has the following basic result.

(Pellikaan, [@pellikaan1], prop. 1.7) The following statements about a function $f \in \int I$ with jacobian ideal $J_f$ are equivalent:\
1) $\dim(I/J_f) <\infty$\
2) $c_{e,I}(f) < \infty$\
3) The singular locus of $f$ is $\Sigma$ and $f$ has only $A_1$-singularities transverse to $\Sigma\setminus\{0\}$.

Note that in this situation $I=rad(J_f)$, which is the same as the saturation of $J_f$ with respect to ${\mathfrak{m}}$. So the jacobian module $I/J_f$ is precisely the module considered in \[imodj\] for the sequence $f_1,f_2,\ldots,f_n$, with $f_i=\partial f/\partial x_i$.\
There is a formula expressing $\dim I/J$ in terms of $c_{e,I}$ and some other invariant that we explain now. The two dual exact sequences $$I/I^2 \stackrel{d}{{\longrightarrow}} \Omega \otimes \mathcal{O}_{\Sigma} {\longrightarrow}\Omega_{\Sigma} {\longrightarrow}0$$ $$0 {\longrightarrow}\Theta_{\Sigma} {\longrightarrow}\Theta \otimes \mathcal{O}_{\Sigma} \stackrel{d^*}{{\longrightarrow}} Hom(I/I^2,\mathcal{O}_{\Sigma})$$ provide important invariants of the curve $\Sigma$.\
$\bullet$ The first tangent homology is $T_1(\Sigma)=ker(d)=\int I/I^2$. In case $I$ is a reduced complete intersection, one has $\int I=I^2$. But for $I=(xy,yz,zx) \subset {{\mathbb C}}\{x,y,z\}$ one has $\int I=(xyz,I^2)$, so $\int I/I^2$ is one-dimensional.\
$\bullet$ The first tangent cohomology if $T^1(\Sigma)=Coker(d^*)$. It is the space of first order infinitesimal deformations of $\Sigma$. The normal module $N=Hom_{\Sigma}(I/I^2,{\cal O}_{\Sigma})$ is identified with the space of embedded deformations of $\Sigma$.\
