Timeline for $R = \mathbb{Z}[ i ] / (5)$ is not an integral domain? Why?
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| Apr 13, 2017 at 12:20 | history | edited | CommunityBot | replaced http://math.stackexchange.com/ with https://math.stackexchange.com/ | |
| Dec 14, 2012 at 18:33 | history | edited | user26857 | CC BY-SA 3.0 | edited body |
| Dec 13, 2012 at 20:49 | comment | added | Cameron Buie | 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. | |
| Dec 13, 2012 at 20:44 | comment | added | user52937 | 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? @Cameron Buie | |
| Dec 13, 2012 at 20:38 | comment | added | user52937 | 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? | |
| Dec 13, 2012 at 20:31 | comment | added | Cameron Buie | You should be able to prove that each ideal of $R$ is principal, and using the result from this related answer in conjunction with Third Isomorphism Theorem, you should be able to prove that both non-trivial ideals of $R$ are prime (maximal) ideals, since $\Bbb Z/\langle 5\rangle$ is an integral domain (a field). | |
| Dec 13, 2012 at 20:28 | comment | added | user52937 | Are they prime ideals? | |
| Dec 13, 2012 at 20:24 | comment | added | Cameron Buie | The ideals of $R$ are precisely those of the form $I/\langle 5\rangle$, where $I$ is an ideal of $\Bbb Z[i]$ containing $(5)$. This is a consequence of what is sometimes referred to as the Correspondence Theorem. Since $\Bbb Z[i]$ is a PID, then the ideals of $\Bbb Z[i]$ that contain $\langle 5\rangle$ are precisely: $\langle 5\rangle$, $\langle 2+i\rangle$, $\langle 2-i\rangle$, and $\langle 1\rangle=\Bbb Z[i]$. Hence, $R$ has only $4$ ideals: $\langle 0\rangle$, $\langle 2+i\rangle/\langle 5\rangle$, $\langle 2-i\rangle/\langle 5\rangle$, and $R$. | |
| Dec 13, 2012 at 20:16 | history | edited | Cameron Buie | CC BY-SA 3.0 | added 588 characters in body |
| Dec 13, 2012 at 20:15 | comment | added | user52937 | What are the ideals of R = Z[i] / (5)? | |
| Dec 13, 2012 at 20:08 | history | answered | Cameron Buie | CC BY-SA 3.0 |