Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

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

share|improve this answer
    
ok, thank you. Do you know what these "particular requirements" might be? –  y_kaplan Jul 20 '10 at 18:49
1  
Things are working in the case $\sigma (x)=x$ or more generally if $\sigma |_{K(x)}$ is an automorphism of $K(x)$. H –  Hagen Jul 21 '10 at 7:33
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.