# If localization ring is a domain, the ring doesn't have to be a domain [duplicate]

Let $$A$$ be a ring and let $$A_{\frak{p}}=S^{-1}A$$ with $$S=A-\frak{p}$$. I know that if $$A_{\frak{p}}$$ is a domain for every $$\frak{p}$$ prime ideal that doesn't mean that $$A$$ is a domain. However, I have a proof of the opposite statement (which is obviously wrong) but I don't know where my mistake is. Could someone please point it out?

Take $$x,y\in A$$, so that $$xy=0$$ in $$A$$. Then take $$s,s'\in S$$. Therefore $$x/s$$ and $$y/s'$$ are elements in $$A_{\frak{p}}$$, and their product is $$(xy)/(ss')=0/(ss')=0$$ (the zero element in $$A_{\frak{p}}$$). But as $$A_{\frak{p}}$$ is a domain, either $$x/s=0$$ (which would mean $$x=0$$ in $$A$$) or $$y/s'=0$$ (which would mean $$y=0$$ in $$A$$). Therefore $$A$$ is an integral domain.

I think there is something about localization wings that I don't quite understand, and therefore I think I am doing something "illegal" in this proof.

Edit I think it might be the fact that $$x/s=0$$ in $$A_{\frak{p}}$$ doesn't mean that $$x=0$$ in $$A$$, but that $$xu=0$$ for some other $$u\in A$$

• The conclusion that $x=0$ because $x/s=0$, does not follow. The conclusion that $x/s=0$ only means that $x$ is annihilated by something in $S$. Apr 29, 2022 at 18:38
• Ah I see you just edited to note that fact. Apr 29, 2022 at 18:39

Consider a very small example to confirm your suspicions. Let $$R=F_2\times F_2$$ where $$F_2$$ is the field of two elements. In fact the localizations of this ring at primes are all fields.
With the prime ideal $$P=\{(0,0),(1,0)\}$$ you can localize at its complement $$S=\{(0,1),(1,1)\}$$ and get that $$RS^{-1}\cong F_2$$.
Of course you have $$(1,0)(0,1)=(0,0)$$. But $$(1,0)/(1,1)\equiv (0,0)/(0,1)\equiv (0,0)/(1,1)$$ without $$(1,0)$$ being zero in $$R$$.