Localization in Noetherian rings Let $R$ be a commutative noetherian ring with no nonzero nilpotents. Is every localization of $R$ at a maximal ideal a field?
 A: For a field $F$, the ring of power series $F [[x]]$ is a Noetherian local domain.
Check against this example.
If you don't like power series, just take any Notherian domain and localize at a nonzero prime and  look at the result as a candidate. Localizations of Noetherian rings are Noetherian.

Bonus:
Among commutative rings, the condition of having all localizations at maximal ideals result in fields characterizes von Neumann regular rings. Such rings do not have nonzero nilpotents, of course, but there are reduced rings that aren't VNR.
A: The statement is true, when we localize such a ring at a minimal prime ideal.
Let $P$ be a minimal prime ideal of $R$. $PR_P$ is the only prime ideal of $R_P$, hence equal to the nil radical. Let $x \in PR_P$. We deduce $x^n=0$ for some $n > 0$. By the definition of localization, there is $s \notin P$ satisfying $sx^n=0$. In particular $sx$ is nilpotent in $R$, thus we deduce $sx=0$ by assumption and therefore $x=0$ in $R_P$. This shows $PR_P=0$, so $R_P$ is a field, since the zero ideal is the maximal ideal.
