }$ of a monomial of maximal total degree $d\le j$. Applying ${\prod }_i(\mathrm{ad}{D}_i{)}^{{\alpha }_i}(\mathrm{ad}{x}_i{)}^{{\beta }_i}$ to $a$ givesIndeed, these commutators differentiate the ordered coefficients: lower-degree monomials vanish, and another degree-$d$ monomial can survive only if all its exponents dominate $(\alpha ,\beta )$, which forces equality. This is scalar extraction by iterated Weyl commutators.

Expanding these iterated commutators writes $\lambda ={\sum }_{\ell }{u}_{\ell }a{v}_{\ell }$ with $u_{\ell },v_{\ell }\in A_j$. If $a$ annihilated $M_j$, every summand would annihilate $V$, contrary to $\lambda V\ne 0$. Therefore the action map is injective:Taking dimensions and comparing polynomial degrees giveswhich proves Bernstein inequality for Weyl algebra modules. The polynomial representation $k[x_1,\dots ,x_n]$ has dimension $n$, so the bound is sharp.

The associated graded ring of the Weyl algebra is a polynomial ring, so ascending filtered-graded transfer of Noetherianity makes $A_n(k)$ a left Noetherian ring. Thus every submodule of a finitely generated module is finitely generated. If $d(M)=n$, every nonzero subquotient has dimension at most $n$ by the exact-sequence formula, and at least $n$ by Bernstein inequality for Weyl algebra modules. Every such subquotient therefore has dimension $n$ and positive integral multiplicity.

In a strictly descending chain $M=M_0⊋M_1⊋\cdots$, each nonzero quotient $M_i/M_{i+1}$ consumes at least one unit of multiplicity, by dimension and multiplicity in a filtered exact sequence. Hence there are at most $m(M)$ strict steps. This proves the descending chain condition, so $M$ is Artinian. In fact this holonomic Weyl algebra module has module length at most $m(M)$, as in multiplicity bounds the length of a holonomic Weyl module.