There is a theorem in Janusz's Algebraic Number Fields stated as follows:
Kummer's Theorem: Let $R$ be a Dedekind ring with quotient field $K$ and $R'$ the integral closure of $R$ in a finite dimensional extension $L$ of $K$. Let $\mathfrak{p}$ be a nonzero prime ideal of $R$. Suppose there is an element $\theta\in L$ such that the integral closure of $R_\mathfrak{p}$ in $L$ is $R_\mathfrak{p}[\theta]$. Let $f(X)$ be the minimum polynomial of $\theta$ over $K$ (so $f(X)$ has coefficients in $R_\mathfrak{p}$). Let $\overline{f}(X)$ denote the polynomial obtained by reducing the coefficients of $f(X)$ modulo $\mathfrak{p}$. Suppose $$\overline{f}(X)=g_1(X)^{a_1}\cdots g_t(X)^{a_t}$$ is the factorization of $\overline{f}(X)$ as the product of distinct irreducible polynomials $g_i(X)$ over $R/\mathfrak{p}$. Then \begin{equation} \mathfrak{p} R'=\mathfrak{P}_1^{a_1}\cdots\mathfrak{P}_t^{a_t} \end{equation} for certain prime ideals $\mathfrak{P}_i$ of $R'$ corresponding one to one with the irreducible factors $g_i(X)$; the relative degree $f_i(\mathfrak{P}_i/R)$ equals the degree of the polynomial $g_i(X)$.
The author then gives the following proof (in which CRT is short for Chinese Remainder Theorem and DVR for discrete valuation ring):
Proof. The factorization of $\mathfrak{p} R'$ is determined locally so we may replace $R$ by $R_\mathfrak{p}$ and $R'$ by $R_S'$ with $S=R-\mathfrak{p}$. In particular we may assume that $R$ is a DVR. By hypothesis we have $R'=R[\theta]$ and this ring is isomorphic to $R[X]/(f(X))$. Hence $R'/\mathfrak{p} R'$ is isomorphic to $R[X]$ modulo the ideal generated by $\mathfrak{p}$ and $f(X)$. If we first divide by $\mathfrak{p}$ and then by $(\overline{f}(X))$ we obtain $$R'/\mathfrak{p} R'\cong \overline{R}[X]/(\overline{f}(X))$$ where $\overline{R}=R/\mathfrak{p}$. The given factorization of $\overline{f}(X)$ along with the CRT now yields $$R'/\mathfrak{p} R'\cong\bigoplus_{i=1}^t\overline{R}[X]/(g_i(X)^{a_i}).$$ We see that $\overline{R}[X]/(g_i(X))$ is a field and the maximal ideals are generated by the $g_i(X)$. It follows that $\mathfrak{p} R'$ has the factorization given in the statement of the theorem and moreover $$R'/\mathfrak{P}_i\cong \overline{R}[X]/(g_i(X))$$ so the relative degree of $\mathfrak{P}_i$ over $R$ equals the degree of $g_i(X)$. This completes the proof.
I understand the proof except for the part that says:
...and the maximal ideals are generated by the $g_i(X)$. It follows that $\mathfrak{p} R'$ has the factorization given in the statement of the theorem and moreover $$R'/\mathfrak{P}_i\cong \overline{R}[X]/(g_i(X))$$
Why does the isomorphism given by CRT imply $\mathfrak{p} R'$ has the factorization given in Kummer's Theorem? The result intuitively makes sense to me, but I can't find the connection. I am also not sure if something like the third isomorphism theorem comes into play to get $R'/\mathfrak{P}_i\cong\overline{R}[X]/(g_i(X))$.
Here is a general statement of CRT that applies to a commutative ring $R$. Let $I,J$ be two proper ideals of $R$ which are coprime, i.e., $I+J=R$. Then $I\cap J=IJ$ and the evident homomorphism $R\to R/I\times R/J$ has kernel $IJ$ and there is an induced isomorphism $R/IJ\cong R/I\times R/J$.
This applies when $I$ maximal and $J$ is a proper ideal not contained in $I$.
This might help you understand the argument.