In Milne's notes on algebraic number theory (https://www.jmilne.org/math/CourseNotes/ANT.pdf), on page 51, Corollary 3.14 and 3.15 both used the argument "use Chinese Remainder Theorem and look at the ideals generated in $A_{p_i}$, and they are the same". I don't get this: to me, CRT has always been something on quotients $A/p_1\cdots p_m\cong A/p_1\times\cdots A/p_m$. I know by CRT you can choose an element satisfying the modulo conditions, but I don't understand how the localization part works.
I think I understand the corollaries in my own ways: for example 3.14, after choosing the $x$, we see that no $p_i$ except $p_1$ divides $(x)$, and $p_i^2$ also doesn't divide $(x)$. So $p_1=(x)$. But I want to understand Milne's argument using localizations.
PS: I am aware of $a=b\iff aA_p=bA_p$ for all $p$. This is not where I have problem with. I don't know why $(x)$ generate the same ideal in $A_{p_i}$, and what is the philosophy/intuition for looking into localizations.
The first corollary says: Let $A$ be an integral domain with only finitely many prime ideals. Then $A$ is a Dedekind domain if and only if it is a principal ideal domain.
Then it says: if $A$ is Dedekind, then it suffices to show the prime ideals are principal. Let $\mathfrak{p}_1,\ldots,\mathfrak{p}_n$ be the finitely many (nonzero) prime ideals of $A$. Pick $x_1\in\mathfrak{p}_1-\mathfrak{p}_n$, and use the Chinese Remainder Theorem to find $x$ such that $x\equiv x_1\pmod{\mathfrak{p}_1^2}$, and $x\equiv 1\pmod{\mathfrak{p}_j}$ for $j=2,\ldots,n$.
The claim is that $\mathfrak{p}_1 = (x)$. To show that, it is enough to show that $\mathfrak{p}_1 A_{\mathfrak{p}_j} = (x)A_{\mathfrak{p}_j}$ for $j=1,\ldots,n$ (this is from Corollary 3.13, a little higher on the page).
If $x\in R$ is not in $\mathfrak{p}_j$, then $\frac{x}{1}$ is a unit in $A_{\mathfrak{p}_j}$. In particular, if $j=2,\ldots,n$, then because $x\notin \mathfrak{p}_j$, then $\frac{x}{1}$ is a unit, so $(x)$ generates the entire ring $A_{\mathfrak{p}_j}$. And since $\mathfrak{p}_1\neq\mathfrak{p}_j$, then the image of $\mathfrak{p}_j$ also contains units (we are using that this is Dedekind here again, as we cannot have $\mathfrak{p}_1$ properly contained in $\mathfrak{p}_j$, whereas in arbitrary integral domains this is possible), so $\mathfrak{p}_1A_{\mathfrak{p}_j} = (x)A_{\mathfrak{p}_j}$ for $j=2,\ldots,n$.
For $j=1$, the only ideals in $A_{\mathfrak{p}_1}$ are the powers of $\mathfrak{p}_1$, and $x\in \mathfrak{p}_1-\mathfrak{p}_1^2$ (since $x-x_1\in\mathfrak{p}_1^2$ and $x_1\in\mathfrak{p}_1-\mathfrak{p}_1^2$). So the ideal $(x)A_{\mathfrak{p}_1}$ properly contains $\mathfrak{p}_1^2A\mathfrak{p}_1$ and is contained in $\mathfrak{p}_1A_{\mathfrak{p}_1}$, so again we have $(x)A\mathfrak{p}_1=\mathfrak{p}_1A_{\mathfrak{p}_1}$.
Since the two ideals extend to equal ideals in every localization, then they are in fact the same ideal.
Basically the argument is that by localizing at $\mathfrak{p}_i$, you can ignore everything except the ideal $\mathfrak{p}_i$. The choice of $x$ makes it clear that if $i\neq 1$ then $x$ generates the whole ring (it's not in $\mathfrak{p}_i$), same as $\mathfrak{p}_1$; and if $i=1$, then $x$ generates $\mathfrak{p}_1$ (because it lies in $\mathfrak{p}_1$ but not in $\mathfrak{p}_1^2$).
The second corollary is similar; by localizing at $A_{\mathfrak{p}_i}$, you just need to look at what happens to $\mathfrak{b}+(a)$ relative to each prime. You know that the extension of $\mathfrak{b}$ is precisely $\mathfrak{p}_i^{r_i}$, and that the extension of $(a)$ is precisely $\mathfrak{p}_i^{s_i}$, so their sum is precisely $\mathfrak{p}_i^{s_i}$.