-3
$\begingroup$

I have a nice solution to the following problem and I thought of writing a paper about it but beforehand, I wanted to ask the problem here to see if this is an easy problem and if you people can solve it easily. If its easy then I will not write a paper about my solution. I know of many people who did not manage to solve the problem so lets see if you can.

Let A be a commutative noetherian algebra. Let M be a finitely generated A-module. For every finitely generated A-module N we define SuppN = V(Ann(N)) \subset Spec(A).

Spec(A) denotes the prime ideals of A.

Ann(N) denotes all the members a of A such that aN=0.

Prove that for any prime ideal p in SuppM, M has a quotient isomorphic to A/p. Meaning there is an A-module N, which is a submodule of M such that M/N is isomorphic to A/p.

Good Luck :)

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

This doesn't sound right. Suppose $A$ is an integral domain, and $M$ a nonzero ideal of $A$. Take $\mathfrak{p}=(0)$. You are claiming that there is a surjective homomorphism $M\rightarrow A$. This implies that $A$ is a direct summand of $M$, which happens if and only if $M$ is principal. So any non-principal ideal gives a counter-example.

Improve this answer
$\endgroup$
7
  • 2
    $\begingroup$ 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. $\endgroup$
    – abx
    Commented Aug 4, 2014 at 14:28
  • 2
    $\begingroup$ No, your quotient $M/N$ is not isomorphic to $A$. Again, think more about asking correctly your question. I down vote. $\endgroup$
    – abx
    Commented Aug 4, 2014 at 14:30
  • 4
    $\begingroup$ OK, enough for me. Please learn some algebra before asking questions here. $\endgroup$
    – abx
    Commented Aug 4, 2014 at 14:33
  • 1
    $\begingroup$ @Student That's not a counterexample to what abx claimed, because your $M$ is not an ideal of $A$. $\endgroup$
    – Jeremy Rickard
    Commented Aug 4, 2014 at 15:12
  • 2
    $\begingroup$ 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 $A$. $\endgroup$
    – Jeremy Rickard
    Commented Aug 4, 2014 at 15:17

Not the answer you're looking for? Browse other questions tagged .