Let $q$ be a power of a prime $p$ and $w$ be an irreducible polynomial of $\mathbb F_q[T]$. Denote by $\mathbb C_w$ the completion of an algebraic closure of $K_w$, the completion of $\mathbb F_q(T)$ for the normalized valuation $v$ ($v(w)=1$). Denote by $K^{\text{sep}}_w$ the separable closure of $K_w$ in $\mathbb C_w$. One knows that the derivative on $\mathbb F_q(T)$ can be extended uniquely in $K^{\text{sep}}_w$ (one still denotes it by $'$). My question: for every $\alpha\in K^{\text{sep}}_w$, does one have $v(\alpha')=v(\alpha)-1$ or $v(\alpha')\ge v(\alpha)$ if $p\mid v(\alpha)$. Obviously, it is true for $\alpha$ in the completion of $\mathbb F_q(T)$.
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1$\begingroup$ The algebraic closure of $\mathbf F_q(T)$ does not admit a unique extension of the valuation $v_P$ on $\mathbf F_q(T)$, just as ${\rm ord}_p$ on $\mathbf Q$ does not have a unique extension to the algebraic closure of $\mathbf Q$ (but ${\rm ord}_p$ does extend uniquely from $\mathbf Q_p$ to its algebraic closure). I would expect the notation $\mathbf C_P$ to mean the completion of the algebraic closure of the completion $(\mathbf F_q(T))_P$. Please reconsider what it is you are asking. $\endgroup$– KConradCommented Feb 3, 2020 at 5:25
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$\begingroup$ It might also be a bit easier to read the question if you picked your notation to avoid using both $P$ and $p$, as the relation $p \mid v_P(\alpha)$ could require extra attention to understand what it is saying (distinguishing a small capital $P$ in $v_P$ from $p$) $\endgroup$– KConradCommented Feb 3, 2020 at 5:27
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$\begingroup$ You write "for every $\alpha \in K_P$ algebraic over $\mathbf F_q(T)$," but that is redundant since you define $K_P$ to be the separable closure of $\mathbf F_q(T)$ (in a certain field), so automatically all elements of $K_P$ are algebraic over $\mathbf F_q(T)$. $\endgroup$– KConradCommented Feb 3, 2020 at 5:29
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1$\begingroup$ Your new notation is confusing: $K_w$ is a completion of $\mathbf F_q(T)$, so the notation $K_w^{\rm sep}$ looks like it should be the separable closure of $K_w$, not the separable closure of $\mathbf F_q(T)$! Would you ever write the separable (= algebraic) closure of $\mathbf Q$ inside $\mathbf C_p$ as $\mathbf Q_p^{\rm sep}$? I hope not. $\endgroup$– KConradCommented Feb 3, 2020 at 6:07
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3$\begingroup$ If $y$ is a root of $y^p-y = 1/T$, then $v(y)=-1/p, v(y')=-2$. $\endgroup$– Felipe VolochCommented Feb 3, 2020 at 9:22
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