TITLE: What is the most simple non-planar Gorenstein curve singularity?
QUESTION [11 upvotes]: Let $R$ be a reduced curve singularity over an algebraically closed field $k$ and $\tilde{R}$ its integral closure in its total ring of fractions.  
The $k$-dimension of $\tilde{R}/R$ is finite.  If we assume $R$ is non-planar and Gorenstein, then how small can this number be?
The ring $R = k[[x,y,z]]/(xy = z^2, z x = y^2)$ is a complete intersection, hence Gorenstein, and the dimension of $\tilde{R}/R$ is $4$.  The question is thus "is $2$ or $3$ possible?"
For the sake of concreteness, let's say that a curve singularity is a $1$-dimensional quotient of $k[[x_1, \dots, x_n]]$ for some $n$.
Edit: I had thought that the $k$-dimension of $\tilde{R}/R$ was widely known as the $\delta$-invariant; I think this the notation Serre uses in Algebraic Groups and Class Fields.  From the comments, it seems this is non-standard, and I have edited accordingly.
As Graham points out, the number $\operatorname{dim}(\tilde{R}/R)$ is also the colength of the conductor ideal.  The number also comes up in computing the (arithmetic) genus of a singular curve.

REPLY [3 votes]: Here is a short geometric proof that if $R$ is Gorenstein and $\tilde{R}/R$ has dimension $\delta \le 3$, then $R$ is planar. 
We can realize $R$ as the local ring of a rational curve $X$ of genus $\delta$.  If $X$ is hyperelliptic (i.e. admit a degree $2$ morphism $f$ to $\mathbb{P}^{1}$), then $X$ embeds into a smoth surface: the ruled surface $\mathbb{P}(\mathcal{E})$ for $\mathcal{E}=f_{*}\mathcal{O}_{X}$.  In particular, the singularities of $X$ are planar.
Otherwise, $X$ is non-hyperelliptic of genus $3$.  But then the canonical map embeds $X$ as a plane quartic curve. In particular, $X$ again embeds in a smooth surface and hence has planar singularities.