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Let $(R,\mathfrak{m})$ be a local ring and $N$ be a $R-$module such that every element of $N$ is killed by a power of $\mathfrak{m}$. Show that $N$ has at least a simple submodule, that is $$soc(N)\ne0.$$

This implies a question : When does a $R-$module $N$ have simple submodule? ( Let $R$ be a commutative unitary ring that need not local)

Can you help me.

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up vote 3 down vote accepted

Let $R$ be a commutative unitary ring that need not be local. Let $N$ be a nonzero $R$-module.

Suppose there exists a nonzero element $x$ of $N$ and a maximal ideal $\mathfrak{m}$ of $R$ such that $x$ is killed by a power of $\mathfrak{m}$. Let $I$ be ann$(x)$. Then $Rx$ is isomorphic to $R/I$. By the assumption, there exists an integer $n > 0$ such that $\mathfrak{m}^n \subset I$. We can assume that $n$ is the least such integer. There exists $a \in \mathfrak{m}^{n-1} - I$. Since $\mathfrak{m}a \subset \mathfrak{m}^n \subset I$, ann($a$ mod $I) = \mathfrak{m}$. Hence $R/I$ contains a submodule isomorphic to $R/\mathfrak{m}$. Hence $Rx$ contains a simple submodule.

Conversely suppose $N$ has a simple submodule $L$. $L$ is of the form $Rx$, where $x$ is a nonzero element of $N$. Let $\mathfrak{m}$ be ann($x$). Since $L$ is isomorphic to $R/\mathfrak{m}$, $\mathfrak{m}$ is maximal. Clearly $\mathfrak{m}x = 0$.

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@ Makoto Kato: Perfect. $\mathfrak{m}^n\subset I\subset \mathfrak{m}$ implies $R/I$ is artinian, hence has a minimal ideal. How do you think about the last question? – tlquyen Nov 9 '12 at 17:47
@tlquyen Since $R$ may not be Noetherian, $R/I$ may not be Artinian. I did not assume that $R$ is a local ring. So I answered your last question. – Makoto Kato Nov 9 '12 at 17:53
I will show that $R/I$ is artinian if $\mathfrak{m}^n\subset I$. – tlquyen Nov 10 '12 at 2:21
I actually proved that $R/I$ is artinian although $R$ may not be noetherian. Do I have a confusion somewhere? – tlquyen Nov 10 '12 at 3:04
@tlquyen Let $k$ be a field. Let $R = k[x_1,x_2,\dots]$ be the polynomial ring with infinitely many indeterminates. Let $\mathfrak{m} = (x_1,x_2,\dots)$. Since $R/\mathfrak{m}$ is isomorphic to $k$, $\mathfrak{m}$ is maximal. Since $x_1$ (mod $\mathfrak{m}^2), x_2$ (mod $\mathfrak{m}^2), \dots$ are linearly independent over $k$, dim$_k \mathfrak{m}/\mathfrak{m}^2 = \infty$. Hence $R/\mathfrak{m}^2$ is not artinian. – Makoto Kato Nov 10 '12 at 3:35

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