Let $k$ be a field, $R=k[t]$ the polynomial ring in one variable, let $S$ be a Noetherian ring flat over $R$, If the fiber $S/tS$ over $t$ is a domain, and $U$ the set of elements of the form $1-ts$ for $s\in S$, then $S[U^{-1}]$ is a domain.(Eisenbud p167 Corollary 6.7)
(The proof is first without loss of generality, assume $S=S[U^{-1}]$. Suppose $I,J\subset S$ are ideals with $IJ=0$, hope to prove $I=0$ or $J=0$. Since $(0:I),(0:(0:I))$ are mutually annihilators, assume $I,J$ are annihilators of each other. Then $IJ\subset (t)$, since $S/(t)$ is domain, assume $J\subset (t)$, then $J=(J:t)t$, thus $I(J:t)t=0$, since $t$ is non-zero-divisor, $(J:t)$ annihilates $I$, thus $(J:t)\subset J$, $J=(J:t)t\subset Jt\subset J$, thus they are equal, by NAK, there exist $S$, such that $(1-ts)J=0$, thus $J=0$.)
Then he remarked by primary decomposition, we can find a single element $f\in U$ such that $S[f^{-1}]$ is a domain.
I thought he meant the zero divisors are finite union of associated primary ideals, which all has finite generators, and can be annihilated by the product of their annihilators.
Then he pointed "However, one may not avoid localization completely, as one sees from the example: " A union of plane and line minus point fibered over $\mathbb{A}^1$, i.e. $k[t]\to k[x,t]\times k[t,t^{-1}]$, the fiber over origin is domain. But all the other fibers are not. This is true, but I can't see how this example illustrates the proposition? What does he mean by "avoid localization"?
And he further pointed such troubles can be avoided by working over graded rings. What does he mean by this? What troubles, how they are avoided for example?