if $P_{1}$ and $P_{2}$ are distinct places of equal degree of the function field F/K, and $\sigma$ is a Kfield 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 $i1$ or $i1$. In both cases $xP_2$ is nonreal and thus $\deg (P_2)=2$. H 

