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.

  • $\begingroup$ Comments about confusion about $\delta$ deleted. $\endgroup$ May 26, 2010 at 19:42
  • $\begingroup$ @jlk: why does delta=4 in your example? $\endgroup$ May 28, 2010 at 4:02
  • $\begingroup$ @Hailong Dao: I computed delta directly from the definitions. The computation is similar to, but harder than, the example of 4 lines in 3-space given in response to Graham's question. Unless I made a mistake, I think the normalization $\tilde{R}$ is isomorphic to the product of 4 power series rings. As a complete $k$-algebra contained in $\tilde{R}$, $R$ is generated by 3 4-tuples of degree 1 monomials. The quotient \tilde{R}/R has basis given by (1,0,0,0), (0,1,0,0), (0,0,1,), and a 4-tuple in which the entries are degree 1 monomials. $\endgroup$
    – jlk
    May 28, 2010 at 20:48
  • $\begingroup$ (continued): If you are curious, I can try to reconstruct the details. The example of 4-lines in 3-space shows that $\delta=4$ can be achieved by a non-planar Gorenstein curve singularity, and this is maybe a nicer examples than $k[[xyz]](xy−z^2, zx=y^22)$. $\endgroup$
    – jlk
    May 28, 2010 at 21:35
  • $\begingroup$ @jlk: I think your example is also an intersection of 4 lines : $(y,z)$ and $y-az, x-a^2z$ with $a^3=1$. $\endgroup$ May 30, 2010 at 3:15

3 Answers 3


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!

  • $\begingroup$ Great! Could you include a reference for Abhyankar's result. I am not familiar with it. $\endgroup$
    – jlk
    May 31, 2010 at 2:14
  • $\begingroup$ Also, what's a reference for the Greither result that $e= V/\mathfrak{m}V$? $\endgroup$
    – jlk
    May 31, 2010 at 2:15
  • 2
    $\begingroup$ If you're an original-reference nerd like me, Abhyankar's result is in Local rings of high embedding dimension, Amer. J. Math. 89 (1967), 1073–1077. Greither's theorem is in On the two generator problem for the ideals of a one-dimensional ring, J. Pure Appl. Algebra 24 (1982), no. 3, 265–276. $\endgroup$ May 31, 2010 at 21:14
  • $\begingroup$ @Hailong Dao: I just noticed you edited your answer to include references. Thanks! $\endgroup$
    – jlk
    Jun 1, 2010 at 0:11

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.

  • $\begingroup$ I am a little confused about Ingredient 1: if $R=k[[t^a,t^b]]$ with $a<b$ then the multiplicity is $a$, while the length of the quotient = the number of integers not in the semigroup generated by $(a,b)$ =$(a-1)(b-1)/2$. $\endgroup$ May 26, 2010 at 21:59
  • $\begingroup$ I'm also a little confused. If we take $R$ to be the subring of the product of 4 power series rings generated by (t,0,0,-t), (0,t,0,-t), (0,0,t,-t) (4 general lines in 3-space), then \tilde{R}/R has basis given by (1,0,0,0), (0,1,0,0), (0,0,1,0), and (t,0,0,0). If m is the maximal ideal, then I am computing that m^{n}/m^{n+1} = 4 for n sufficiently large. I think this means the multiplicity is 4. Am I using the wrong definition of multiplicity or something? $\endgroup$
    – jlk
    May 26, 2010 at 22:20
  • $\begingroup$ btw, is "degree" the standard term for the dimension of \tilde{R}/R? $\endgroup$
    – jlk
    May 26, 2010 at 22:21
  • $\begingroup$ Hmm, that's a problem. Hailong's right. I think I confused $\tilde{R}/\mathfrak{m}$ with $\tilde{R}/\mathfrak{m}\tilde{R}$. Drat, it was such a clean and satisfying answer too. jlk, I don't know a standard term for that dimension -- it's never come up in my experience before. Why is it useful/interesting? $\endgroup$ May 26, 2010 at 23:51
  • $\begingroup$ @Graham, The number $\operatorname{dim} \tilde{R}/R$ is a basic invariant of the curve. The Gorenstein relation is, as you probably know, equivalent to the equation $2 \cdot \operatorname{dim} \tilde{R}/R = \operatorname{dim} \tilde{R}/I$, where $I$ is the conductor ideal. $\endgroup$
    – jlk
    May 27, 2010 at 2:29

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.


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