Direct-Sum Decomposition of an Artinian module Let $R$ be a commutative Noetherian ring. Suppose $M$ is a finitely-generated non-zero Artinian $R$-module. 
Question: How can we prove that there are maximal ideals $m_1 , m_2 , \ldots , m_n$ such that :
$$M = \Gamma_{m_1} (M)\oplus\Gamma_{m_2} (M)\oplus\cdots\oplus \Gamma_{m_n} (M)$$
where $\Gamma_{I} (M) = \bigcup_{\ell\geq0} \{x \in M : \,  I^{\ell} x =0\}$?
 A: First of all, since $M$ is Artinian and Noetherian, its Krull dimension is zero. Hence all associated primes of $M$ are maximal ideals of $R$. Moreover, these finitely many associated primes - let's call them $m_1,\dots,m_n$ - constitute the entire support of $M$. The above remarks imply that $S := R/Ann(M)$ is Artinian ring and $Spec(S)=\left\{\bar{m}_1,\dots,\bar{m}_n\right\}$, where $\bar{m}_i$ is the image of $m_i$ in $S$.
Now let us prove the statement for the case where our module is $S$. We first prove that the sum
$\Gamma_{\bar{m}_1}(S) + \cdots + \Gamma_{\bar{m}_n}(S)$ is direct. So suppose
$0 \neq \xi \in \Gamma_{\bar{m}_i}(S) \cap (\Gamma_{\bar{m}_1}(S)+\cdots+\Gamma_{\bar{m}_{i-1}}(S))$. Then the annihilator of $\xi$ contains a power of $\bar{m}_i$. Since $\xi \neq 0$, $Ann(\xi)$ must be inside some maximal ideal, which contains a power of $\bar{m}_i$. We conclude that $Ann(\xi) \subset \bar{m}_i$. But also
$Ann(\xi)$ contains $\bar{m}_1^{j_1}\cdots \bar{m}_{i-1}^{j_{i-1}}$ for some powers $j_{s}$. This implies that $\bar{m}_i \supset \bar{m}_1^{j_1}\cdots \bar{m}_{i-1}^{j_{i-1}}$, which is a contradiction.
Next we show that $S = \Gamma_{\bar{m}_1}(S) \oplus \cdots \oplus \Gamma_{\bar{m}_n}(S)$.
Since $S$ is Artinian, every descending chain $\bar{m}_1 \supset \bar{m}_1^2 \supset \bar{m}_1^3 \supset \cdots$ must stabilize after finitely many steps. Hence there exist
positive integers $\ell_i$ such that $\bar{m}_i^{\ell_i} = \bar{m}_i^{\ell_i+1}$. Localizing at $\bar{m}_i$, Nakayama's Lemma gives $\bar{m}_i^{\ell_i} = 0$ in $S_{\bar{m}_i}$. Thus there must exist $\bar{s}_i \not\in \bar{m}_i$, such that 
$\bar{m}_i^{\ell_i} \bar{s}_i = 0$ or in other words $\bar{s}_i \in \Gamma_{\bar{m}_i}(S)$. 
Now define $\bar{s} = \bar{s}_1+\cdots+\bar{s}_n$. We will show that $\bar{s}$ is a unit of $S$. Suppose not. Then without loss of generality we can assume that $\bar{s} \in \bar{m}_1$. Now by definition $\bar{m}_1^{\ell_1} \bar{s}_1 =0$ and so $\bar{s}^{\ell_1} \bar{s}_1 = 0$. Note also that $\bar{s}_i \bar{s}_j \in \Gamma_{\bar{m}_i}(S) \cap \Gamma_{\bar{m}_j}(S)$ and so $\bar{s}_i \bar{s}_j =0$ whenever $i \neq j$. This implies 
that $\bar{s}^{\ell_1} = \bar{s}_1^{\ell_1}+\bar{s}_2^{\ell_1}+\cdots + \bar{s}_n^{\ell_1}$. Then the equality $\bar{s}^{\ell_1} \bar{s}_1 = 0$ implies that
$\bar{s}_1^{\ell_1+1}=0$. Since $\bar{s} \in \bar{m}_1$, we must also have that $\bar{s}^{\ell_1+1} \in \bar{m}_1$. 
But $\bar{s}^{\ell_1+1} = \bar{s}_1^{\ell_1+1}+\bar{s}_2^{\ell_1+1}+\cdots + \bar{s}_n^{\ell_1+1}=\bar{s}_2^{\ell_1+1}+\cdots + \bar{s}_n^{\ell_1+1}=(\bar{s}_2+\cdots+\bar{s}_n)^{\ell_1+1}$ and so we must have that
$(\bar{s}_2+\cdots+\bar{s}_n)^{\ell_1+1} \in \bar{m}_1$. This in turn implies
that $\bar{s}_2+\cdots+\bar{s}_n \in \bar{m}_1$. But since our hypothesis is 
that $\bar{s}_1+\cdots+\bar{s}_n \in \bar{m}_1$, this implies 
that $\bar{s}_1 \in \bar{m}_1$, which is a contradiction. 
Since $\bar{s}$ is a unit of $S$, there must exist some $\bar{t} \in S$ 
such that $\bar{t} \bar{s} = \bar{1}$. Equivalently, there must exist an
element $r \in Ann(M)$ such that $1 = ts +r$.
Now take any $x \in M$. Then $x = 1 \cdot x = (ts+r)x = (ts)x=
(ts_1)x+\cdots + (ts_n) x$. It remains to prove that 
$(ts_i)x \in \Gamma_{m_i}(M)$. We know that $m_i^{\ell_i} s_i \subset Ann(M)$
and so $m_i^{\ell_i} ts_i \subset Ann(M)$. It follows that 
$m_i^{\ell_i} (ts_i)x=0 $ and the proof is concluded.
