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.

The question is very simple: Do you know an easy proof of the following: Let $I$ be an injective $R$-module and $a\subset R$ an ideal. Then the localized module $ I_a$ is again injective.

Of course we are working in commutative algebra. I know a long proof using cohomology with compact suport, where you just have to prove that the kernel of the localization morphism, or the $a$-torsion of $I$, $\Gamma_a(I)=\{m\in I | a^t m=0 \text{ for some } t\in\mathbb{N}\}$ is injective.

share|improve this question
    
By the way, wikipedia says that Rotman gave a wrong proof in his book Introduction to homological algebra... xd –  Miguel Dec 26 '12 at 16:45
1  
Have you taken a look at theorem 4.88 in page 200 of the second edition of Rotman's book? He proves that if $R$ is noetherian, then any localization of an injective $R$-module is an injective module. He then acknowledges in the following remark that the assertion is false if $R$ is not noetherian, and points to a counterexample of Dade (the one pointed out in the wikipedia article). –  lentic catachresis Dec 26 '12 at 16:54
    
Ok I was always thinking in the noetherian case, yes. Anyway, when you talk about local cohomology, the rings are noehterian, and the proof is not wrong, you can have a look at it at Brodman's book in local cohomology. Bruno: I had a different edition of Rotman's book and haven't found the proof, thanks! –  Miguel Dec 28 '12 at 0:11
    
@Miguel How should we know that you are thinking in the noetherian case? Moreover, local cohomology functors can be defined for any commutative ring. Last, but not least, I said that the proof must be wrong without the noetherian condition. (Anyway, still don't know what means $I_a$.) –  user26857 Dec 28 '12 at 17:33
    
As I said, $I_a$ is the localized module. Given an ideal $a\in R$, you can localize any $R$-module, just considering the multiplicatively closed system $R-a$. Take a look at Atiyah McDonnald! –  Miguel Dec 30 '12 at 13:13

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.