Let $\phi \colon A \to B$ be a flat homomorphism of rings (commutative, with unit). Let $R$ be the total ring of fractions of $A$ (obtained by inverting all nonzerodivisors), and let $S$ be the total ring of fractions of $B$. Since $B$ is flat over $A$, nonzerodivisors of $A$ map to nonzerodivisors of $B$, and so we obtain a natural extension $\phi \colon R \to S$. Let $r \in R$ be a "rational function on $\operatorname{Spec} A$". Let $I, J$ be the ideals of denominators of $r$ in $A, B$, respectively:
```
\begin{align*}
I &= \{ a \in A : a r \in A \} \\
J &= \{ b \in B : b r \in B \}.
\end{align*}
```

Is $J = B \phi(I)$?

The statement amounts to "Flat ring homomorphisms respect ideals of denominators." I know this holds if $B$ is a localization of $A$ (although my proof is rather messy; there's a much nicer proof in the case of inverting a single element given on page 34 of this document from the stacks project). On the other hand, ideals of denominators cannot respect pullbacks in general, since a general pullback need not take a rational function to a rational function. It seems to me that counterexamples in the flat case would probably need to involve a rational function with a non-principal ideal of denominators, and I don't have a great store of such examples to test.

**Motivation:** The ideal of denominators of a rational function $f$ vanishes precisely where $f$ is not defined.

`$0 \to I \stackrel{\binom{1}{1}}{\longrightarrow} A \oplus A \stackrel{\scriptscriptstyle(1,-r)}{\longrightarrow} R$`

. – Charles Staats Sep 6 '11 at 18:06