# Example of an overring of an integral domain which is not a ring of quotients?

Hi!

I'm trying to make headway on a question for my undergraduate honors thesis, specifically the question of which rings of integer-valued polynomials if any satisfy the QR-property; that is, the property that all overrings are rings of quotients with respect to some multiplicative subsets.

But, this occurred to, which I'm not exactly proud of given the topic of my thesis and the length of time I've worked on it: I can't think of a concrete example of an integral domain that does not satisfy the QR-property and a corresponding overring of it that is not a ring of quotients of that domain. Does anyone here have any good examples, preferably something somewhat concrete I can share with beginning students (since I'll be presenting this at an undergraduate conference)? I'm honestly unsure if this question is too elementary, and if it is, I apologize.

Thanks so much!

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$\mathbb Q\left[X,Y,X/Y\right]$ is an overring of $\mathbb Q\left[X,Y\right]$, yet not a ring of fractions thereof (for example, because $\mathbb Q\left[X,Y,X/Y\right] \cong \mathbb Q\left[Z,Y\right]$ via the isomorphism sending $Z$ and $Y$ to $X/Y$ and $Y$, respectively, and therefore the only invertible elements of $\mathbb Q\left[X,Y,X/Y\right]$ are constants). Or are you looking for some special kinds of rings? –  darij grinberg Apr 10 '13 at 3:03
$R=k[x^2,x^3]$ has $k[x]$ as an overring (assuming "overring" means a ring that lies between $R$ and its field of quotients). –  Steven Landsburg Apr 10 '13 at 3:11
Those are great examples. I particularly like @darij grinberg's example. Thanks! –  Reeve Apr 10 '13 at 4:53
By the way, yes, that is exactly what I mean by overring @Steven Landsburg. –  Reeve Apr 10 '13 at 5:15
In $\S 22.2.2$ of my commutative algebra notes I discuss overrings of Dedekind domains. In particular I discuss a beautiful theorem of Oscar Goldman: in a Dedekind domain $R$, every fractional ideal has some positive integer power which is principal -- in other words $\operatorname{Pic} R$ is a torsion group -- if and only if every overring is a localization.
The proof is quite explicit: if $\operatorname{Pic} R$ is not torsion, there is some prime ideal $\mathfrak{p}$ of $R$ no power of which is principal, and then the argument shows that $R^{\mathfrak{p}} = \bigcap_{\mathfrak{q} \in \operatorname{MaxSpec} R, \ \mathfrak{q} \neq \mathfrak{p}} R_{\mathfrak{q}}$ is not a localization: indeed, it is a proper overring of $R$ with the same group of units.