Lemma 10.161.10 (0AE0)—The Stacks project

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

Comments (3)

Comment #9808 by Jonas on October 19, 2024 at 06:16

Minor correction: in the last sentence of the proof, $y$ should be replaced with $D(a)^{i+1}y$ .

Comment #9928 by Branislav Sobot on January 10, 2025 at 04:39

I am not sure what you mean by derivative here? If you mean " $A$ -derivative", then you would automatically have $D(a)=0$ , so I guess not. If you mean that only the Lebniz rule should be satisfied, then I don't see why you can replace $a$ with $f^pa$ . My guess is that you want this to be $A^p$ -derivative, where $A^p\sseq A$ is the subring of all $p$ -powers, but should be in the statement. Also, in the statement it should be $D:K\rightarrow K$ instead of $D:A\rightarrow A$ .

Comment #9930 by Branislav Sobot on January 10, 2025 at 10:05

Sorry, I guess you do want just the Leibniz rule since then automatically $D(b^p)=0$ for all $b\in R$ . However, still $D:R\rightarrow R$ is the statement is not correct I think

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