separability is required for integral closures to be finitely generated

The parent theorem assumed that $L$ was a separable extension of $K$. Here is an example that shows that separability is in fact a necessary condition for the result to hold.

Let $k=(\mathbb{Z}/p\mathbb{Z})(b_1,b_2,\mathrm{\dots })$ where the $b_i$ are indeterminates.

Let $B$ be the subring of $k[[t]]$ (power series in $t$ over $k$) consisting of

$$\left\{\sum _{i=0}^{\mathrm{\infty }}{c}_i{t}^i\right\}$$

such that $\{c_0,c_1,c_2,\mathrm{\dots }\}$ generates a finite extension of $(\mathbb{Z}/p\mathbb{Z})(b_1^p,b_2^p,\mathrm{\dots })$. Then every element of $B$ is a power of $t$ times a unit (first factor out the largest power of $t$. What’s left is $a(1+c_{1}t+c_2{t}^2+\mathrm{\cdots })$; its inverse is $a^{-1}(1-c_{1}t-\mathrm{\cdots })$). Hence the ideals of $B$ are powers of $t$, so $B$ is a PID (in fact, it is a DVR).

Now, let $u=b_0+b_{1}t+b_2{t}^2+\mathrm{\cdots }$. Now, $u\notin B$ because it uses all of the $b_i$ and thus the coefficients do not define a finite extension of $(\mathbb{Z}/p\mathbb{Z})(b_1^p,b_2^p,\mathrm{\dots })$. However, $u$ is integral over $B$: $b_i^p\in B\Rightarrow u^p\in B$ which implies that the degree of $u$ over the field of fractions is $p$. Hence a basis for $K(u)/K$ is $\{1,u,u^2,\mathrm{\dots },u^{p-1}\}$. There are other elements integral over $B$:

$$b_1+b_{2}t+b_3{t}^2+\mathrm{\cdots }=\frac{u-b_0}{t}=\frac{-b_0}{t}+\frac{1}{t}u$$

$$b_2+b_{3}t+b_4{t}^2+\mathrm{\cdots }=\frac{u-b_0-b_{1}t}{t^2}=\frac{-b_0-b_{1}t}{t^2}+\frac{1}{t^2}u$$

$$\mathrm{⋮}$$

Clearly the denominators are getting bigger, so the integral closure of $B$ cannot be finitely generated as a $B$-module.