Timeline for Isomorphic quotient of a Module over Noetherian commutative algebra
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 4, 2014 at 15:17 | comment | added | Jeremy Rickard | abx's answer is correct. For an explicit example, let $A=\mathbb{C}[x,y]$ be the ring of polynomials in two variables and let $M$ be the ideal consisting of polynomials with zero constant term. Then $M$ does not have a quotient module isomorphic to $A$. | |
| Aug 4, 2014 at 15:12 | comment | added | Jeremy Rickard | @Student That's not a counterexample to what abx claimed, because your $M$ is not an ideal of $A$. | |
| Aug 4, 2014 at 14:52 | comment | added | Guy L. | Well here is a counter example to your argument that M is principal. Take M=A^2. And a surjective morphism defined by (1,0)->1. (0,1)->0. Then M is not principal although there is a surjective morphism. | |
| Aug 4, 2014 at 14:37 | comment | added | Guy L. | Please learn some manners. | |
| Aug 4, 2014 at 14:33 | comment | added | abx | OK, enough for me. Please learn some algebra before asking questions here. | |
| Aug 4, 2014 at 14:30 | comment | added | abx | No, your quotient $M/N$ is not isomorphic to $A$. Again, think more about asking correctly your question. I down vote. | |
| Aug 4, 2014 at 14:28 | comment | added | abx | In your post you say that $M$ has a quotient isomorphic to $A/\mathfrak{p}$. $Am$ is not a quotient... Maybe you should think more about your question. | |
| Aug 4, 2014 at 14:16 | history | answered | abx | CC BY-SA 3.0 |