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if $P_{1}$ and $P_{2}$ are distinct places of equal degree of the function field F/K, and $\sigma$ is a K-field automorphism, such that $\sigma(P_{1})=P_{2}$. then, does $\deg (P_{1}\cap K(x))=\deg (P_{2}\cap K(x))$, where K(x) is the rational function field? in particular, is this true over the hermitian function field?

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No, not in general, that is not without particular requirements for $x$:

take $F=\mathbb{R}(y)$, the rational function field in one variable over the reals. Then the equation $\sigma (y)=y+1$ determines an automorphism of $F/\mathbb{R}$.

Let $P_1$ be the place associated to the polynomial $y^2+1$; then $\deg (P_1)=2$.

Let $P_2 := \sigma (P_1)$; then $P_2$ is associated to the polynomial $y^2+2y+2$ and (automatically) $\deg (P_2)=2$.

Let $x := y^2+1$; then $[F:\mathbb{R}(x)]=2$ and $P_1|_{\mathbb{R}(x)}$ has degree $1$.

On the other hand $yP_2 $ either equals $i-1$ or $-i-1$. In both cases $xP_2$ is non-real and thus $\deg (P_2)=2$.

H

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  • $\begingroup$ ok, thank you. Do you know what these "particular requirements" might be? $\endgroup$
    – y_kaplan
    Jul 20, 2010 at 18:49
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    $\begingroup$ Things are working in the case $\sigma (x)=x$ or more generally if $\sigma |_{K(x)}$ is an automorphism of $K(x)$. H $\endgroup$
    – Hagen
    Jul 21, 2010 at 7:33

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