Let $A$ be a ring. Suppose that, for each prime ideal $\mathfrak{p}$, the local ring $A_{\mathfrak{p}}$ has no nilpotent elements $\neq 0$. Show that $A$ has no nilpotent element $\neq 0$. If each $A_{\mathfrak{p}}$ is an integral domain, is $A$ necessarily an integral domain?
I have found an answer to the second question. Consider $A=\mathbb{Z_{4}}$, the only prime ideal of $A$ is $\{\bar{0},\bar{2}\}$ then for $S:=A\setminus\mathfrak{p}$ we have $S^{-1}A=\{\frac{0}{1},\frac{1}{1},\frac{2}{1},\frac{1}{3}\}$ which is an integral domain, but $A$ is not. (The property of being Integral Domain is not local.)
For the first part, I tried to proceed as follows: Let there exist $(0\neq)a \in A$ such that $a^n=0$, again $1 \in S$ $\implies$ $\frac{a}{1} \in S^{-1}A$ such that $(\frac{a}{1})^n=\frac{a^n}{1}=\frac{0}{1}$. To arrive at a contradiction it remains to show that $\frac{a}{1}\neq 0.$ If $\frac{a}{1}=\frac{0}{1}$ then there exist $t \in A\setminus \mathfrak{p}$ such that $ta=0$.
I cannot think of any way out from here. Please help.
Is the converse true? (i.e Is "the existence of a nonzero nilpotent element in $A$" a local property?)
Best Answer
You are on the right track, and you're correct that you're done if you can show $\frac{a}{1}\ne 0,$ i.e. there is no $t\in A\setminus\mathfrak p$ such that $ta=0.$ But keep in mind you only need to find one prime ideal $\mathfrak p$ for which this holds. So you can take $\mathfrak p$ to be a maximal extension of the ideal $\{b\in A: ba=0\}.$