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Let $R$ be an integral domain and let $S$ be a ring with $R \le S \le \text{Frac}(R)$ (fraction field).

Question: Is there a multiplicatively closed subset $U \subseteq R\setminus \{0\}$ such that $S=R[U^{-1}]$ ?

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closed as off-topic by user26857, Watson, gebruiker, Jon Mark Perry, MXYMXY Mar 12 at 16:39

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In the case that $R$ is a PID this is proven in my answer here. – user38268 Jan 26 '13 at 11:41
up vote 12 down vote accepted

There are lots of examples arising from integral closures. For example, $k[x^2,x^3] \subseteq k[x]$ with field of fractions $k(x)$ is no localization since $k[x^2,x^3]^* = k^* = k[x]^*$.

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+1 for clarity, conciseness and generalizability. – Georges Elencwajg Jan 26 '13 at 13:12
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@BenjaLim It's the group of units. The argument is that since the units of $R = k[x^2,x^3]$ are the same as the units of $S = k[x]$, the ring $S$ cannot arise from adjoining fractions to $R$, as this would enlarge the set of units. (At least, that's what I understood.) – k.stm Jan 26 '13 at 14:09
    
@K.Stm. But unless I am misunderstanding martin's example, shouldn't we want a ring containing $k[x]$ and not one that $k[x]$ contains? – user38268 Jan 26 '13 at 14:11
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@BenjaLim The field of fractions of $k[x^2,x^3]$ is $k(x)$ as well, since $x^2/x^3 = 1/x$, so $k[x^2,x^3]$ is our base ring $R$ and $k[x]$ is the extension $S$. – k.stm Jan 26 '13 at 14:13
    
@BejaLim Me too! I really learned something from it. : - ) – k.stm Jan 26 '13 at 14:26

I've shown here that the ring $S=K[X]+YK(X)[Y]$ is not noetherian. Note that we have $K[X,Y]=R\subset S\subset Q(R)=K(X,Y)$. Since $R$ is noetherian, $S$ can't be a localization of $R$.

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Let $U$ be the multiplicative set $\{r\in R\mid \frac 1r\in S\}$, which is the "obvious" (and maximal) candidate. Then clearly $R[U^{-1}]\subseteq S$. But is $S\subseteq R[U^{-1}]$? Can we conclude $\frac1b\in S$ from $\frac ab\in S$? Note how we would e.g. conclude $\frac13\in S$ from $\frac23\in S$: We'd use $2\cdot \frac23-1=\frac13$. Such is not possible in general.

Let $R=\mathbb Z[X]$ and $S=R[\frac X2]$. Then if we assume $S=R[U^{-1}]$, there must be $g\in U$, $f\in R$ such that $\frac X2=\frac fg$, i.e. $2f=Xg$. Since also $\frac 1g\in S\subseteq \mathbb Q[X]$, necessarily $g$ is constant, i.e. $g=2k$ with $k\in\mathbb Z\setminus \{0\}$. but $\frac1{2k}\notin S$ because the image of the homomorphism $S\to\mathbb Q$ induced by $X\mapsto 0$ is in $\mathbb Z$.

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Do you mean $S=\mathbb{Z}[X/2]$? – Martin Brandenburg Jan 26 '13 at 12:04
    
@MartinBrandenburg: Since $X\in\Bbb Z[\frac X2]$ it makes no difference with what Hagen says. I suppose he just wanted to emphasise that $S\supset R$. – Marc van Leeuwen Jan 26 '13 at 17:33

Another class of counterexamples (to which Martin's example belongs to) is:

If $R \lneqq S \le \text{Frac}(R)$ is finitely generated as $R$-module, then $S$ is no localization of $R$.

Proof: Suppose $S=R[U^{-1}]$ is generated by $s_i \in S\;(i=1,\ldots,n)$. By taking a common denominator, we can write $s_i=r_i/u$ with $r_i \in R, u \in U$. Thus $S$ is generated by $1/u$ over $R$. Therefore we find $r \in R$ such that $1/u^2=r/u$, i.e. $1/u=r$ is a unit in $R$. This yields $S=R(1/u) = Rr=R$. q.e.d.

In combination with BenjaLim's comment we obtain the corollary:

If $R$ is a PID, then no ring $R \lneqq S \le \text{Frac}(R)$ is finitely generated over $R$.

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