# Why doesn't the condition $a/s=b/t$ iff $at=bs$ suffice when defining localization of commutative rings?

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.

• This has been discussed in many prior answers (most of which are among this search). See also this viewpoint. Jul 14 at 13:39

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 ^_^

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 $$1\cdot (aus-sua)=0$$.