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Jun 15, 2010 at 13:13 comment added Boyarsky The operation of dividing $f$ by the value $f(x)$ at a distinguished point $x$ does not commute with the operation of $\sigma$. So this proof is not quite right. Instead, writing $\sigma(f) = c_ {\sigma} f$ for all $\sigma \in G_K$, where $c_ {\sigma} \in \overline{K}^{\times}$, we see that $\sigma \mapsto c_ {\sigma}$ lies in ${\rm{H}}^1(K, \mathbf{G}_ m) = 1$, so $c_ {\sigma} = \sigma(a)/a$ for some $a \in \overline{K}^{\times}$. Then $(1/a)f$ is Galois-invariant and hence defined over $K$.
Jan 5, 2010 at 22:01 vote accept H. Hasson
Jan 5, 2010 at 22:01 comment added H. Hasson I changed it in the body of the question. Aha, I see. Oh, excellent. Wonderful.
Jan 5, 2010 at 21:47 history edited Ilya Nikokoshev CC BY-SA 2.5
correction
Jan 5, 2010 at 21:40 comment added H. Hasson Oh, darn. Of course that's what I meant. Up to constant.
Jan 5, 2010 at 21:37 history answered Ilya Nikokoshev CC BY-SA 2.5