0
$\begingroup$

Let $R$ be a commutative Noetherian ring and $M$ a finitely generated $R$-module. Let $I$ an ideal of $R$. We have $$0:_MI = \cap_x(0:_Mx),$$ where $x$ runs a set of generators of $I$.

Now set $S = R[T]$ with $T$ is a variable. We have $M\otimes S = M[T]$ and $IS = I[T]$ and $$0:_{M[T]}IS = (0:_MI)[T]$$ by the flat extension property.

Question: Does there exist an element $X \in IS$ such that $$0:_{M[T]}IS = (0:_{M[T]}X)?$$

In my problem we can assume that $M$ is also Artinian (so has finite length).

$\endgroup$
4
  • 3
    $\begingroup$ "colon operation" hurts my eyes. $\endgroup$
    – johndoe
    May 14, 2014 at 16:00
  • 3
    $\begingroup$ @johndoe: " 'colon operation' hurts my eyes". Only your eyes? I want to know the number for your proctologist. $\endgroup$ May 14, 2014 at 18:35
  • $\begingroup$ I don't think this works in general. Let $R=M=k[X,Y]/(X^2, Y^2) = k[x,y]$ (where $k$ is a field, and lower case $x$ and $y$ are the homomorphic images of $X$, $Y$ respectively). Let $I = (x,y) = $the maximal ideal of $R$. Then $(0:I) = (xy)$, but I don't think there is any $g\in I[T]=IS$ such that $(0:_{R[T]}g) = xyS$. $\endgroup$ May 15, 2014 at 3:53
  • $\begingroup$ @Pham But hold on; I'm not sure my counterexample really works (which is why I wrote it as a comment instead of an answer). Does it? If so, I'll repost it as an answer and you can checkmark it.. $\endgroup$ May 16, 2014 at 2:41

1 Answer 1

0
$\begingroup$

The question is not true by the example of Neil Epstein as above.

Edit: Let me finish this question.

We give a generalization for the example of Neil Epstein.

Let $(R, \mathfrak{m})$ be a Gorenstein local ring of dimension $0$ and the embedded dimension $\ell(\mathfrak{m}/\mathfrak{m}^2)>1$. We show that $M = R$ and $I = \mathfrak{m}$ are not satisfying my question.

Indeed, assume that there exist an element $x \in \mathfrak{m}S$ such that $$0:_S \mathfrak{m}S = 0:_Sx.$$ By localization at $\mathfrak{m}S$ we get a Gorenstein local ring, says $(A, \mathfrak{n})$, such that $$0:_A \mathfrak{n} = 0:_Ax$$ for some $x$. Since the extension $R \to A$ is flat we have $\ell(\mathfrak{m}/\mathfrak{m}^2)>1$. So $x \subsetneq \mathfrak{n}$. By the duality of Gorenstein ring of dimension $0$ we have $0:_A \mathfrak{n} \subsetneq 0:_Ax$, a contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.