definitions of $I$-torsion functor $\Gamma_I$

Question

Let $R$ be a commutative ring, $M$ an $R$-module and $I$ be an ideal in $R$.
Bruns-Herzog, Brodmann-Sharp and many other authors define $I$-torsion functor $\Gamma_I$ as: $$\Gamma_I(M)=\bigcup_{n\in N} (0:_MI^n).$$ But some also define it as: $$\Gamma_I(M)=\{m\in M: \text{supp}\ Rm\subseteq V(I) \}.$$

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I have two questions about definitions:

Question 1. Is these definitions equivalent? Does one results the other?
Question 2. What preponderance each one has?

Thank you.

Answer 1

They are equivalent if $R$ is noetherian.

The second definition is equivalent to $\sqrt{I} \subset \sqrt{Ann(m)}$ due to $Supp(Rm)=V(Ann(m))$ and the formal Nullstellensatz.

So let $m \in \Gamma_I(M)$ according to the first definition. We have $I^n \subset Ann(m)$ for some $n$ by definition. This implies $\sqrt{I} \subset \sqrt{Ann(m)}$.

On the other hand let $\sqrt{I} \subset \sqrt{Ann(m)}$. We clearly find some $n$ with $a^nm=0$ for any $a \in I$. But to deduce $I^nm=0$ for some $n$, we need the noetherian property for $R$ (and this is clearly sufficient, since we have then $\sqrt{Ann(m)}^n \subset Ann(m)$ for some $n$ and thus $I^n \subset Ann(m)$).

Here is a counter example for a non-noetherian Ring:

Let $R=k[x_1, x_2, \dotsc]$ be the polynomial ring in infinitely many indeterminates. Let $I=(x_1, x_2, \dotsc)$ and $M=R/(x_1,x_2^2,x_3^3,\dotsc)$. Consider $1 \in M$. We have $I^n \cdot 1 \neq 0$ for any $n \geq 1$, so $1 \notin \Gamma_I(M)$ according to the first definition. But we have $Supp(R \cdot 1) = V(J) = V(\sqrt{J})=V(I)$, so $1 \in \Gamma_I(M)$ according to the second definition.

Answer 2

The two functors are equal if the ideal $I$ is of finite type, but not in general (as illustrated by MooS's example). However, they may be equal if $I$ is not of finite type.

If $I$ is not of finite type, then the first functor need not be a radical, but the second functor is a radical.