Calculating the localization at a prime ideal of $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/q\mathbb{Z}$ I’m following some notes on how to calculate the localization $(R \setminus P)^{-1}R$ where $P$ is a prime ideal of $R:=\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/q\mathbb{Z}$ (where $p$ and $q$ are different primes) and there’s a step I don’t understand. Here is what it says on my notes (I translated it but I hope it’s clear):

(We already know that the prime ideals of $R$ are ${0} \times
 \mathbb{Z}/q\mathbb{Z}$ and $\mathbb{Z}/p\mathbb{Z} \times {0}$). Let
$i_P: R \rightarrow R_P$ be the localization morphism at a prime ideal
$P$. Let’s find the kernel of $i_P$ for $P={0} \times
 \mathbb{Z}/q\mathbb{Z}$. We know that for all $r \in R$,
$i_Pr=0$ iff $\exists s \notin P$ such that $sr=0$. But $R \setminus P
= (\mathbb{Z}_p \setminus {0}) \times \mathbb{Z}_q$ and then $i_P$ factorizes into $\widetilde{i_P}:R/P \rightarrow R_P$.

I don’t understand the last sentence. Why is the kernel of $i_P$ equal to the prime ideal $P$ (assuming they just applied the first isomorphism theorem there). Shouldn’t it be $\{0\}$? (they just said  $i_Pr=0$ iff $\exists s \mathbf{\notin} P$ such that $sr=0$, and I don’t see why $sr=0$ should hold for $s \in P$).
 A: (For $r\in P$) $sr=0$ doesn't hold for $s\in P$;  but rather for $s\notin P$.  That's for $s\in R\setminus P$.
You can check that that's true precisely when $r\in P$.  Thus $P$ is the kernel of $i_p$ (the localization morphism).
The kernel is $P$ because $R$ has zero divisors.   After all $R\cong \Bbb Z_{pq}$, and $pq$ is composite.
So, in summary,  $(R\setminus P)^{-1}R\cong R/P$ (by the first isomorphism theorem).
This is apparently a standard construction in commutative algebra.
A: For a commutative ring $R$ (not necessarily with $1$) and multiplicative subset $S$, the standard map $\varphi_S\colon R\to S^{-1}R$ is the map that sends $r\in R$ to $\frac{rs}{s}\in S^{-1}R$, for some $s\in S$. The map is well-defined: if $t\in S$ is any other element then $\frac{rs}{s} = \frac{rt}{t}$, because $s((rs)t - (rt)s) = 0$. (Recall that $\frac{a}{s}=\frac{b}{t}$ if and only if there exists $s'\in S$ such that $s'(at-bs) = 0$).
Here, we have $R=(\mathbb{Z}/p\mathbb{Z})\times(\mathbb{Z}/q\mathbb{Z})$. We let $P=\{0\}\times(\mathbb{Z}/q\mathbb{Z})$, and let $S=R\setminus P$, which is well-known to be a multiplicative set. We let $i_P$ denote the canonical map $R\to S^{-1}R$.
Suppose $r\in R$. When is $i_P(r)$ the zero element of $S^{-1}R$? By definition, $i_P(r) = \frac{rs}{s}$ for some $s\in S$ (that is, some $s\notin P$). So $\frac{rs}{s}=\frac{0}{s}$ if and only if there exists $t\in S$ such that $t(rss-0s)=0$, if and only if $trss=0$, if and only if $(tss)r=0$. Now, $tss\in S$. So if $r\in\ker(i_P)$, then there exists $u\in S$ such that $ur=0$. Conversely, if there exists $u\in S$ such that $ur=0$, then $i_P(r) = \frac{ru}{u} = \frac{0}{u}$ is the zero element of $S^{-1}R$, so $r\in\ker(i_P)$. Thus,
$$\begin{align*}
r\in\ker(i_P) &\iff \text{there exists }s\in S\text{ such that }sr=0\\
&\iff\text{there exists }s\in R\setminus P\text{ such that }sr=0\\
&\iff\text{there exists }s\notin P\text{ such that }sr=0.
\end{align*}$$
Now, since $P=\{0\}\times(\mathbb{Z}/q\mathbb{Z})$, we see that an element $(a,b)\in R$ is in $P$ if and only if $a=0$, and hence $(a,b)\notin P$ if and only if $a\neq 0$.
What elements $(x,y)\in R$ satisfy that there exists $(a,b)\notin P$ such that $(a,b)(x,y)=(0,0)$? Because $(a,b)(x,y)=(ax,by)$, we see that $(a,b)(x,y)=(0,0)$ requires either $a=0$ or $x=0$. But if $(a,b)\notin P$, theN $a\neq 0$, so we must have $x=0$. Thus, if $(x,y)\in\ker(i_P)$, then $x=0$, hence $x\in P$.
Conversely, if $(x,y)\in P$, then $x=0$, and then we can take $(a,b)=(1,0)\notin P$, and we have $(1,0)(0,y) = (0,0)$, so $(0,y)\in\ker(i_P)$.
Thus, we conclude that $\ker(i_P)$ is precisely $P$.
To finish off, we note that the map $i_P$ is, in this instance, surjective (for arbitrary $R$ and $S$ the map $\varphi_S$ need not be surjective; for example, the localization of $\mathbb{Z}$ at $(0)$ is $\mathbb{Q}$, and the map only gives you the integers).  Indeed, an arbitrary element of $(R\setminus P)^{-1}R$ is of the form $\frac{(x,y)}{(a,b)}$ with $a\neq 0$. We want to show that there exists $(r,s)\in R$ such that $i_P(r,s) =\frac{(x,y)}{(a,b)}$. We can take
$$i_P(r,s) = \frac{(r,s)(1,0)}{(1,0)} = \frac{(r,0)}{(1,0)}$$
since $(1,0)\notin P$. Now, $\frac{(x,y)}{(a,b)} = \frac{(r,0)}{(1,0)}$ if and only if there exists $s\notin P$ such that $s( (x,y)(1,0) - (r,0)(a,b)) = (0,0)$, if and only if $s(x-ra,0) = (0,0)$. Now, $s$ must be of the form $(u,v)$ with $u\neq 0$. So this will occur if and only if we can find $r$ such that $ra=x$ in $\mathbb{Z}/p\mathbb{Z}$, where $a\neq 0$. But since $a\neq 0$, this can always be done (because $\mathbb{Z}/p\mathbb{Z}$ is a field). So $i_P$ is surjective.
By the First Isomorphism Theorem, $S^{-1}R$ (the image of $i_P$) is isomorphic to $R/\ker(i_P)$, which is $R/P$.
