Lemma 10.53.5. Any ring with finitely many maximal ideals and locally nilpotent Jacobson radical is the product of its localizations at its maximal ideals. Also, all primes are maximal.
In Lemma 10.53.5, where did they apply localisation? I see they've used the Chinese Remainder Theorem, using which they've extracted some idempotents $(e_i)_i$.
Consider $e_j \neq e_k$, both not in $\mathfrak m_i$. I suppose we get $\frac{e_j}{1} = \frac{e_j e_k}{e_k} = \frac{0}{e_k} = 0$. So localisation at $\mathfrak m_i$ kills all $e_j \neq e_i$.
How do you show that the subset $R e_i$ gets left alone by this localisation?
Consider $\frac{x e_i}{1} = \frac{0 e_i}{1}$. By the definition of localisation, we get that $t(x e_i) = 0$ for some $t \not \in \mathfrak{m}_i$. I'd like to somehow cancel the $t$ to get $x e_i = 0$, but I'm not sure how to justify it.
Take the isomorphism $R\cong \prod_i Re_i$ as the starting point. The maximal ideal $\mathfrak{m}_i$ on the left corresponds to the maximal ideal $\mathfrak{M}_i=Re_1\times\cdots\times Re_{i-1}\times \mathfrak{m}_ie_i\times Re_{i+1}\times\cdots\times Re_n$ on the right. So $R_{\mathfrak{m}_i}\cong\left(\prod_j Re_j\right)_{\mathfrak{M}_i}$. But now I claim that $\left(\prod_j Re_j\right)_{\mathfrak{M}_i}\cong Re_i$. Indeed, we have a natural map $\phi:Re_i\to \left(\prod_j Re_j\right)_{\mathfrak{M}_i}$. If an element $re_i$ is mapped to $0$ then by definition of localisation there exists $se_i\in Re_i\setminus \mathfrak{m}_ie_i$ such that $re_ise_i=0$. But $\mathfrak{m}_ie_i$ is the unique maximal ideal of $Re_i$, so $se_i$ is a unit, and hence $re_i=0$. Hence the map is injective.
To see surjectivity, take an arbitrary element $x/y\in\left(\prod_j Re_j\right)_{\mathfrak{M}_i}$. Let $\tilde{e}_i$ be the element of $\prod_j Re_j$ with $(\tilde{e}_i)_j=0$ if $j\neq 0$ and $(\tilde{e}_i)_i=e_i$. Then $\tilde{e}_i\notin\mathfrak{M}_i$, so $x/y=(x\tilde{e}_i)/(y\tilde{e}_i)=\phi(x_ie_i)/\phi(y_ie_i)$. But note that $y_ie_i\notin\mathfrak{m}_ie_i$ by definition, so $y_ie_i$ is a unit of $Re_i$, and hence $x/y=\phi((xe_i)(ye_i)^{-1})$. Hence $\phi$ is also surjective.
Therefore, we obtain $R_{\mathfrak{m}_i}\cong Re_i$ and conclude.