* 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 $M/Mt$.



Depth Zero Ideals in the Homogenized Weyl Algebra
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](http://www.ams.org/mathscinet/search/publdoc.html?pg1=IID&s1=42575&vfpref=html&r=81&mx-pid=399074) of Bruns. It is even implemented here (be warned that the statement misses the at least depth $1$ part):


<http://www.math.uiuc.edu/Macaulay2/doc/Macaulay2-1.2/share/doc/Macaulay2/Bruns/html/>


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. 


