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Suppose $M$ is module and $N$ a submodule over some commutative ring $R$. If $x\neq 0$, and $(x)\cap N=0$, why does this imply $(x)$ is isomorphic to some submodule of $M/N$? (Let's also assume that $M\neq 0$ and $N$ is a proper submodule.)

It was a quick detail in a passage I was reading, and I can't recall why it is.

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What about the case $M=0$? Do you require $N$ to be a proper submodule? –  Alex Becker Feb 19 '12 at 10:33
    
@AlexBecker Sure, this is related to associated primes, so I'll say $M\neq 0$ and $M/N\neq 0$ as well. –  ricky rubio Feb 19 '12 at 10:45
    
My ring theory is a little rusty, to say the least, but if you let $K$ be the submodule generated by $x$ and $N$, then doesn't the "3rd isomorphism theorem" tell you that $(x) \cong K/N$? –  William DeMeo Feb 19 '12 at 11:06
    
@williamdemeo How would that follow exactly? I don't see why $(x)\cap N=0$ is needed in that case. –  ricky rubio Feb 19 '12 at 11:08
    
@ricky Because the isomorphism theorems tell you that $(x)/((x)\cap N) \cong K/N$. If it's still not clear, let me know and I'll elaborate in a proper answer. –  William DeMeo Feb 19 '12 at 11:14
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The general theorem is this (Hungerford p. 179):

If $B$ and $C$ are submodules of a module $A$ over a ring $R$ then there is an $R$-module isomorphism

$$B/B\cap C \cong (B+C)/C.$$

That $+$ notation looks weird to me, but i'm not a ring theorist :) I would prefer to use $\langle B, C\rangle$ or $B\vee C$ and to view it as the "join" because the theorem is "obvious" if you draw the lattice of submodules, and because the theorem holds more generally, of course (though for groups make sure $C$ is normal in $\langle B, C\rangle$! :)

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Here's a sketch of a quick proof:

If $\pi: M \to M/N$ denotes the canonical projection and we set $\overline{x} = \pi(x)$, then $\text{im}(\pi|_{(x)}) = (\overline{x})$, and the latter is a submodule of $M/N$. Now, $\text{ker}(\pi|_{(x)}) = \text{ker}(\pi) \cap (x) = N \cap (x) = 0$, so $\pi|_{(x)}: (x) \to (\overline{x})$ is an isomorphism.

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