Let $\Omega$ be the completion of an algebraic closure of $\mathbb F_q\left(\left(\frac1T\right)\right)$ for the valuation $-\deg$. Let $K/\mathbb F_q(T)$ be a algebraic extension of finite dimension $n$ and unseparability degree $u$. Denote by $|.|_{\infty_i}$ ($1\le i\le g$) be the normalized absolute values of $K$ above prolonging the infinite absolute value of $\mathbb F_q(T)$. Can one assert that for all $x\in K$, one has $$\prod_{i=1}^g|x|_{\infty_i}\le q^{nu\deg(x)}.$$ Obviously, one has the equality for $x\in \mathbb F_q(T)$. Thanks in advance for any answer.
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$\begingroup$ What is the rôle of $\Omega$ in your question? $\endgroup$– LSpiceCommented Jul 4, 2019 at 0:03
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1$\begingroup$ $\Omega$ is where the $-\deg$ valuation lives. I could have taken only an algebraic closure of $\mathbb F_q(\left(\left(\frac 1T\right)\right)$ if you prefer. $\endgroup$– joaopaCommented Jul 4, 2019 at 0:28
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1$\begingroup$ I just meant that, after defining $\Omega$, it seems that you do not use it again. $\endgroup$– LSpiceCommented Jul 4, 2019 at 0:52
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$\begingroup$ I use it through $\deg$. $\endgroup$– joaopaCommented Jul 4, 2019 at 13:01
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$\begingroup$ $\deg(x)$ is not well defined the way you suggest. $K$ may have several different embeddings into $\Omega$. You may want to define $\deg(x) = [K:\mathbb{F}_q(x)]$ but that's a different question. $\endgroup$– Felipe VolochCommented Jul 5, 2019 at 14:55
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