# The Stacks Project

## Tag: 07PH

This tag has label more-algebra-lemma-find-D and it points to

The corresponding content:

Lemma 14.40.4. 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 9.146.10 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 9.109.7 to factor $R \to B$ as $$R \subset R[y_1, \ldots, y_d] \subset B' \subset B$$ 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 $f.f.(B') = f.f.(B)$. If we can find a derivation $D' : B' \to B'$ with $D'(f') \not = 0$, then Lemma 14.40.1 guarantees that $D = g^MD'$ extends to $S$ 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 exteion of $A = k[[x_1, \ldots, x_n]][y_1, \ldots, y_m]$.

Note that $\text{d}f$ is not zero in $\Omega_{f.f.(B)/\mathbf{F}_p}$, see Algebra, Lemma 9.144.1. We apply Lemma 14.38.5 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 $\Omega_{f.f.(B)/f.f.(A')}$. Thus we can choose a $f.f.(A')$-derivation $D' : f.f.(B) \to f.f.(B)$ 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. $\square$

\begin{lemma}
\label{lemma-find-D}
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.
\end{lemma}

\begin{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
\ref{algebra-lemma-complete-local-Noetherian-domain-finite-over-regular}
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 \ref{algebra-lemma-Noether-normalization-over-a-domain}
to factor $R \to B$ as
$$R \subset R[y_1, \ldots, y_d] \subset B' \subset B$$
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
$f.f.(B') = f.f.(B)$. If we can find a derivation
$D' : B' \to B'$ with $D'(f') \not = 0$, then
Lemma \ref{lemma-derivation-extends}
guarantees that $D = g^MD'$ extends to $S$ 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 exteion of $A = k[[x_1, \ldots, x_n]][y_1, \ldots, y_m]$.

\medskip\noindent
Note that $\text{d}f$ is not zero in $\Omega_{f.f.(B)/\mathbf{F}_p}$, see
Algebra, Lemma \ref{algebra-lemma-derivative-zero-pth-power}.
We apply Lemma \ref{lemma-power-series-ring-subfields} 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 $\Omega_{f.f.(B)/f.f.(A')}$.
Thus we can choose a $f.f.(A')$-derivation $D' : f.f.(B) \to f.f.(B)$
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.
\end{proof}


To cite this tag (see How to reference tags), use:

\cite[\href{http://stacks.math.columbia.edu/tag/07PH}{Tag 07PH}]{stacks-project}


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).