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Let $A$ be the power series ring $\mathbb{C}[[x,y]]$.

Assume we are given two ideals $I,J$ of finite length in $A$ such that:

  • $xJ\subseteq I\subseteq J$

Is it possible to find ideals of finite length $B,C$ in $A$ such that:

  • $xC\subseteq B\subseteq C$
  • $B\subset I$ and $I/B\cong R/J$
  • $C\subset J$ and $J/C\cong R/I$

Here are my thoughts so far: we see that $x$ annihilates $J/I$ so that this is a module of finite length over $\mathbb{C}[[y]]$, hence also some power of $y$ annihilates $J/I$. I thought maybe we should write $I=<f_1,\ldots,f_n>$ and $J=<h_1,\ldots,h_m>$ and see if we can use these generators to construct/find the wanted ideals, but this seems quite hard.

Is it even possible to find such ideals $B$ and $C$? Any ideas how to approach this problem are welcome. One could also ask more genrally:

Assume we are given $n$ ideals $I_r$ of finite length in $A$ such that:

  • $xI_r\subseteq I_1\subseteq I_r$ for all $r\geq 2$
  • for all $r\geq 2$: $xI_r\subseteq I_s\subseteq I_r$ for all $s\geq r+1$

Is it possible to find ideals of finite length $J_r\subset I_r$ in $A$ for $r=1\ldots n$ which satisfy:

  • $xJ_r\subseteq J_1\subseteq J_r$ for all $r\geq 2$
  • for all $r\geq 2$: $xJ_r\subseteq J_s\subseteq J_r$ for all $s\geq r+1$
  • $I_r/J_r\cong R/I_{r-1}$ ($r\geq 2$) and $I_1/J_1\cong R/I_n$
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  • $\begingroup$ What do you mean when you speak of an ideal in $\mathbb{C}[[x,y]]$ of finite length? Do you mean finite colength, i.e., the quotient module has finite length? $\endgroup$ Jun 16, 2015 at 14:07
  • $\begingroup$ @Jason Starr: Yes, exactly. An ideal $I$ of finite length in $A$ is an ideal such that $A/I$ has finite length. $\endgroup$
    – Bernie
    Jun 16, 2015 at 15:18
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    $\begingroup$ In general, no. Take $I=J=(x,y)^2$. Then your third bulleted requirement want a surjection from $J$ to $R/I=R/J$. An elementary calculation will show that this is not possible. $\endgroup$
    – Mohan
    Jun 17, 2015 at 0:17
  • $\begingroup$ @Mohan: Do you mean something like this: every morphism $f: J \rightarrow R/J$ must have image in $(x,y)/J$, because we can make things like $[y]f(x^2)=f(yx^2)=[x]f(xy)$ and so on? So no morphism can be surjective in this case. $\endgroup$
    – Bernie
    Jun 17, 2015 at 11:35
  • $\begingroup$ Yes, the image of any such $f$ will be contained in $(x,y)$ and thus will not be onto. $\endgroup$
    – Mohan
    Jun 17, 2015 at 12:39

1 Answer 1

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As I commented, in general, the answer is no. For an easy example, take $I=J=(x,y)^2$. Then you require a surjection $f:J\to R/I$. Letting $f(x^2)=a, f(xy)=b, f(y^2)=c$, since $y.x^2=x.xy$, we get $ya=xb$ and similarly, $yb=xc$, using $y.xy=x.y^2$. Since $ya, xb\in (x,y)/I$, which is a $k$ vector space with basis $x,y$, we see that $a,b\in (x,y)$. Similarly, we get $b,c\in (x,y)$. Thus $a,b,c\in (x,y)/I$ and then the map $f$ can not be onto.

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