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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?


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    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

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You don't need to know that $5 = (2-i)(2+i)$ is a prime factorization; all you need is that the two factors are not units.

One way to see that $2-i$ is not a unit is by computation:

$ \mathbb{Z}[i] / (2-i) \xrightarrow{i \to x} \mathbb{Z}[x] / (x^2 + 1, 2-x) \xrightarrow{x \to 2} \mathbb{Z} / (2^2 + 1) \cong \mathbb{Z} / 5 $

(all arrows are isomorphisms). The result isn't the zero ring, so $2-i$ is not a unit. The fact the result is a domain does additionally prove that $2-i$ is prime, though.

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If you'll check out this answer, you'll get an idea how to prove that $\Bbb Z[i]$ is a Euclidean domain. Every Euclidean domain is a PID, and in a PID, the "prime" and "irreducible" elements are the same. Also, in a general ring $R$ with a non-$0$ ideal $I$, we have that $R/I$ is an integral domain if and only if $I$ is a prime ideal of $R$.

Since you know that $2\pm i$ are irreducible in $\Bbb Z[i]$, then in particular, $5=(2+i)(2-i)$ is not irreducible, so not prime. Thus, $\langle 5\rangle$ is not a prime ideal of $\Bbb Z[i]$, and so $\Bbb Z[i]/\langle 5\rangle$ is not an integral domain.

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    I suspect that the $M_1$ and $M_2$ to do the trick will be the two non-trivial ideals of $R$, but I'm not where$I$can play with that and see if it's right. Give it a try, though.2012-12-13
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$\mathbb{Z}[i]$ is a Euclidian domain, so in particular it is also a UFD. We know in UFDs that an element is prime if and only if it is irreducible. If you pull out your trusty copy of Dummit and Foote, you can see on p. 291 that the list of irreducibles in the Gaussian integers includes:

1) $1+i$ and $1-i$

2) Prime integers such that $p \equiv 3\mod 4$

3) If $p \equiv 1\mod 4$ then $p$ can be uniquely expressed (up to signs of $a$ and $b$, and interchanging) as $p = a^2+b^2$ for integers $a$ and $b$, and the elements $a+ib$ and $a-ib$ are irreducible.

An ideal $I$ in a ring $R$ is prime if and only if $R/I$ is an integral domain. You have already noticed that $5$ is not an irreducible element, so $(5)$ is not prime, therefore $\mathbb{Z}[i]/(5)$ is not an integral domain.

Edit: Just noticed you had asked other questions too. Getting to them now.

-What are the ideals in $R$: They're in bijective correspondence with ideals in $\mathbb{Z}[i]$ containing the ideal $(5)$. These should be generated by elements which divide $5$ in $\mathbb{Z}[i]$

-Every ideal of $R$ is principal. See this answer https://math.stackexchange.com/a/90014/29076

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    For modules over R=Z[i]/(5) , I am trying to prove the so-called classification theorem: There should exist modules M_1 , M_2 such that any finitely generated module M over R is isomorphic to the direct sum M_1^r ⊕ M_2^s , where M_1^r is the direct sum of r copies of module M_1, and similarly for M_2 . Do you have any ideas how to do this?2012-12-13