## Tag `0AE0`

Chapter 10: Commutative Algebra > Section 10.155: Japanese rings

Lemma 10.155.10. Let $R$ be a Noetherian normal domain with fraction field $K$ of characteristic $p > 0$. Let $a \in K$ be an element such that there exists a derivation $D : R \to R$ with $D(a) \not = 0$. Then the integral closure of $R$ in $L = K[x]/(x^p - a)$ is finite over $R$.

Proof.After replacing $x$ by $fx$ and $a$ by $f^pa$ for some $f \in R$ we may assume $a \in R$. Hence also $D(a) \in R$. We will show by induction on $i \leq p - 1$ that if $$ y = a_0 + a_1x + \ldots + a_i x^i,\quad a_j \in K $$ is integral over $R$, then $D(a)^i a_j \in R$. Thus the integral closure is contained in the finite $R$-module with basis $D(a)^{-p + 1}x^j$, $j = 0, \ldots, p - 1$. Since $R$ is Noetherian this proves the lemma.If $i = 0$, then $y = a_0$ is integral over $R$ if and only if $a_0 \in R$ and the statement is true. Suppose the statement holds for some $i < p - 1$ and suppose that $$ y = a_0 + a_1x + \ldots + a_{i + 1} x^{i + 1},\quad a_j \in K $$ is integral over $R$. Then $$ y^p = a_0^p + a_1^p a + \ldots + a_{i + 1}^pa^{i + 1} $$ is an element of $R$ (as it is in $K$ and integral over $R$). Applying $D$ we obtain $$ (a_1^p + 2a_2^p a + \ldots + (i + 1)a_{i + 1}^p a^i)D(a) $$ is in $R$. Hence it follows that $$ D(a)a_1 + 2D(a) a_2 x + \ldots + (i + 1)D(a) a_{i + 1} x^i $$ is integral over $R$. By induction we find $D(a)^{i + 1}a_j \in R$ for $j = 1, \ldots, i + 1$. (Here we use that $1, \ldots, i + 1$ are invertible.) Hence $D(a)^{i + 1}a_0$ is also in $R$ because it is the difference of $y$ and $\sum_{j > 0} D(a)^{i + 1}a_jx^j$ which are integral over $R$ (since $x$ is integral over $R$ as $a \in R$). $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 42718–42725 (see updates for more information).

```
\begin{lemma}
\label{lemma-Noetherian-normal-domain-insep-extension}
Let $R$ be a Noetherian normal domain with fraction field $K$
of characteristic $p > 0$.
Let $a \in K$ be an element such that there exists a derivation
$D : R \to R$ with $D(a) \not = 0$. Then the integral closure
of $R$ in $L = K[x]/(x^p - a)$ is finite over $R$.
\end{lemma}
\begin{proof}
After replacing $x$ by $fx$ and $a$ by $f^pa$ for some $f \in R$
we may assume $a \in R$. Hence also $D(a) \in R$. We will show
by induction on $i \leq p - 1$ that if
$$
y = a_0 + a_1x + \ldots + a_i x^i,\quad a_j \in K
$$
is integral over $R$, then $D(a)^i a_j \in R$. Thus the integral
closure is contained in the finite $R$-module with basis
$D(a)^{-p + 1}x^j$, $j = 0, \ldots, p - 1$. Since $R$ is Noetherian
this proves the lemma.
\medskip\noindent
If $i = 0$, then $y = a_0$ is integral over $R$ if and only if $a_0 \in R$
and the statement is true. Suppose the statement holds for some $i < p - 1$
and suppose that
$$
y = a_0 + a_1x + \ldots + a_{i + 1} x^{i + 1},\quad a_j \in K
$$
is integral over $R$. Then
$$
y^p = a_0^p + a_1^p a + \ldots + a_{i + 1}^pa^{i + 1}
$$
is an element of $R$ (as it is in $K$ and integral over $R$). Applying
$D$ we obtain
$$
(a_1^p + 2a_2^p a + \ldots + (i + 1)a_{i + 1}^p a^i)D(a)
$$
is in $R$. Hence it follows that
$$
D(a)a_1 + 2D(a) a_2 x + \ldots + (i + 1)D(a) a_{i + 1} x^i
$$
is integral over $R$. By induction we find $D(a)^{i + 1}a_j \in R$
for $j = 1, \ldots, i + 1$. (Here we use that $1, \ldots, i + 1$
are invertible.) Hence $D(a)^{i + 1}a_0$ is also in $R$ because it
is the difference of $y$ and $\sum_{j > 0} D(a)^{i + 1}a_jx^j$ which
are integral over $R$ (since $x$ is integral over $R$ as $a \in R$).
\end{proof}
```

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