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?
Best Answer
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.