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Given a hypersurface $D = h^{-1}(0)$ for some polynomial $h \in \mathbb{C} [x,y,z]$ I want to be able to use Macaulay 2 to tell if it's a free divisor or not.

What I've got so far;

Let $h_{p}$ be the reduced equation for $h$. For $D$ to be a free divisor I need $Der(-logD) := \{ \delta : \delta(h_{p}) \in (h_{p})\}$ to be a locally free $\mathcal{O}_{\mathbb{C} ^{n}}$-module, where $\delta$ is a logarithmic vector field. So for a vector field $\delta = a \frac{\partial}{\partial{x}} + b \frac{\partial}{\partial{y}} + c \frac{\partial}{\partial{z}}$ I need $a \frac{\partial{h}}{\partial{x}} + b \frac{\partial{h}}{\partial{y}} + c \frac{\partial{h}}{\partial{z}} = r h_{p}$ for some $r$.

So I think If I remove the bottom row of ones that I get when when using Macaulay 2 to calculate the kernel of the map given by the matrix $\lbrack \frac{\partial{h}}{\partial{x}} , \frac{\partial{h}}{\partial{y}} , \frac{\partial{h}}{\partial{z}} , -h_{p} \rbrack$ I'll get the module $Der(-logD)$.

But I'm not sure how to then check if its a free module, I am slightly confused too as I've been told to use the resolution function in Macaulay 2 but when ever I try I can't seem to get it to work.

I've only just started using Macaulay 2 and I am fairly new to free divisors too so sorry for any errors and thanks in advance for any help.

Rodrigo de Azevedo
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Geraint Jones
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  • Well, a free module have no relations among its generators. What happens when you try to make a resolution of $Der(-\log D)$? – Fredrik Meyer Oct 23 '13 at 09:34
  • When I calculate the kernel of the matrix $[\frac{\partial{h}}{\partial{x}} , \frac{\partial{h}}{\partial{y}} , \frac{\partial{h}}{\partial{z}} , -h ]$ I get the message (in this case I'm using h=xyz): -- warning: experimental computation over inexact field begun -- results not reliable (one warning given per session) and then I get $Der(-logD)$ by getting the generators of this then by rewriting it without the bottom row of ones and calculating its image. Then when I use resolution on this I get, (in next comment) – Geraint Jones Oct 23 '13 at 15:40
  • $= 0 \ : \ R^{3} <--------------------- R^{3} \ : \ 0$ then on the next row $\begin{bmatrix} x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z \end{bmatrix} $ then the next row is $1 \ : \ <---- \ : \ 1$ and the final row is just $0$ – Geraint Jones Oct 23 '13 at 15:56
  • Sorry for the formatting It was the best I could do. – Geraint Jones Oct 23 '13 at 15:57
  • You get the error message because you're working over $\mathbb C$, try using R=QQ[x,y,z] instead. Let K=ker M. Then write "prune K". – Fredrik Meyer Oct 23 '13 at 16:40
  • Thanks that seems to work, does it matter that I'm doing it for $Ker ( [ \frac{\partial{h}}{\partial{x}} , \frac{\partial{h}}{\partial{y}} , \frac{\partial{h}}{\partial{z}} , -h] )$ instead of $Der(-logD)$ ? – Geraint Jones Oct 23 '13 at 19:33

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