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Is the localization of a reduced ring (no nilpotents) still reduced?

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    By the way, this is a question which (in most American universities at least) is at the beginning graduate level. As Mariano's answer shows, it has an answer which is straightforward and consists of little more than understanding the definitions. Thus I think it would have been better for the OP to solve it on his/her own.2010-10-28

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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.

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    @Javi Recall that in the definition of a localization, $a/b=c/d$ iff $t(ad-bc)=0$ for some $t\in S$. Here take $c=0,d=1$.2018-09-21
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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.

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    I see. I'm afraid I need to consult my books.2010-10-28