Let $R$ be the local ring of a complex curve singularity. (Can assume the singularity planar, the ring locally analytic or formal.) Let $\bar{R}$ be the normalization, let $R\subset R'\subset \bar{R}$ be an intermediate extension, corresponding to the factorization of the normalization map: $Spec(\bar{R})\to Spec{R'}\to Spec{R}$.

The relative conductor: $I^{cd}_{R'/R}:=(r\in R: r R' \subset R)$ is an ideal in $R$ (and in $R'$).

  1. The conductor for the normalization $I^{cd}_{\bar{R}/R}$ is well studied. Any reference for the relative conductor $I^{cd}_{R'/R}$?

    Specific questions:

  2. The relative conductor can be defined as $I^{cd}_{R'/R}=R:R'$. (Just another way to write the same thing.) Can we also say: $R'=R:I^{cd}_{R'/R}$? (i.e. $R:(R:R')=R'$). In words: for a given conductor $I$ take the maximal extension, whose conductor is $I$. Will this reproduce the initial extension?

  3. For a general ideal $I\subset R$, not a conductor of some extension, define $R'=R:I\subset\bar{R}$. (i.e. the maximal subring of the integral closure such that $R'I\subset R$). Then $I^{cd}_{R'/R}$ contains $I$, but in general is bigger. Any conditions on $I$ to ensure: $R:(R:I)=I$?

  4. Can this be somehow generalized to higher dimensions? At least, is the following always true: $R:(R:(R:I))=R:I$ ?

Any reference?

  • $\begingroup$ Dear Admin, I tried to fix Latex, have checked the source twice. Still, the question is not shown neatly. (Why??) Could you make the needed correction? $\endgroup$ Commented Feb 12, 2011 at 4:05
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    $\begingroup$ @qui-vadis: sometimes more complex math formulas throw off the rendering. You can fix this (as I did) by enclosing the math formulas (outisde the $ symbols) in ` signs. See the FAQ. $\endgroup$ Commented Feb 12, 2011 at 5:27

3 Answers 3


For 1. You could try looking at various exercises in the Swanson-Huneke book on Integral Closure. There might be something there. In particular, see chapter 12 (titled, the conductor).

For 2. 3. 4. Another way to identify the conductor (or relative conductor) is to consider $\text{Hom}_{R}(R', R)$. This module always has a map to $R$ (evaluation at $1$) and the image is the conductor. However, that map is injective, so that $\text{Hom}$ can be viewed as the conductor itself, in fact $\text{Hom}_{R}(R', R)$ can be identified with $R :_R R'$.

Now we can play to sort of games you want to play -- sometimes. Assuming $R$ is S2 and Gorenstein in codimension 1, which any plane curve singularity is, you can apply, say, Theorem 1.9 in Hartshorne's ``Generalized divisors on Gorenstein schemes''. In that case, applying $\text{Hom}_R(\cdot, R)$ twice to $R'$ will get you back something isomorphic to $R'$ by that cited result. In particular, in the Gorenstein in codim 1 case, I think you get the formulas you are hoping for.

Now, curves are always S2, but not always Gorenstein = Gorenstein in codimension 1. For example, Sándor's ring $B$ is not Gorenstein I think. Let me assume $r = m = 3$ for simplicity. Now mod out $R = k[x^3, x^6, x^7, \dots]_{\mathfrak{p}}$ by $(x^3) = (x^3, x^6, x^9, x^{10}, \dots)$. One gets an Artinian module generated by the images of $1, x^7, x^8$. This isn't Gorentsein (see for example Bruns-Herzog, Exercise 3.2.15 or notice that the socle isn't 1-dimensional), so $R$ wasn't either.

If you are not Gorenstein, you might be able to still get a lot of mileage out of applying $\text{Hom}(\cdot, \omega_R)$ instead, see for example Hartshorne's ``Generalized divisors and Biliaison".

  • $\begingroup$ You can also see Sandor's B is not Gorenstein by observing that dim (as a vector space) C[[t]]/B = m+r-2 and dim B / Conductor = 2. For the ring to be Gorenstein these must be equal, and $m \ge r \ge 3$ ensures this cannot happen. $\endgroup$ Commented Jun 6, 2011 at 12:26

For your question 1), I'd say that probably the best reference is anything on fractional ideals.

For question 2): I don't think this is true.

Let $m,r\in \mathbb N$, $m\geq r\geq 3$ and $A=k[t]$.

Let $B:=k[t^m,t^{m+r},t^{m+r+1},t^{m+r+2},\dots]\subset A$ and $B\subset B':=k[t^2,t^3]\subset A$. Let $\mathfrak p=At\cap B$ and $\mathfrak p'=At\cap B'$. Finally let $R=B_{\mathfrak p}$ and $R'=B'_{\mathfrak p'}$ Obviously $\overline R=k[t]_{(t)}$.

It is easy to see that $I=(R:R')=\overline Rt^{m+r}\cap R=(t^{m+r},t^{m+r+1},\dots)$, but this is actually an $\overline R$ ideal, so $(R:I)=\overline R$.

As for conditions on when your condition holds, I don't see a clear one. I can tell you certain patterns that makes it clear how these can fail for subrings of $k[t]$, but those seem a little ad hoc.


Ok, so it has been 7 years, but I do have something new to add to the answers by Karl and Sándor.

All your questions are about whether some modules/ideals are reflexive. That is because for a fractional ideal $I$ (including any birational extension of $R$ or it's conductor), you can identify $R:I$ with $I^*:=Hom_R(I,R)$. Then Question 2 asks if $R'$ is reflexive, Question 3 asks if $I$ is reflexive, and Question 4 asks if $I^*$ is reflexive.

(a small quibble: in Question 3) you wrote that the $R'= R:I$ is a subring, but I am not sure, this is a ring. Perhaps you meant $R'=I:I$?)

Anyhow, as Karl's answer indicated, if $R$ is Gorenstein then any torsion-free module is reflexive. So if $R$ is Gorenstein, for instance if it is a planar curve, the answers are yes to all questions.

When $R$ is not Gorenstein, surprisingly reflexive modules are not very well-understood. Recently I wrote a paper on this topic with a couple of collaborators, you can find it here: https://arxiv.org/abs/2101.02641 and the references there.

You can use the results there to understand some of the bad examples offered by Sándor. For instance if $R=k[[t^3,t^6,t^7...]]=k[[t^3,t^7,t^8]]=k[[x,y,z]]/P$, then the conductor is $c=(x^6,x^7,..)= (x^2,y,z)$. Theorem 3.5 of the above paper says that any reflexive ideal would be isomorphic to an ideal containing the conductor, so there are only two proper choices, $c$ and the maximal ideal $(x,y,z)$. Both are reflexive, and their endormorphism rings are the only reflexive birational extensions.

The answer to Question 4 is also yes in higher dimensions, you just need $R$ to be $S_1$ and generically Gorenstein, see Lemma 2.5 of the same paper.


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