• Let $\mathcal{D}$ be the $n$th Weyl algebra $ \mathcal{D} :=k[x_1,...,x_n,\partial_1,...,\partial_n] $, where $\partial_ix_i-x_i\partial_i=1$.
  • Let $\widetilde{\mathcal{D}}$ be its Rees algebra, which is $ \mathcal{D} :=k[t, x_1,...,x_n,\partial_1,...,\partial_n] $, where $\partial_ix_i-x_i\partial_i=t$, and $t$ is central.
  • Let $\mathcal{O}_X$ denote the polynomial algebra $k[x_1,...,x_n]$, which is a left $\widetilde{\mathcal{D}}$-module, where $t$ and all the $\partial_i$ act by zero. NOTE: this is different than the homogenization of the standard $\mathcal{D}$-module structure on $\mathcal{O}_X$.

The question I am interested in is, how many generators does a left ideal $M$ in $\widetilde{\mathcal{D}}$ need before $Hom_{\widetilde{\mathcal{D}}}(\mathcal{O}_X,\widetilde{\mathcal{D}}/M)$ can be non-zero? My conjecture is that $M$ needs at least $n+1$ generators. NOTE: Savvy Weyl algebra veterans will know every left ideal in $\mathcal{D}$ can be generated by two elements; however, this is not true in $\widetilde{\mathcal{D}}$. There can be ideals generated by $n+1$ elements and no fewer.

The functor $Hom_{\widetilde{\mathcal{D}}}(\mathcal{O}_X,-)$ acts as a relative analog of the more familiar functor $Hom_R(k,-)$ (where $R=k[t,y_1,...y_n]$). Therefore, the above question is analogous to asking "how many generators must an ideal $I\subseteq R$ have before $R/I$ can have depth zero?" The answer here is $n+1$, which follows from Thm 13.4, pg 98 of Matsumura (essentially a souped up version of the Hauptidealsatz).

In the noncommutative case, if you try to make this work with the $\widetilde{\mathcal{D}}$-module $k$ (where $t$, $x_i$ and $\partial_i$ all act by zero), it doesn't work. The natural conjecture would be that $Hom_{\widetilde{D}}(k,\widetilde{\mathcal{D}}/M)\neq 0$ implies $M$ had at least $2n+1$ generators, except this fails even for the first Weyl algebra and $M=\widetilde{\mathcal{D}}x_1+\widetilde{\mathcal{D}}\partial_1$ (since $\widetilde{\mathcal{D}}/M=k$).

However, it seems that things might work right for the relative module $O_X$, based on a fair amount of experimentation. It is easily true in the first Weyl algebra. Oh, and equivalent condition is to ask when $Hom_{\overline{\mathcal{D}}}(\mathcal{O}_X,\overline{M})\neq 0 $, where $\overline{\mathcal{D}}=\widetilde{\mathcal{D}}/t$, and $\overline{M}$ is $M/Mt$.


1 Answer 1


I am not sure I understand the analogue correctly, but in the commutative case, one can get to depth zero with 3 generators. That is because any second syzygy of a module of depth at least $1$ is isomorphic to a second syzygy of a 3-generated ideal by a result of Bruns. It is even implemented here (be warned that the statement misses the at least depth $1$ part):


You can take the N to be second syzygy of $m=(t,y_1,...,y_n)$, so $N$ has depth 3. Produce a three generated ideal $I$ such that $syz^2(I)\cong N$. So $depth I =3-2=1$, and $depth R/I=0$.

I think Theorem 13.4 shows that $dim R/I=0$ implies $I$ is at least $n+1$-generated.

  • $\begingroup$ I am fairly certain your answer is correct. In fact, it can be used to produce counterexamples to my conjecture, for all $n>2$. The frustrating thing is that the reason I want this fact is the $n=2$ case, and I am as interested in a counterexample as I am in a proof... and I can find neither. $\endgroup$ Dec 27, 2009 at 0:11

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