5
$\begingroup$

Let $A$ be a commutative ring with identity. Let $S \subseteq A$ be a multiplicatively closed set. Then the localization of $A$ by $S$ is defined as $S^{-1}A = \frac{A \times S}{\sim}$ where $(a,s) \sim (b,t) \iff \exists u \in S,$ such that $u(at-bs)=0$.

My question is, why do we have a condition $\exists u\in S$ and not $\forall u \in S$?

Since in case of construction of $\mathbb{Q}$ from $\mathbb{Z}$, my intuitions says , if we want $\frac{a}{s}=\frac{b}{t}$ then we better make sure $\frac{a}{s}=\frac{b}{t}=\frac{ua}{us} $ for all $u\in S:= \mathbb{Z}-\{0\}$ and hence $u(at-bs)=0$ for every $u \in S$. Why don't we translate this to general rings?

I can see that this has something to do with $\mathbb{Z}$ not allowing any zero divisors. But I am unable to see clearly.

$\endgroup$
1
  • 1
    $\begingroup$ This has been discussed in many prior answers (most of which are among this search). See also this viewpoint. $\endgroup$ Commented Jul 14, 2022 at 13:39

2 Answers 2

6
$\begingroup$

Localization is much easier conceptually when $R$ is a domain, as in that case we can work with the simpler equivalence relation $\frac{r_1}{s_1} \sim \frac{r_2}{s_2}$ if and only if $s_2 r_1 = s_1 r_2$.

Indeed, you can check that this is an equivalence relation provided $S$ doesn't contain any zero divisors.

The question, then, is what goes wrong in the case that $S$ does have a zero divisor. Well, everything works fine except for transitivity. If we say $\frac{r_1}{s_1} \sim \frac{r_2}{s_2} \sim \frac{r_3}{s_3}$, then we compute $r_1 s_3 s_2 = r_3 s_1 s_2$. Again, if $s_2$ is not a zero divisor then we can cancel it and see $\frac{r_1}{s_1} \sim \frac{r_3}{s_3}$, but if $s_2$ is a zero divisor then we have an issue.

Thankfully, this issue is easily repaired! If $s_2$ is a zero divisor then we can derive $s_2 (r_1 s_3 - r_3 s_1) = 0$ by rearranging the last equality. From here it's an easy check that we get an equivalence relation if we make move the goalposts a little bit to account for this.

Then, like all good mathematicians, we cover our tracks. We can't let people know that we wanted a simple equivalence relation and failed. So instead we say that all along we wanted our equivalence relation to be $\frac{r_1}{s_1} \sim \frac{r_2}{s_2}$ if and only if, for some $s'$, we have $s' (r_1 s_2 - r_2 s_1) = 0$. And would you look at that, everything works out nicely!


More conceptually, we now know that the issue comes when $S$ contains a zero divisor. So let's just... kill all of the witnesses to zero-divisor-ness in $S$. That is, let $I$ be the ideal of elements $r$ so that $rs = 0$ for some $s \in S$.

Then it's easy to see that in $R/I$, the subset $S/I$ is still multiplicative, so we can use the easier definition in $R/I$. It's not hard to check that the composite

$$R \to R/I \to R/I[(S/I)^{-1}]$$

satisfies the desired universal property of localizations.

From here it's not hard to check (using the definition of $I$) that the easier definition in $R/I$ lifts to the more complicated definition in $R$.


I hope this helps ^_^

$\endgroup$
0
3
$\begingroup$

Since $1\in S$ by definition of a multiplicatively closed set, the condition $u(at-bs)=0$ for all $u\in S$ is simply equivalent to $at-bs=0$. That indeed looks like a more intuitive definition to $\frac{a}{s}=\frac{b}{t}$. However, in general that might not be an equivalence relation. (it is an equivalence relation when $A$ is an integral domain, but in general it might not be transitive). So we have to replace it with something else, close as possible.

Note that the equality $\frac{a}{s}=\frac{ua}{us}$ still holds in $S^{-1}A$, actually for all $u\in A$. Because we have $1\cdot (aus-sua)=0$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .