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Let $X$ be a smooth projective curve of genus 3 over an algebraically closed field of characteristic 0. How do I show that any curve like this is hyperelliptic or a plane curve of degree 4? Why is $K(X)$ isomorphic to $k(t)[y]$ for $y^{2}=f$ with $f \in k[t]$?

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Under your hypotheses, the Riemann–Roch theorem and the embedding theorem are applicable. If I recall correctly this theorem is a case bash involving the canonical divisor and the aforementioned theorems. –  Zhen Lin Dec 31 '11 at 1:14

1 Answer 1

up vote 6 down vote accepted

The canonical divisor has degree $4 = 2g-2$ and $3$ (the genus) linearly independent global sections. If the canonical divisor is very ample, then the imbedding it defines must land on a plane quartic. If not, then the curve is hyperelliptic. (Recall that a curve is hyperelliptic if and only if the canonical bundle is not very ample.)

The second question is equivalent to asking whether the curve is a double cover of $\mathbb{P}^1$ (or equivalently, that the corresponding field is a degree two extension of $k(t)$), and that's the definition of hyperellipticity.

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