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Given a $1$ dimensional domain $R$ with fraction field $K$. Suppose $\mathfrak{m}$ is a maximal ideal of $R$. Then, I need to show that $R_m$ is a maximal subring of $K$ in the sense that if $x\in K-R_\mathfrak{m}$, then, $R_m[x]=K$.

1)Is this true? How would you go about showing this?

2)If this is true, can any of the hypothesis be relaxed (domain, $1$ dimensional,"at a maximal ideal")?

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    Write $x$ as $a/b$ with $a\in R$ and $b\in R\setminus\mathfrak m$, and see what $R_\mathfrak m[x]$ is.2010-11-15
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    @Mariano: If $x\in K$ then, in your description, $b\in R-{0}$ right?2010-11-15
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    I write $\setminus$ (which is gotten with `\setminus` in $\LaTeX$) to denote set difference. You seem to be writing $-$ for the same thing.2010-11-15
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    @Mariano: Yes I got that. I am saying for an $x$ in $K$ and not in $R$ the "denominator" should only be restricted to be nonzero whereas you wrote that the denominator should be restricted to be in the complement of the maximal ideal.2010-11-15
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    Yes, that is a typo, but I cannot edit the comment now :(2010-11-15
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    I still don't get it. I need to show, $1/c \in R_m[x]$ for any nonzero $c \in R$. I don't see why this is true.2010-11-15
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    @Tim: Why do you think it need be true? What is the source of the problem?2010-11-15
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    This is from lemma 3.8, page 32 in the following notes http://websites.math.leidenuniv.nl/algebra/ant.pdf. Maybe it also needs an additional hypothesis that $R_m$ is a DVR.2010-11-15
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    @Tim: Yes, that suffices. Since someone just explained nicely how it fails generally I believe you are all set now.2010-11-15

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No, I am afraid this maximality is not true.

Let us always suppose that $R$ is a domain: else there is no fraction field (although there are substitutes).

First, the result is certainly not true for higher dimensional rings. For example take $R=\mathbb C[x,y] , m=(x,y) , K=\mathbb C(x,y)$ . Then you have strict inclusions

$$\mathbb C[x,y]_m \subset \mathbb C[x,y]_m [1/x] \subset \mathbb C (x,y)$$

Hence we see that the result in your case is not purely formal and we must use dimension one. But unfortunately the result may not be true even then, which answers question 1) negatively.

Indeed, just observe in the case $R=\mathbb C[[t^2,t^3]], \; m=(t^2,t^3), \; R=R_m$ the strict inclusions $$R=\mathbb C[[t^2,t^3]] \subset \mathbb C [[t]]\subset \mathbb C ((t))$$

To end on a cheerful note, your result is indeed true if $R_m$ is a discrete valuation ring. Here is the proof. Take $a\in K\setminus R_m$ and consider $R_m\subset R_m [a]\subset K$ ( the first inclusion is strict). Since $a$ is not in $R_m$, we know that $1/a$ *is* in $R_m$: this is characteristic of valuation rings. Hence $1/a$ is a fortiori in $R_m[a]$ so that $1\in R[a]$ and we have proved $R[a]=K$.

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    This is awesome. I figured about the DVR case, but was still wondering in general. Thanks for the neat examples.2010-11-15