# The Stacks Project

## Tag 0AE0

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}

There are no comments yet for this tag.

There are also 3 comments on Section 10.155: Commutative Algebra.

## Add a comment on tag 0AE0

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).