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There is a one-to-one correspondence between isomorphism classes of smooth absolutely irreducible curves $X/\mathbb{k}$ and isomorphism classes of fields $\mathbb{K}$ of transcendence degree $1$ over $\mathbb{k}$ such that $\mathbb{K}\cap \bar{\mathbb{k}} = \mathbb{k}$ (For example - Theorem 2.5.28 in Algebraic Geometry Codes - https://www.math.umass.edu/~hajir/m499c/tvn-book.pdf)

What is the genus of a curve in terms of function fields?

UPD: I see it is rather difficult. So let $\mathbb{k}$ be a finite field and genus = 1. What is the answer now?

KCd
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Alexey Milovanov
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    http://mathoverflow.net/questions/152/how-do-you-see-the-genus-of-a-curve-just-looking-at-its-function-field – Matt Jul 14 '13 at 20:16
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    I see you are interested in codes. In that case it may interest you that I learned this stuff from Stichtenoth. He gives (for $|k|<\infty$) an adelic description which gives a very quick route to genus, divisor spaces and Riemann-Roch. IIRC that approach was first described by Chevalley. Getting to R-R is, of course, not the whole story. My study group benefitted from using other material on the side that doesn't completely ignore the geometric point of view either :-) – Jyrki Lahtonen Jul 14 '13 at 21:20
  • As others have mentioned, this correspondence is in fact quite a deep and important result in (arithmetic) algebraic geometry: you would need to read a substantial portion of a basic text on algebraic curves to see a complete proof. I'm not sure what your precise question is though: do you mean, given the function field $\mathbb{K}$, what is the genus? (I'm confused because your updated question takes it the other way around.) – Pete L. Clark Jul 14 '13 at 21:52
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    If so, there' a (relatively!) easy answer: given $\mathbb{K}$, you can use discrete valuations which are trivial on $\mathbb{k}$ to define divisors $D$ and the $\mathbb{k}$-vector space $\mathcal{L}(D)$ of (all nonzero rational functions $f$ with $\operatorname{div} f \geq -D$) $\cup {0}$. Then it is (truly) easy to show that there is at most one natural number $g$ such that for all effective divisors $D$ with $\operatorname{deg} D \geq 2g-2$, then $\operatorname{dim} \mathcal{L}(D) = \operatorname{deg} D - g + 1$. That's the genus. Riemann proved that such a $g$ exists! – Pete L. Clark Jul 14 '13 at 21:58
  • I think you need your fields $\mathbb{K}$ to be finitely generated over $\mathbb{k}$ if you want a bijective correspondence. – Axel Boldt Nov 29 '14 at 00:59

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