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We know that any hyperelliptic curve over a field with characteristic not equal to $2$ has an affine model given by $y^2 = f(x)$, with $deg(f) = 2g+1$ or $2g+2$.

Can we always find a model such that $deg(f)=2g+1$?

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If you find a model with $\text{deg}(f) = 2g + 2$, send one of the roots of $f$ to infinity. –  Qiaochu Yuan May 23 '12 at 15:18
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up vote 5 down vote accepted

Over an algebraically closed field, yes. Let the roots of $f$ in $\mathbb{P}^1$ be $r_1$, $r_2$, ..., $r_{2g+2}$, and apply a Mobius transformation taking $r_{2g+2}$ to $\infty$ as Qiaochu says.

Over a nonalgebraically closed field, not necessarily. Let $g \geq 2$ for simplicity, so the hyperelliptic involution $\tau: (x,y) \mapsto (x, -y)$ is an intrinsic property of the curve. Then the curve has such a model if and only if at least one of the fixed points of $\tau$ is defined over the ground field.

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For $g=1$ [or even $g=0$] a counterexample is $f = -(x^4+1)$ [or $f = -(x^2+1)$] over ${\bf R}$, or over any field contained in ${\bf R}$ such as ${\bf Q}$. The involution is no longer intrinsic, but there are no rational points, whereas a model with $\deg(f) = 2g+1$ would have a rational point at infinity. –  Noam D. Elkies May 23 '12 at 16:17
    
Thanks for the explanation! –  expmat May 23 '12 at 20:24
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