Timeline for How to find ideals of finite length in a power series ring with special properties?

9 events

when toggle format	what		by	license	comment
Jun 18, 2015 at 9:58	vote	accept	Bernie
Jun 17, 2015 at 17:15	answer	added	Mohan		timeline score: 1
Jun 17, 2015 at 13:46	comment	added	Bernie		@Mohan: If you turn your comment into a quick answer, then I can accept the answer, so the question in this form is resolved :-)
Jun 17, 2015 at 12:39	comment	added	Mohan		Yes, the image of any such $f$ will be contained in $(x,y)$ and thus will not be onto.
Jun 17, 2015 at 11:35	comment	added	Bernie		@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.
Jun 17, 2015 at 0:17	comment	added	Mohan		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.
Jun 16, 2015 at 15:18	comment	added	Bernie		@Jason Starr: Yes, exactly. An ideal $I$ of finite length in $A$ is an ideal such that $A/I$ has finite length.
Jun 16, 2015 at 14:07	comment	added	Jason Starr		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?
Jun 16, 2015 at 14:04	history	asked	Bernie	CC BY-SA 3.0