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Let $R$ be a commutative ring with $1$, and let $M$ be a cyclic $R$-module with generator $m$ (so that $M=Rm$). Suppose that $M = N_1 \bigoplus N_2$ for some submodules $N_1$ and $N_2$ of $M$. Let $n_1 \in N_1$ and $n_2 \in N_2$ such that $m = n_1 + n_2$. Show that if $R$ is commutative, then $\mathrm{Ann}(n_1) + \mathrm{Ann}(n_2) = R$.

user26857
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user10024395
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    Don't ask people not to ask you to show your effort. Regardless of how much strain you personally feel like you put in, you will still come across as someone who hasn't tried anything. You will then be less likely to get help. There is a minimal amount of effort EVERYONE can do: copy the statement of a possibly relevant definition or theorem from the book and then explain that you didn't have any good ideas about how to proceed. – Barry Smith Apr 20 '15 at 12:45

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$am=0\Rightarrow an_1+an_2=0\Rightarrow an_1=an_2=0$ (why?). So $\mathrm{Ann}(m)=\mathrm{Ann}(n_1)\cap\mathrm{Ann}(n_2)$. Moreover, $N_i=Rn_i$ for $i=1,2$. We have $R/\mathrm{Ann}(m)\simeq Rm$, $Rm=N_1\dotplus N_2$, and $N_1\dotplus N_2\simeq R/\mathrm{Ann}(n_1)\oplus R/\mathrm{Ann}(n_2)$, therefore $$R/\mathrm{Ann}(n_1)\cap\mathrm{Ann}(n_2)\simeq R/\mathrm{Ann}(n_1)\oplus R/\mathrm{Ann}(n_2).$$

This shows that your question is in fact a converse to CRT, and for a proof see this answer.

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user26857
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    I can't see how it is related. There is no gcd involved here – 123 Apr 20 '15 at 13:21
  • Can you explain how this will tell me $Ann(n_1) + Ann(n_2) = R$? I can't see how CRT will help me. – user10024395 Apr 20 '15 at 13:24
  • The link proves that $ A+B = R$ iff $R/AB \cong R/A \times R/B$ and u go on to say that $R/I \times R/J $is cyclic iff $I+J = R$. I think what I need here is the second part, but I don't understand why the first part implies the second part. Is R/AB cyclic? – user10024395 Apr 20 '15 at 13:36
  • @Aha The proof for the first part applies here no matter that we have $A\cap B$ instead of $AB$. (Just tensor $R/A\cap B\simeq R/A\times R/B$ by $R/M$, where $M$ is a maximal ideal containing $A+B$.) – user26857 Apr 20 '15 at 13:38
  • why is $N_1 + N_2 \cong R/Ann(n_1) \bigoplus R/Ann(n_2)$? – user10024395 Apr 20 '15 at 14:24
  • @Aha Actually it is $N_1\dotplus N_2$ (the internal direct sum; if you look carefully can see a dot over that plus) and this is isomorphic to the external direct sum $N_1\oplus N_2$. Then I've used $N_i=Rn_i\simeq R/Ann(n_i)$. – user26857 Apr 20 '15 at 17:43