Is the localization of a reduced ring (no nilpotents) still reduced?
Does localization preserve reducedness?
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2Interestingly, while localization preserves reducedness (as Mariano has shown below), the somewhat analogous and in theory equally "inoffensive" operation of completion in general does not. One example is $\mathbb{C}[x]/(x^2+1)$ completed at $(x)$; then $x^2+1$ will have $n$th roots in the completion (by the binomial expansion). Nonetheless, if you complete most rings you encounter daily at ideals contained in the Jacobson radical, then reducedness will be preserved. (The condition you need is called excellence.) – 2010-10-28
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2@Akhil: don't believe the hype -- although they are roughly analogous, completion is a significantly more powerful and less completely benign construction than localization. There are lots of tricky things that can happen if you look at completion in the general case. – 2010-10-28
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3By 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
2 Answers
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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1(I have to say that doing this proof by contradiction is very silly…) – 2017-08-18
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0Are you assuming that $A$ is a domain? If not, I don't see why $ta=0$ implies $a/b=0$. – 2017-10-11
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1@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
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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2This is not right: $r^n/s^n$ may be zero without $r^n$ being zero. An example, localize $k[x,y]/(xy)$ at $S=\{x^i:i\geq0\}$ and look at $y/x$. – 2010-10-28
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0@Mariano Suárez-Alvarez I must misunderstand something. I thought the zero element of the localization was 0/s. Can you point me to a reference to understand this better? – 2010-10-28
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0@WWring: any textbook on commutative algebra which gives the construction of localization! That *is* the zero element: your problem is with the way equality is defined. – 2010-10-28
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0I see. I'm afraid I need to consult my books. – 2010-10-28