Prove that $\Gamma_I(\frac{M}{\Gamma_I(M)})=0$

Question

I was trying to prove this theorem (problem):

Suppose that $R$ is a commutative ring with identity, $I\unlhd R$, and $M$ an $R$-module. We define: $$\Gamma_I(M)=\bigcup_{n\geq0}\operatorname{Ann}_M(I^n)$$ in which for each natural $n\geq 0$: $$\operatorname{Ann}_M(I^n)=\{x\in M\;;\;I^nx=0\}.$$ Prove $$\Gamma_I\left(\frac{M}{\Gamma_I(M)}\right)=0.$$

Note that $\Gamma_I(\cdot)$ will be defined for any $R$-module naturally. I can show $\Gamma_I(M)\leq M$, but couldn't prove the latter claim. please show me how to prove this theorem.

Thanks in advance.

Answer 1

This is not necessarily true, unless there is some finiteness condition like $R$ being noetherian. I will give a proof assuming $I$ is finitely generated, and a counterexample otherwise.

Let $N=\frac{M}{\Gamma_I(M)}$, and $x\in \Gamma_I(N)$. That means $\exists n \geq 0$, such that $x\in \text{Ann}_N(I^n)$. By definition, this means $I^nx=0$ in $N$. The problem is to show $x=0$.

Let $y\in M$ be an element that maps to $x$ under the quotient homomorphism $M \rightarrow N$. Then $x=0$ if and only if $y \in \Gamma_I(M)$. Therefore the problem is to prove if $I^n y \subset \Gamma_I(M)$, then $y\in \Gamma_I(M)$.

$I^n y\subset \Gamma_I(M)$ means for every $r\in I^n$, there exists some $k$ (depending on $r$) such that $ry \in \text{Ann}_M(I^k)$, i.e. $I^kry = 0$. We want to show $I^m y=0$ for some $m \geq 0$.

If we know that $I^n$ is finitely generated (for example if $R$ is noetherian), we may proceed as follows. Pick generators $r_1,\dots ,r_s$ for $I^n$. For each $r_i$, there exists $k_i \geq 0$ such that $I^{k_i}r_i y = 0$. Let $k=\max_{1 \leq i \leq s} k_i$, so that $I^kr_iy = 0$ for every $i$. Then for every $r\in I^n$, $r=a_1 r_1 + \cdots + a_s r_s$ for some $r_i \in R$, and $$I^kry = I^k(a_1 r_1 + \cdots +a_s r_s)y = a_1 I^kr_1y + \cdots +a_s I^kr_s y=a_1.0+\cdots +a_s.0=0.$$

Then $I^kry=0$ for every $r\in I^n$, so $I^{k+n}y=I^k I^ny = 0$. Thus $y\in \text{Ann}_M(I^{n+k})\subset \Gamma_I(M)$, and so $x=0$ in $\frac{M}{\Gamma_I(M)}$, and we are done.

For a counterexample in general, let $R=\prod_{n=3}^\infty \mathbb{Z}/2^n\mathbb{Z}$, and let $R_0$ to be the ideal $\oplus_{n=3}^\infty \mathbb{Z}/2^n\mathbb{Z}$ of eventually-zero vectors in $R$. Take $M=R$, and $I=2R_0$. The element $y=(1,1,1,...)$ in $M$ does not belong to $\Gamma_I(M)$, since for any $n\geq 0$, $I^n y=I^n \neq 0$. But in fact $Iy \subset \Gamma_I(M)$. Indeed, if $r=(r_1,r_2,\dots ,r_s,0,0,...) \in I$, then $ry =r$ is annihilated by $I^{k}=2^kR_0$ whenever $2^k \geq s+2$, so that $ry \in \text{Ann}_{I^k}(M)$, and $Iy\subset \Gamma_I(M)$. In this case $\Gamma_I(M)=R_0$, and two vectors $y_1,y_2\in M$ map to the same element in $M/\Gamma_I(M)$ if and only if they are equal almost everywhere. Since any vector in $M$, once multiplied with an element of $I$ is zero almost everywhere, we have

$$\Gamma_I\left(\frac{M}{\Gamma_I(M)}\right)= \text{Ann}_I\left(\frac{M}{\Gamma_I(M)}\right)=\frac{M}{\Gamma_I(M)}=R/R_0\neq 0.$$

Answer 2

In a comment to Prometheus' answer, Martin mentioned another definition of a torsion functor having the desired property. Let me explain this.

For a ring $R$, an ideal $\mathfrak{a}\subseteq R$ and an $R$-module $M$ we define $$\Gamma'_{\mathfrak{a}}(M):=\{x\in M\mid{\rm Supp}(Rx)\subseteq{\rm Var}(\mathfrak{a})\}.$$ This can be extended to a subfunctor of the identity functor on the category of $R$-modules. It contains $\Gamma_{\mathfrak{a}}$ as defined in the question. The two torsion functors coincide in case $\mathfrak{a}$ is of finite type, but not necessarily in general.

Now, it can be shown that $\Gamma'_{\mathfrak{a}}$ is a so-called radical, i.e., $\Gamma'_{\mathfrak{a}}(M/\Gamma'_{\mathfrak{a}}(M))=0$. (It has also further nice properties that $\Gamma_{\mathfrak{a}}$ loses over non-noetherian rings.)