What is the most simple non-planar Gorenstein curve singularity? - MathOverflow

What is the most simple non-planar Gorenstein curve singularity?

2010-05-26T04:56:04Z

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.

Answer by Graham Leuschke for What is the most simple non-planar Gorenstein curve singularity?

2010-05-26T19:34:10Z

edit: this answer is garbage (or, rather, answers a question that the asker did not ask). I leave it here because Hailong's answer refers to some of its ingredients.

$3$ is the least possible.

Ingredient 1: For a one-dimensional complete local ring $R$ with integral closure $\tilde R$, the $k$-dimension of $\tilde{R}/R$ is one less than the multiplicity $e(R)$ of the ring (this is false: I confused $\tilde{R}/\mathfrak{m}$ with $\tilde{R}/\mathfrak{m}\tilde{R}$). This is because $e(R) = \dim_k (\tilde {R}/\mathfrak{m}\tilde{R})$, which is due to Greither in 1982.

Ingredient 2: There is an inequality due to Abhyankar for the multiplicity of a CM local ring: $$e(R) \geq \mu_R(\mathfrak{m}) - \dim R + 1$$ where $\mu$ denotes the minimal number of generators.

Ingredient 3: It's relatively easy to see that for a Gorenstein local ring that is not a hypersurface (e.g. a one-dimensional non-planar Gorenstein local ring) we can do one better than Abhyankar's bound. Namely, if we had equality in Abhyankar's bound, then $\mathfrak m^2 = \mathbf{x}\mathfrak m$ for some minimal reduction $\mathbf{x}$ of the maximal ideal. Count lengths in $\bar{R} = R/(\mathbf{x})$, remembering that it has one-dimensional socle since $R$ is Gorenstein, to see that $\dim R = \mu_R(\mathfrak m) -1$. Therefore we have $$e(R) \geq \mu_R(\mathfrak{m}) - \dim R + 2$$ for a Gorenstein non-hypersurface $R$.

Applying this formula with $\mu_R(\mathfrak m) \geq 3$ and $\dim R = 1$, we get that the multiplicity is at least $4$, so the degree is at least 3.

Answer by Hailong Dao for What is the most simple non-planar Gorenstein curve singularity?

2010-05-30T04:59:16Z

I think Graham's answer already gave most of what you need to prove that $4$ is the smallest possible. Let $V$ be the integral closure of $R$, $n$ be the embedding dimension of $R$, and $e=e(R)$ be the multiplicity.

Claim: If $R=k[[x_1,\cdots,x_n]]/I$ is Gorenstein and $n$ is at least $3$, then $\dim_k(V/R)\geq e$.

Proof: Let $m$ be the maximal ideal of $R$. As Graham pointed out, we have $e = \dim_k(V/mV)$. So:

$$\dim_k(V/R) =\dim_k(V/mR)-\dim_k(R/mR) \geq \dim_k(V/mV)-1=e-1$$

We need to rule out the equality. If equality happens, then one must have $mV=mR$. This shows that $m$ is the conductor of $R$. As you already knew, since $R$ is Gorenstein, one must then have $\dim_k(V/R)=\dim_k(R/m)=1$. The inequality now gives $e\leq 2$. Abhyankar's inequality (part 2 of Graham's answer) gives $n\leq 2$, so $R$ is planar, contradiction.

Now, one needs to show that for $R$ non-planar, $e\geq 4$. You could use part $3$ of Graham's answer, or arguing as follows: if $n\geq 4$ we are done by Abhyankar inequality. If $n=3$, a Gorenstein quotient of $k[[x,y,z]]$ must be a complete intersetion, and so $I=(f,g)$, each of minimal degree at least $2$ since $R$ is not planar, thus $e$ must be at least $4$.

By the way, one could construct a domain $R$ such that $\dim_k(V/R)=4$ as follows: Take $R=k[[t^4,t^5,t^6]]$. The semigroup generated by $(4,5,6)$ is symmetric, so $R$ is Gorenstein. The Frobenius number is $7$, and $V/R$ is generated by $t,t^2,t^3,t^7$.

EDIT (references, per OP's request): Abhyankar inequality is standard, for example see Exercise 4.6.14 of Bruns-Herzog "Cohen-Macaulay rings", second edition (Link to the exact page). Or see exercise 11.10 of Huneke-Swanson book, also available for free here. Or Google "rings with minimal multiplicity".

(The original references are now available thanks to Graham, see his comment below)

As for $e=\dim_k(V/mV)$, I could not find a convenient reference, but here is a sketch of proof using the above reference: First, using the additivity and reduction formula (Theorem 11.2.4 of Huneke-Swanson) to reduce to the domain case. Assume that $R$ is now a complete domain, then $V=k[[t]]$, and $R$ is a subring of $V$. Let $x\in m$ be an element with smallest minimal degree. Then $mV=xV$ ($V$ is a DVR), and it is not hard to see that $e=$ the minimal degree of $x$ $=\lambda(V/xV)$ (see Exercise 4.6.18 of Bruns-Herzog, same page as the link above).

Alternatively, one can use the fact that: $$e(m,V) = \text{rank}_RV.e(m,R) = e $$ The second inequality is because $V$ is birational to $R$ so $\text{rank}_RV=1$. The left hand side can be easily computed by definition to be length of $V/xV$, which equals $\dim_k(V/mV)$. (use $m^nV=x^nV$ since $V$ is a DVR)

Fun exercise!

Answer by jlk for What is the most simple non-planar Gorenstein curve singularity?

2012-10-06T15:22:37Z

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.