on the relative conductor of curve singularity and quotient of ideals - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T00:51:36Z http://mathoverflow.net/feeds/question/55180 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55180/on-the-relative-conductor-of-curve-singularity-and-quotient-of-ideals on the relative conductor of curve singularity and quotient of ideals Dmitry Kerner 2011-02-12T03:53:38Z 2011-06-06T08:37:10Z <p>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: $Spec(\bar{R})\to Spec{R'}\to Spec{R}$.</p> <p>The relative conductor: <code>$I^{cd}_{R'/R}:=(r\in R: r R' \subset R)$</code> is an ideal in $R$ (and in $R'$).</p> <ol> <li><p>The conductor for the normalization <code>$I^{cd}_{\bar{R}/R}$</code> is well studied. Any reference for the relative conductor <code>$I^{cd}_{R'/R}$</code>?</p> <p>Specific questions:</p></li> <li><p>The relative conductor can be defined as <code>$I^{cd}_{R'/R}=R:R'$</code>. (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?</p></li> <li><p>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: $R:(R:I)=I$?</p></li> <li><p>Can this be somehow generalized to higher dimensions? At least, is the following always true: $R:(R:(R:I))=R:I$ ?</p></li> </ol> <p>Any reference?</p> http://mathoverflow.net/questions/55180/on-the-relative-conductor-of-curve-singularity-and-quotient-of-ideals/55189#55189 Answer by Sándor Kovács for on the relative conductor of curve singularity and quotient of ideals Sándor Kovács 2011-02-12T09:15:41Z 2011-02-12T22:08:17Z <p>For your question 1), I'd say that probably the best reference is anything on fractional ideals.</p> <p>For question 2): I don't think this is true.</p> <p>Let $m,r\in \mathbb N$, $m\geq r\geq 3$ and $A=k[t]$.</p> <p>Let <code>$B:=k[t^m,t^{m+r},t^{m+r+1},t^{m+r+2},\dots]\subset A$</code> and <code>$B\subset B':=k[t^2,t^3]\subset A$</code>. Let $\mathfrak p=At\cap B$ and $\mathfrak p'=At\cap B'$. Finally let $R=B_{\mathfrak p}$ and $R'=B'_{\mathfrak p'}$ Obviously <code>$\overline R=k[t]_{(t)}$</code>. </p> <p>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 $(R:I)=\overline R$.</p> <p>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.</p> http://mathoverflow.net/questions/55180/on-the-relative-conductor-of-curve-singularity-and-quotient-of-ideals/55207#55207 Answer by Karl Schwede for on the relative conductor of curve singularity and quotient of ideals Karl Schwede 2011-02-12T14:50:27Z 2011-02-12T19:26:12Z <p>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). </p> <p>For 2. 3. 4. Another way to identify the conductor (or relative conductor) is to consider <code>$\text{Hom}_{R}(R', R)$</code>. 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 $R :_R R'$.</p> <p>Now we can play to sort of games you want to play -- sometimes. Assuming $R$ is S2 and Gorenstein in codimension 1, which any <em>plane</em> 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.</p> <p>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. </p> <p>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". </p> http://mathoverflow.net/questions/55180/on-the-relative-conductor-of-curve-singularity-and-quotient-of-ideals/67021#67021 Answer by GiulioP for on the relative conductor of curve singularity and quotient of ideals GiulioP 2011-06-06T08:37:10Z 2011-06-06T08:37:10Z <p>just to be sure, when you write $R:I$ is that the column ideal taken inside the quotient field $K$ of $R$ (I assume $R$ to be a domain), that is $(x\in K:xI\subset R)$ ? in the case of $(R:R´)$ it is the same to take that ideal in $R$ or in $K$. But in the case of the ideal $I$ that is not the same:$(x\in R:xI\subset R)=R$ since $I$ is an ideal in $R$. (I just started recently to work with conductors... I´ll post a question on them soon.... someway related to this question)</p>