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Can we localize rings with zero divisors? Can those zero divisors be in the denominator?

I thought defining $$\frac{a}{b}=\frac{c}{d} \text{ iff }t(ad-bc)=0 \text{ where $b,d,t$ belong to the same multiplicative system}$$

accommodated for that little detail. But my professor thinks not. I am confused.

Any help would be greatly appreciated.

user26857
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  • let (R,m) be artinian. then all elements of m are z.d. and we can localize at m. but in this case denominator is from n.z.d –  Jun 26 '14 at 06:44
  • I think you only need to rule out nilpotent elements from the multiplicative system for then you would also need to include zero and that would spoil the fun. As remarked by user26857 we can localize for example $R=\Bbb{Z}/6\Bbb{Z}$ w.r.t. the multiplicative system $S={1,3}$. This has the effect of making $\frac21$ equal to zero, so we end up with $S^{-1}R\cong\Bbb{Z}/2\Bbb{Z}$. – Jyrki Lahtonen Jun 26 '14 at 07:08
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    There's nothing wrong with localising with respect to $0$ – then you get the trivial ring. – Zhen Lin Jun 26 '14 at 07:39
  • Usually we exclude $0$ from being an element of a multiplicative system (as part of the definition). This means that while we can have zero-divisors in the system, we can only have "one of each pair", meaning that if $xy = 0$ and $x$ is in the system, then $y$ cannot be in the system. (As remarked by @ZhenLin this is just to avoid degenerate cases). – Tobias Kildetoft Jun 26 '14 at 09:20

2 Answers2

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Yes, you definitely can localize by sets which contain zero divisors, and the inclusion of the factor $t$ into the definition is there to ensure precisely that. That is, if we don't include $t$ into the definition, then we can still localize by sets without zeros divisors, but then we cannot localize by sets with zero divisors, because then the relation between fractions in $S^{-1}A$ is no longer an equivalence relation,as Hagen Knaf has pointed out. So localization is in fact defined the way it is because we want to localize by sets which may contain zero divisors.

silvascientist
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Your definition is correct and even necessary: if you don't include the factor $t$ into the definition, then you will in general not get an equivalence relation between pairs (a,b) of ring elements. This however is necessary to define the notion of a fraction a/b.

Hagen Knaf
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