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Homeseparability is required for integral closures to be finitely generated

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# 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 $k=(\mathbb{Z}/p\mathbb{Z})(b_{1},b_{2},\ldots)$ where the $b_{i}$ are indeterminates.

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

$\left\{\sum_{{i=0}}^{{\infty}}c_{i}t^{i}\right\}$ |

such that $\{c_{0},c_{1},c_{2},\ldots\}$ generates a finite extension of $(\mathbb{Z}/p\mathbb{Z})(b^{p}_{1},b^{p}_{2},\ldots)$. 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 $a(1+c_{1}t+c_{2}t^{2}+\cdots)$; its inverse is $a^{{-1}}(1-c_{1}t-\cdots)$). Hence the ideals of $B$ are powers of $t$, so $B$ is a PID (in fact, it is a DVR).

Now, let $u=b_{0}+b_{1}t+b_{2}t^{2}+\cdots$. Now, $u\notin B$ because it uses all of the $b_{i}$ and thus the coefficients do not define a finite extension of $(\mathbb{Z}/p\mathbb{Z})(b^{p}_{1},b^{p}_{2},\ldots)$. However, $u$ is integral over $B$: $b^{p}_{i}\in B\Rightarrow u^{p}\in B$ which implies that the degree of $u$ over the field of fractions is $p$. Hence a basis for $K(u)/K$ is $\{1,u,u^{2},\ldots,u^{{p-1}}\}$. There are other elements integral over $B$:

$\displaystyle b_{1}+b_{2}t+b_{3}t^{2}+\cdots=\frac{u-b_{0}}{t}=\frac{-b_{0}}{t% }+\frac{1}{t}u$ | ||

$\displaystyle b_{2}+b_{3}t+b_{4}t^{2}+\cdots=\frac{u-b_{0}-b_{1}t}{t^{2}}=% \frac{-b_{0}-b_{1}t}{t^{2}}+\frac{1}{t^{2}}u$ | ||

$\displaystyle\vdots$ |

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

## Mathematics Subject Classification

13B21*no label found*12F05

*no label found*

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