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

It is obvious that $R$ is not an integral domain, and any ideal in $R$ is principal.

Now I want to prove the following 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 ⊕ M_2^s$, where $M_1^r$ is the direct sum of r copies of module $M_1$, and similarly for $M_2$ .

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I've removed algebra tag, since we don't use algebra tag anymore, see meta for details. –  Martin Sleziak Dec 14 '12 at 8:09

2 Answers 2

$R=\mathbb Z[i]/(5)$ is isomorphic to $\mathbb Z[i]/(2-i)\times\mathbb Z[i]/(2+i)$. But $\mathbb Z[i]/(2-i)$ and $\mathbb Z[i]/(2+i)$ are finite fields with $5$ elements and this allows us to say that $R\simeq\mathbb F_5\times\mathbb F_5$. Thus any finitely generated $R$-module is a finitely generated $\mathbb F_5\times\mathbb F_5$-module. Let $M$ be a finitely generated $\mathbb F_5\times\mathbb F_5$-module. Then $M=(1,0)M\oplus(0,1)M$ and $(1,0)M$ and $(0,1)M$ are f.g. $\mathbb F_5$-modules. As a consequence, $M\simeq\mathbb F_5^r\times\mathbb F_5^s$

Remark. This question rise a harder problem: what can be said about the structure of finitely generated modules over principal ideal rings (which are not necessarily domains!)? For a good start see this topic.

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To begin with, we can smash apart the ring: $$\mathbb{Z}[i]/5 \cong \mathbb{Z}[x]/(5, x^2 + 1) \cong \mathbb{F}_5[x]/(x^2 + 1) \cong \mathbb{F}_5[x]/(x-3) \times \mathbb{F}_5[x]/(x - 2)$$

So our ring is just 2 points (i.e. it has 2 primes ideals, $\mathfrak{p}, \mathfrak{q}$ corresponding to $(\mathbb{F}_5, 0)$ and $(0, \mathbb{F}_5)$), any module is isomorphic to the direct sum of its stalks over the points, which are just $\mathbb{F}_5$ vector spaces with $x$ acting as multiplication by $2$ and $3$.

This is kinda not the greatest answer...I can probably give a more hands on argument, if you haven't seen localization and stuff before, lemme know! All the best, uncookedfalcon

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Direct sum of rings doesn't make sense... But yes this is the right idea. –  Dylan Wilson Dec 14 '12 at 3:40
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And you should probably explain to Mike what you mean by the "ring" being "two points" :) –  Dylan Wilson Dec 14 '12 at 3:43

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