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).



Colon operation after adjoint variables
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. 


