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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. –  Mariano Suárez-Alvarez Nov 15 '10 at 15:18
    
@Mariano: If $x\in K$ then, in your description, $b\in R-{0}$ right? –  Timothy Wagner Nov 15 '10 at 15:19
    
I write $\setminus$ (which is gotten with \setminus in $\LaTeX$) to denote set difference. You seem to be writing $-$ for the same thing. –  Mariano Suárez-Alvarez Nov 15 '10 at 15:21
    
@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. –  Timothy Wagner Nov 15 '10 at 15:23
    
Yes, that is a typo, but I cannot edit the comment now :( –  Mariano Suárez-Alvarez Nov 15 '10 at 15:25
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1 Answer

up vote 1 down vote accepted

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. –  Timothy Wagner Nov 15 '10 at 19:00
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