Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 85 / 6

This part proves $(c)\Rightarrow (a)$. Write the Weyl algebra generators as $x_1,\dots ,x_n,D_1,\dots ,D_n$, with $[D_i,x_j]={\delta }_{ij}$ and the other generator commutators zero. The ordered monomials $x^{\alpha }{D}^{\beta }$ form a basis over any field: one may construct the algebra by successively adjoining the commuting derivations $D_i$ to $k[x_1,\dots ,x_n]$, with multiplication $D_{i}f=f{D}_i+\partial f/\partial x_i$, obtaining unique ordered forms. This also makes $A_n(k)$ infinite-dimensional for $n\ge 1$.

Assume $\mathrm{char}k=0$ and let $I$ be a nonzero two-sided ideal. Choose a nonzero element $a=\sum c_{\alpha \beta }{x}^{\alpha }{D}^{\beta }\in I$, and a nonzero term of maximal total degree, say $({\alpha }_0,{\beta }_0)$. The relations giveApply $(\mathrm{ad}D{)}^{{\alpha }_0}(\mathrm{ad}x{)}^{{\beta }_0}$ to $a$. Every lower-degree term vanishes. A term of the same degree survives only if all its exponents dominate $({\alpha }_0,{\beta }_0)$, which forces equality. Thus the resulting element of $I$ isThe factorials are nonzero in characteristic zero. This is a nonzero scalar, so $1\in I$ and $I=A_n(k)$. Therefore $A_n(k)$ is simple in characteristic zero. This scalar extraction by iterated Weyl commutators supplies the full simplicity proof.

This part proves $(a)\Rightarrow (b)$. If $M\ne 0$ were a finite-dimensional unital module, its action would give a homomorphismThe kernel is a two-sided ideal and is proper because the identity acts as the identity on $M$. A simple ring therefore makes the homomorphism injective. This is impossible: $A_n(k)$ has the infinitely many linearly independent ordered monomials established in part (a), whereas the endomorphism algebra has dimension $({\dim }_{k}M{)}^2$. Thus simplicity excludes every nonzero finite-dimensional module.

There is also a direct characteristic-zero check: the relation $[D_1,x_1]=1$ acts as a matrix identity. Taking its trace gives $0={\dim }_{k}M$ in $k$. In characteristic zero this forces $M=0$. This agrees with the Weyl algebra has no nonzero finite-dimensional modules criterion, but the injectivity argument proves the requested implication without presupposing characteristic zero.

For the contrapositive of $(b)\Rightarrow (c)$, suppose $\mathrm{char}k=p>0$. ConsiderLet $x_i$ act by multiplication and $D_i$ by formal differentiation. Differentiation preserves the defining ideal, since $D_j(x_i^{p}f)=x_i^p{D}_{j}f$ in characteristic $p$. The relations $[D_i,x_j]={\delta }_{ij}$ hold before quotienting and hence hold on $M$. Its basis consists of the monomials $x^{\alpha }$ with $0\le {\alpha }_i<p$, so ${\dim }_{k}M=p^n>0$. This is the explicit finite-dimensional module for the Weyl algebra in positive characteristic.

Therefore condition (b) fails in positive characteristic and implies characteristic zero. Combining this with the preceding two implications proves $(a)⟺(b)⟺(c)$. The same representation also has nonzero proper kernel, since the infinite-dimensional Weyl algebra cannot inject into a finite-dimensional endomorphism algebra, confirming its failure of simplicity in positive characteristic.

The Bernstein inequality for Weyl algebra modules requires characteristic zero: for a nonzero finitely generated left or right module $M$ over $A_n(k)$,Equivalently, its Bernstein growth dimension of a Weyl algebra module is at least $n$. Without the characteristic-zero hypothesis this is false, as the module in part (c) has dimension zero in the growth sense. The proof below works over any characteristic-zero field, not just an algebraically closed one.

For a left module, choose a nonzero finite-dimensional generating subspace $M_0$ and put $M_j=F_j{A}_n{M}_0$, where $F_j$ is the Bernstein filtration. Ordered monomials giveMultiplication defines the linear action mapWe prove this map injective by induction on $j$, which is the essential faithful finite-step action of a Weyl algebra argument. For $j=0$, an element is a scalar, and a scalar killing $M_0\ne 0$ is zero. Suppose $a\in F_j$ kills $M_j$. For any generator $z$ and $m\in M_{j-1}$, both $azm$ and $zam$ vanish because $zm\in M_j$ and $m\in M_j$. Thus $[a,z]\in F_{j-1}$ kills $M_{j-1}$, so induction gives $[a,z]=0$ for every generator. By the explicit commutator formulas in part (a), commuting with all $D_i$ forces all positive $x_i$ exponents of $a$ to vanish, and commuting with all $x_i$ forces all positive $D_i$ exponents to vanish. Characteristic zero is used to cancel the nonzero integer exponents. Hence $a$ is scalar and, because it kills $M_0$, is zero. This proves injectivity.

Taking dimensions yieldsThe left side is asymptotic to $j^{2n}/(2n)!$. Consequently ${\dim }_k{M}_{2j}\ge C{j}^n$ for all sufficiently large $j$ and some $C>0$. Taking logarithms and the limsup in the definition of Gelfand–Kirillov dimension of a module proves the claimed lower bound. For a good Bernstein filtration, the associated graded module is finite over $k[x_1,\dots ,x_n,{\xi }_1,\dots ,{\xi }_n]$; its cumulative Hilbert function is eventually polynomial, and the same comparison says its degree is at least $n$.

For right modules, use the opposite ring isomorphism $A_n(k)\cong A_n(k{)}^{\mathrm{op}}$ sending $x_i$ to $x_i$ and $D_i$ to $-D_i$. It respects the Bernstein filtration, so the identical estimate applies. The bound is sharp: the polynomial module $k[x_1,\dots ,x_n]$, generated by $1$, has $M_j$ equal to the polynomials of degree at most $j$, of dimension $\left(\genfrac{}{}{0ex}{}{j+n}{n}\right)$. Its Gelfand–Kirillov dimension of a module is exactly $n$.