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Let $K$ be a number field and let $\pi$ be an element in $K$. Assume that $\pi$ is not contained in a subfield of $K$.

Consider the curve $y^2 = x^{2g+1}+\pi$. This defines (after homogenization and normalization) a hyperelliptic curve of genus $g$ over $K$.

Why is this curve not defined over a smaller field $k\subset K$?

That is, why isn't there some transformation of the equation $y^2 = x^{2g+1}+\pi$ such that the coefficients lie in a smaller field?

I might be wrong about this though. That is, maybe these curves ARE defined over a smaller field. I would just like to know how to prove the correct statement rigorously.

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    Your new hypothesis is still not strong enough to prevent the curve from being defined over a smaller field. For instance, if $\pi$ is of the form $x^{2g+1} \pi'$ with $\pi' \in k$, then a change of variables gives a defining equation with coefficients in $k$. I might recommend the hypothesis: suppose that there is a prime $\mathfrak{p}$ of $k$ such that $\mathfrak{p} Z_K = (\pi)^{[K:k]}$. Here the curve is not obviously defined over $k$...2012-07-30
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    Pete, I don't see this change of variables. Its early in the morning ... maybe this is the reason,2012-07-31
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    @Hagen: I don't think I phrased my comment especially well. Let me try again: as $\pi$ varies over $K$, the family of curves $y^2 = x^{2g+1} + \pi$ are all **twists** of the same hyperelliptic curve $y^2 = x^{2+1} + 1$ over the algebraic closure. But the isomorphism class of the twist depends only on the class of $\pi$ in $K^{\times}/K^{\times 2g+1}$. So if $\pi$ is not in $k$ but there is some perfect $2g+1$ power $a^{2g+1}$ ($x$ was an unfortunate name for this, perhaps!) such that $\pi a^{2g+1} \in k$, then the curve can be defined over $k$. This can certainly happen.2012-08-03
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    @Hagen: By the way, since you didn't use an @ in your comment, I wasn't notified of it.2012-08-03

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