Let $R$ be an integral domain (in particular commutative ring) with identity and let $X$ be a sub field of $R^{frac}$ (fraction field of $R$). Is it true that there exists a sub ring $A$ of $R$ such that $X=A^{frac}$?

  • $\begingroup$ But this is true when $R$ is UFD (or PID). $\endgroup$ – Martin Brandenburg May 29 '12 at 12:23
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    $\begingroup$ @MartinBrandenburg I'm confused, in wxu's counterexample the ring is a UFD. $\endgroup$ – JSchlather May 29 '12 at 13:04

It is not true. Consider $\mathbb{Q}[x,y]\subset \mathbb{Q}(x,y)$, let $K=\mathbb{Q}(\frac{x}{y})$. If there exists a subring $A$ of $\mathbb{Q}[x,y]$, such that the fractional field of $A$ is $K$, then $A\subset \mathbb{Q}[x,y]\cap K=\mathbb{Q}$. Contradiction.

Edit: Here is a proof for $\mathbb{Q}[x,y]\cap K=\mathbb{Q}$. Suppose $f(x,y)\in \mathbb{Q}[x,y]\cap K$, then $f(x,y)=\frac{g(x/y)}{h(x/y)}$ with gcd$(g(t),h(t))=1$. Then we have $f(x,y)h(x/y)=g(x/y)$. If $h(t)$ is not constant, say $\alpha\in \overline{\mathbb{Q}}$ is a root of $h$, then let $x=\alpha, y=1$, we have $0=g(\alpha)\neq 0$. so $h(t)$ is a constant. Hence $f(x,y)=g(x/y)$, now comparing the degree of $y$, it is easy to see $g(x/y)$ must be a constant. We are done.

  • $\begingroup$ There we go, nice and simple. $\endgroup$ – rschwieb May 29 '12 at 11:37
  • $\begingroup$ @rschwieb, thank you~ $\endgroup$ – wxu May 29 '12 at 12:10
  • $\begingroup$ Quite ingenious, wxu! $\endgroup$ – Georges Elencwajg May 29 '12 at 15:39

wxu has given a nice example in above to show the claim is incorrect. To clarify that it is far from being true, the following example show it is not true even if the domain is a principal ideal domain. Let $k$ be any field, $A$ be the polynomial ring $k[X]$ and $F$ be the field $k(X)$ of rational fuctions. For any positive integers $m$ and $n$, take the subfield $X=k(x^m+x^{-n})$. Then we can check that $X\cap A = k$ by some computation. Therefore, this will be another counter-example in which $A$ is even a principal ideal doamin.


This is a comment for ding8191. Let me interpret that $k[x]\cap k(x^m+x^{-n})=k$ for any positive integers $m,n$ with the same method.

Suppose a polyonmial $f(x)=\frac{g(x^m+x^{-n})}{h(x^m+x^{-n})}$ with $g,h$ having no common factor. Then $f(x)h(x^m+x^{-n})=g(x^m+x^{-n})$. Then if $h$ is not a constant, say $\alpha\in \overline{k}$ is a root of $h$, since the equation $x^m+x^{-n}=\alpha$ can be solved in $\overline{k}$, say $\beta$ is root of it. Then let $x=\beta$, we get $0=g(\alpha)\neq 0$, contradiction. Thus $h$ is a constant. Hence we may assume $f(x)=g(x^m+x^{-n})$, in this case, $f$ must be constant too! We are done.

With this method, we can find more subfield $K$ of $k(x)$, such that $k[x]\cap K=k$.

Aside: If $n=m$, consider the ring map $k(x)\to k(x)$ sending $x$ to $x^{-1}$, we can immediately see that $k[x]\cap k(x^n+x^{-n})=k$.

In fact, we can decide all subfield $K\subseteq k(x)$ such that $K\cap k[x]=k$.

Claim: Let $r(x)\in k(x)$. If $r(x)\notin k[x]$ and $r(x)^{-1}\notin k[x]$, then $K=k(r(x))$ is a subfield of $k(x)$ such that $K\cap k[x]=k$. Convesely, if $K\cap k[x]=k$, then $K=k$ or $K=k(r(x))$ for some $r(x)\in k(x)$ such that $r(x)\notin k[x]$ and $r(x)^{-1}\notin k[x]$.

Proof. By luroth theorem, every $k$-subfield of $k(x)$ is of the form $k(r(x))$ for some $r(x)\in k(x)$. So it suffices to prove the statement that if $r(x)\notin k[x]$ and $r(x)^{-1}\notin k[x]$, then $k(r(x))\cap k[x]=k$.

Suppose $f(x)h(r(x))=g(r(x))$, with $(g,h)=1,h\neq 0$. Then we can find an $\beta\in \overline{k}$ such that $r(\beta)=\alpha$ is a root of $h(t)$ if $h$ is not a constant. Contradiction. Thus $h$ is a nonzero constant. In this case, if $g$ is not a constant, then $r(x)$ is integral over $k[x]$, but $k[x]$ is a normal ring. Hence $r(x)\in k[x]$. Contradiction again.

If either $r(x)\in k[x]$ or $r(x)^{-1}\in k[x]$, it is obvious that $k(r(x))\cap k[x]\neq k$ if $K\neq k$. We are done!


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