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Let $M$ be a module over a commutative ring $R$, $\mathfrak a$ is an ideal of $R$.

Define $\Gamma_\mathfrak a(M)=\lbrace m\in M\mid\mathfrak a^tm=0 \text{ for some } t\in \mathbb{N}\rbrace$.

Then $\Gamma_\mathfrak a$ is a functor (from the category $R-\operatorname{mod}$ to $R-\operatorname{mod}$). Some textbook in commutative algebra claimed that $\Gamma_\mathfrak a$ is left exact but not right exact, but I could not prove that.

Please help me to prove and better help me to find an example that $\Gamma_\mathfrak a$ is not right exact.

Edit. Thanks to the answer of YACP, but I think I should make my question more precise:

Suppose that I have an exact sequence of the form $0\rightarrow N\rightarrow M\rightarrow P\rightarrow 0$. After taking the action of $\Gamma_{\mathfrak{a}}$, why we can not get a short exact sequence $0\rightarrow \Gamma_{\mathfrak{a}}(N)\rightarrow \Gamma_{\mathfrak{a}}(M)\rightarrow \Gamma_{\mathfrak{a}}(P)\rightarrow 0$ ? Why is it not exact at $\Gamma_{\mathfrak{a}}(P)\rightarrow 0$?

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Take $0\to\mathfrak{a}\to R\to R/\mathfrak{a}\to 0$, where $R$ is an integral domain and $\mathfrak{a}\subset R$ a nonzero ideal. Then $\Gamma_\mathfrak a(\mathfrak a)=\Gamma_\mathfrak a(R)=0$ and $\Gamma_\mathfrak a(R/\mathfrak a)=R/\mathfrak a$.

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Could you please make it more precise ? What is $\Gamma_{\mathfrak{a}}(\mathfrak{a})$, $\Gamma_{\mathfrak{a}}(R/\mathfrak{a})$ ? – user51685 Dec 4 '12 at 15:06
So, could you please show me an example that $\Gamma_{\mathfrak{a}}$ send an exact sequence of the form : $0\rightarrow N\rightarrow M\rightarrow P\rightarrow 0$ to $0\rightarrow \Gamma_{\mathfrak{a}}(N)\rightarrow\Gamma_{\mathfrak{a}}(M)\rightarrow \Gamma_{\mathfrak{a}}(P)$ ? That is, the exact sequence can not be right exact ? – user51685 Dec 4 '12 at 15:16
I think your answer gives me the exact sequence(after taking the action of $\Gamma_\mathfrak a$ : $0\rightarrow R/\mathfrak{a} \rightarrow 0$, which do not give us the above form. – user51685 Dec 4 '12 at 15:27
I want an example of exact sequence : $0\rightarrow N\rightarrow M\rightarrow P\rightarrow 0$ such that after taking the action of $\Gamma_{\mathfrak{a}}$, the module $\Gamma_{\mathfrak{a}}(N), \Gamma_{\mathfrak{a}}(M), \Gamma_{\mathfrak{a}}(P)$ are different from $0$ and from that we can see that after taking the action of $\Gamma_{\mathfrak{a}}$, we can not get the exact sequence $0\rightarrow \Gamma_{\mathfrak{a}}(N)\rightarrow \Gamma_{\mathfrak{a}}(M)\rightarrow \Gamma_{\mathfrak{a}}(P)\rightarrow 0$ . – user51685 Dec 4 '12 at 15:59
@user51685 $0\rightarrow R/\mathfrak{a} \rightarrow 0$ is clearly not exact: the kernel of the right morphism is $R/\mathfrak{a}$ but the image of the left morphism is $\mathfrak{a}/\mathfrak{a}$. ( YACP clearly intended $\mathfrak{a}$ to be unequal to $R$ so that $R/\mathfrak{a}$ is nontrivial.) – rschwieb Dec 4 '12 at 16:16

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