I was seeing properties of rings that are preserved when we pass to the localizations or that can be characterized locally. I was wondering if the following question in this vein has an affirmative answer:
If $R_P$ is an integral domain for every prime $P$, then is $R$ reduced?
It is true. Suppose that $r \in R$. Suppose that $r^n = 0$. We want to show that $r = 0$. For every prime $P$ we have $(\frac{r}{1})^n = \frac{r^n}{1} =\frac01$. So, for every prime $P$ there is $v_P \in R \setminus P$ such that $v_Pr=0$. Consider the ideal $J$ generated by all the elements $\{v_P\}$. One can show that this ideal is equal to $R$. For completeness sake we will do this. Assume that this $J \neq R$. Then there is maximal ideal $M$ such that $J \subseteq M$. But $M$ is prime, so there is $v_M \in J$ not in $M$, a contradiction. Thus $J = R$. So we can show $1$ annihilates $r$ and therefore $r = 0$.