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Let $A$ be the ring $\mathbb{Z}/60$. What does it look like under the localization at the prime (2)? Everything that is not divisible by 2 should be a unit, so one might thing this looks like $\mathbb{Z}/4$ - but on the other hand, rings always inject into their localizations...

I'm asking because Vakil in his notes says that the stalk at (2) of this ring is $\mathbb{Z}/4$. But this doesn't make sense to me because of the above confusion.

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Dear @Wouter, It is not true that rings always inject into their localizations. If $R$ is a ring and $S$ a multiplicative set, then the kernel of $R\rightarrow S^{-1}R$ is the set of $r\in R$ for which there exists some $s\in S$ with $sr=0$. So if $S$ contains zero divisors, $R$ will not inject into $S^{-1}R$ (unless $R=0$). If there are no zero divisors, that is, if $R$ is a domain, then it is true that $R$ injects into all of its (non-zero) localizations. – Keenan Kidwell Dec 14 '12 at 13:26
A nice exercise says that any ring of fractions of $\mathbb Z_n$ is isomorphic to some $\mathbb Z_d$ with $d\mid n$. – user26857 Dec 14 '12 at 15:23
up vote 2 down vote accepted

but on the other hand, rings always inject into their localizations...

This is false: you've been spoiled by the case of integral domains! Actually this can even fail for integral domains: $ 0^{-1} \mathbb{Z}$ is the zero ring, since this forces $1 = 0 \cdot 0^{-1} = 0$.

In the more general case, localizations have the effect of discarding any "components" of your ring that are away from your ideal (this sentence makes more sense in the geometric picture). In this case, the components are $\mathbb{Z} / 4$, $\mathbb{Z} / 3$, and $\mathbb{Z} / 5$. So, localization at $(2)$ has the effect of focusing only on the $\mathbb{Z} / 4$ case, and discarding the other two components.

If you understand the example above of inverting zero, this might be more clear if you write it as $$ \left(\mathbb{Z}/60 \mathbb{Z}\right)_{(2)} = \{3^{-1}, 5^{-1}\} \mathbb{Z} / 60 \mathbb{Z} $$

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I see my error, but how does your definition correspond to the formal definition of a localization? Isn't 3/2 an element of the localization, for instance? – Wouter Zeldenthuis Dec 14 '12 at 13:45
No. $2$ cannot appear in the denominator of a fraction in the localisation at $(2)$, because we are not inverting the elements in $(2)$! – Zhen Lin Dec 14 '12 at 13:58
Sorry, I meant 2/3... – Wouter Zeldenthuis Dec 14 '12 at 14:09
Yes, $2/3$ is an element of the localization; it equals $2$. – Hurkyl Dec 14 '12 at 14:16

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