Lemma 10.161.10. 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 $a \in R$). $\square$

There are also:

• 7 comment(s) on Section 10.161: Japanese rings

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 toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0AE0. Beware of the difference between the letter 'O' and the digit '0'.