Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.
  • 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 $M/Mt$.

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

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):

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.

share|improve this answer
    
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. –  Greg Muller Dec 27 '09 at 0:11
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.