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Let $R$ be a commutative ring with identity, and $S$ a multiplicative subset of $R$ not containing $0$. Which step in my reasoning is wrong?

  1. Every element of $S$ is a unit in $S^{-1}R$.
  2. Every element of $S$ that's a zero divisor in $R$ is a zero divisor in $S^{-1}R$.
  3. But zero divisors can't be units. Therefore it's not possible to localize $R$ at $S$.
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    For such a subset you get $S^{-1}R = \{0\}$, so 1,2,3 are true. Simply you get $1=0$ in your ring :) – 2017-02-19
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    @N.H. Maybe I'm misunderstanding you, but according to Wikipedia, $S^{-1}R=\{0\}$ if and only if $0\in S$. – 2017-02-19
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    You are right I am sorry, I did answer too quickly. – 2017-02-19
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    The only related thing in [the Wikipedia article](https://en.wikipedia.org/wiki/Localization_(algebra)) that I found says that $R\to S^{-1}R$ is injective if and only if $S$ does not contain any zero divisors. – 2017-02-19
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    Ok let me try a second time. Assume you have $as = 0$ with $a \in A$ and $s \in S$. In the localization, $(a,1) \sim (as,s) \sim (0,1)$. Thus $a = 0$ and $s$ is not a zero divisors anymore in $S^{-1}A$. – 2017-02-19
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    @JyrkiLahtonen I'm sorry, I was talking about [this](https://en.wikipedia.org/wiki/Localization_of_a_ring#Properties) article. – 2017-02-19
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    Oh! Sorry, Jack. I should have guessed that there might be several WP articles on this. – 2017-02-19

2 Answers 2

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Consider the localization of $R=\Bbb{Z}_6$ with respect to the multiplicative set $S=\{1,3\}$. In $S^{-1}R$ we have $2/1=0/1$ because $3\cdot(2\cdot1-0\cdot1)=0$. It follows soon that $S^{-1}R\simeq \Bbb{Z}_2$. You see that $3/1=1$ is invertible in that ring.

The error is that even if an element $a\in R$ is a zero divisor in $R$, it may happen that $a/1$ is not a zero divisor. This is because it may happen that for all $b\in R$ such that $ab=0$ we get $b/1=0$ in $S^{-1}R$.

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    My mental model this whole time was that $S^{-1}R$ is the simplest extension of $R$ in which every element of $S$ is a unit. But in fact, $S^{-1}R$ might not even be an extension of $R$. – 2017-02-19
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    Yeah, this kind of a localization is a bit non-intuitive. I had to think about it for a while. I guess algebraic geometers meet such beasts daily :-) – 2017-02-19
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    Although, the inclusion $R\to S^{-1}R$ has kernel $\{a\in R\mid \exists s\in S,\ as = 0\}$, so localizing at a set $S$ with zero divisors in it can be thought of as a two-stage process of first quotienting out zero divisors, and then taking the "minimal extension" approach I mentioned above. – 2017-02-19
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Step 2. is false:
Consider $R=\mathbb R[X,Y]/(X\cdot Y)=\mathbb R[x,y]$ and $S=\{1,x,x^2,x^3,\cdots\}$.
We have $S^{-1}R\cong \mathbb R[x,\frac 1x]$. The element $x\in R$, which was a zero-divisor in $R$ (since $xy=0,\; y\neq 0$), is no longer a zero-divisor in $S^{-1}R$ (since that last ring is a domain).