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I am asked to show that if $M$ is an essential extension of $N$, then $S^{-1}M$ is an essential extension of $S^{-1}N$.

So this is how I approached the problem:

Let $U\neq (0)$ be a submodule of $S^{-1}M$. Then we can write $U = S^{-1}V$ for some submodule $V$ of $M$. Since $U\neq (0)$, $V\neq (0)$. Since $M$ is an essential extension of $N$, and $V$ is non-zero submodule of $M$, $V\cap N\neq (0)$. So choose $x\neq 0$ in $V\cap N$.

My problem is now trying to get a non-zero element of $U\cap S^{-1}N$. I initially thought I could just use $x/1$ but I do not know that this is non-zero. So now I am stuck.

Any suggestions?

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    We need to know a few things in order to get started. You are working with modules over some ring $R$. Can $R$ be any ring, or does it have some special properties? Secondly, what about the multiplicatively closed set $S$? Is it arbitrary, or are the elements of $S$ of a particular type, say regular elements of $R$? I'm not sure the result is true arbitrarily. Goodearl and Jordan have addressed this issue in their paper "Localizations of Essential Extensions", Proc. Edinburgh Mathematical Society (1988) 31, 243-247. See in particular Theorem 4. Perhaps someone has more up-to-date info on this.2012-02-29
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    I'm not sure why I didn't include this. My apologies. $R$ is a Noetherian commutative ring with identity, and $S$ is multiplicatively closed in $R$. That is, $1\in S$, and $a,b\in S$ force $ab\in S$. Anyway, this is due in an hour so I'll have to wait and see the solutions. Sorry again for posting incomplete hypothesis. That is embarassing. :S2012-02-29
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    No problem. In your case, if $S$ consists of regular elements of $R$, then the result is certainly true, and Goodearl and Jordan treat this case in detail, without the commutativity assumption.. When you get the solution, I'd be interested in seeing how it goes.2012-02-29

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