Does localization preserve reducedness?

Solution 1:

Let $A$ be a ring, $S\subset A$ a multiplicatively closed subset, and suppose that $0\neq a/b\in A_S$ is nilpotent. Then there exists $n$ such that $(a/b)^n=0$, i.e., such that there exists $t\in S$ with $ta^n=0$. But then $ta$ is nilpotent in $A$. If it is zero, then $a/b=0$ in $A_S$, which it isn't.

Solution 2:

EDIT: This argument is incorrect but I feel others could learn from Mariano Suárez-Alvarez's comments so I've made the post CW.

If I understand correctly you are asking does reduce ring imply reduced localization.

Argue by contrapositive, assume that localization is not reduced, i.e. contains nilpotents. Since elements of the localization of our ring are of the form $\frac{r}{s}$ where s is a subset of our ring that does not contain 0. Then choose a nilpotent element in the localization $(\frac{r}{s})^{n}=0$ Since 0 is not in S it must be the case that $r^{n}=0$, i.e. r is nilpotent. Hence the original ring is nilpotent.