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Let $K=\mathbb{Q}(x,y)$ be a function field of genus at least 2, with defining equation $f(x,y)=0$ (say, absolutely irreducible and with coefficients not divisible by $p$), and let $k$ be the mod-$p$ reduced function field (arising by reducing the coefficients of $f$). I would like to conclude that $K$ has trivial automorphism group from the fact that $k$ does. Under what extra conditions is this true?

(I'm thinking of something like $g(k)=g(K)$, maybe combined with $p$ being sufficiently large. Of course, $Aut(K)$ should more generally embed into $Aut(k)$ for almost all $p$, but I'd like to have at least an explicit lower bound, e.g. depending on the genus).

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It will be true if the genus remains the same. –  ulrich Nov 19 '13 at 6:23
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@ulrich: Instead of a comment, wouldn't it be more useful to give a complete answer? Via a reference, or a complete argument if the proof is easy. –  Peter Mueller Nov 19 '13 at 9:29
    
The automorphisms of a smooth projective curve $X$ of genus at least $2$ over a field $F$ act faithfully on $H^1_{et}(X_{\bar{F}},\mathbb{Q}_l)$. If the genus is preserved, $H^1$ is also preserved (by smooth and proper base change), so the claim follows by functoriality. (There is probably a more elementary proof.) –  ulrich Nov 19 '13 at 12:12

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