Lemma 15.47.4. Let $p$ be a prime number. Let $B$ be a domain with $p = 0$ in $B$. Let $f \in B$ be an element which is not a $p$th power in the fraction field of $B$. If $B$ is of finite type over a Noetherian complete local ring, then there exists a derivation $D : B \to B$ such that $D(f)$ is not zero.

**Proof.**
Let $R$ be a Noetherian complete local ring such that there exists a finite type ring map $R \to B$. Of course we may replace $R$ by its image in $B$, hence we may assume $R$ is a domain of characteristic $p > 0$ (as well as Noetherian complete local). By Algebra, Lemma 10.154.11 we can write $R$ as a finite extension of $k[[x_1, \ldots , x_ n]]$ for some field $k$ and integer $n$. Hence we may replace $R$ by $k[[x_1, \ldots , x_ n]]$. Next, we use Algebra, Lemma 10.114.7 to factor $R \to B$ as

with $B'$ finite over $R[y_1, \ldots , y_ d]$ and $B'_ g \cong B_ g$ for some nonzero $g \in R$. Note that $f' = g^{pN} f \in B'$ for some large integer $N$. It is clear that $f'$ is not a $p$th power in the fraction field of $B'$. If we can find a derivation $D' : B' \to B'$ with $D'(f') \not= 0$, then Lemma 15.47.1 guarantees that $D = g^ MD'$ extends to $B$ for some $M > 0$. Then $D(f) = g^ ND'(f) = g^ MD'(g^{-pN}f') = g^{M - pN}D'(f')$ is nonzero. Thus it suffices to prove the lemma in case $B$ is a finite extension of $A = k[[x_1, \ldots , x_ n]][y_1, \ldots , y_ m]$.

Assume $B$ is a finite extension of $A = k[[x_1, \ldots , x_ n]][y_1, \ldots , y_ m]$. Denote $L$ the fraction field of $B$. Note that $\text{d}f$ is not zero in $\Omega _{L/\mathbf{F}_ p}$, see Algebra, Lemma 10.152.2. We apply Lemma 15.45.5 to find a subfield $k' \subset k$ of finite index such that with $A' = k'[[x_1^ p, \ldots , x_ n^ p]][y_1^ p, \ldots , y_ m^ p]$ the element $\text{d}f$ does not map to zero in $\Omega _{L/K'}$ where $K'$ is the fraction field of $A'$. Thus we can choose a $K'$-derivation $D' : L \to L$ with $D'(f) \not= 0$. Since $A' \subset A$ and $A \subset B$ are finite by construction we see that $A' \subset B$ is finite. Choose $b_1, \ldots , b_ t \in B$ which generate $B$ as an $A'$-module. Then $D'(b_ i) = f_ i/g_ i$ for some $f_ i, g_ i \in B$ with $g_ i \not= 0$. Setting $D = g_1 \ldots g_ t D'$ we win. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

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.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)