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Let $A$ be a ring finitely generated over $\mathbb{Z}$. Let $D$ be a derivation on $A$. Let $D^p$ be the composition of $D$ with itself, $p$ times. We suppose that $D^p(x)$ belongs to the ideal $pA$, for every prime number $p$. Moreover, we suppose that there exist $T$ in $A$ such that $D(T)=1$.

Let $B$ be a subring of $A$ such that $B$ contains $T$. Let $K(B)$ be the field of fractions of $B$. We suppose that $K(A)|K(B)$ is algebraic. My question is if $D$ becomes an operator over $K(B)$ for a arbitrary $B$ with this property. It is $D(K(B))\subset K(B)$. Is posible that there exist a derivation whit the property of my question?

I do that question because if $K$ is a field of function of some component of $A/pA$. Let $L$ be the space of solutions of $D(x)=0$ in characteristic $p$. Then the condition over $D$ imply that $K|L$ is a purelly inseparable and $K=L(T)$. So if $K_1$ is such that $K|K_1$ is separable then $K_1=(L\cap K_1)(T)$. Therefore $D(K_1)$ is in $K_1$. Using the Nakayama's lemma $D(K_1)\subset K_1$ for every field such that $K(A)|K_1$ be algebraic and $T$ belongs to $K_1$. Therefore I have a derivation with that property.

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