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Let $R = \mathbb{Z}[ i ] / (5)$ .

How should I prove that $5 = (2+i) (2-i)$ is a prime factorization in $\mathbb{Z}[i]$? Can we deduce from this that R is not an integral domain? How?

I know that we can prove any ideal in R is principal.

Now I want to prove the classification theorem for modules over $R$ :

There exist modules $M_1, M_2$ such that any finitely generated module $M$ over $R$ is isomorphic to the direct sum $M_1^r \oplus M_2^s$, where $M_1^r$ is the direct sum of $r$ copies of module $M_1$, and similarly for $M_2$.

I notice that $R$ is not an PID...........

Do you have any ideas how to prove this?


  • 3
    What have you tried so far? Do you know about the Chinese Remainder Theorem in (say) a PID? This tells you something about the structure of $R/(p_1 p_2)$ where $p_1$ and $p_2$ are nonassociate prime elements.2012-12-13
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    What is the Chinese Remainder Theorem in a PID?2012-12-13
  • 0
    [Wikipedia](http://en.wikipedia.org/wiki/Chinese_remainder_theorem#Statement_for_principal_ideal_domains) might be a good start.2012-12-14
  • 0
    Why did you post the same question splitted in two? Duplicate of http://math.stackexchange.com/questions/258421/the-classification-theorem-for-modules-over-mathbbzi-5/258869#258869 and http://math.stackexchange.com/questions/258523/why-is-every-ideal-in-mathbb-z-i-5-principal2012-12-14

3 Answers 3