Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\Gamma_{\mathfrak{a}}(P)\rightarrow 0$?

share|improve this question
    
What's the difference between the original question and the edited one? –  user26857 Dec 5 '12 at 18:20
    
No, I just want to make it more clear. –  user51685 Dec 6 '12 at 3:45

1 Answer 1

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 $\Gamma_\mathfrak a(R/\mathfrak a)=R/\mathfrak a$.

share|improve this answer
    
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
2  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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