How do you see the genus of a curve, just looking at its function field?

Yuhao asked in the 20-questions seminar:

The genus of a curve is a birational invariant; the function field of a curve determines it up to birational equivelance.

How do you see the genus directly from the field?

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Let $k$ be the ground field.

The Kahler differentials $\Omega_{K/k}$ are the $K$ vector space generated by formal symbols $dg$ subject to $d(f+g)=df+dg$, $d(fg) = f dg + g df$ and $da=0$ for $a \in k$. This is a one dimensional $K$ vector space.

Let $\omega$ be a differential. For any valuation $v$ on $K$, let $t$ be such that $v(t)=1$. We say that $\omega$ is regular at v if $v(\omega/dt) \geq 0$. We say that $\omega$ is regular if it is regular at every valuation of $K/k$. Then the space of regular differentials is a $k$ vector space of dimension $g$.

FURTHER THOUGHTS

Both Ben's answer and mine used the set of valuations of $K/k$. This essentially means that we used the ground field $k$. A valuation of $K/k$ is defined as a valuation of $K$ which is trivial on $k$; conversely, the ground field can be recovered from the valuations that respect it by the formula $k = \bigcap_{v} v^{-1}(\mathbb{R}_{\geq 0})$.

Here is a cautionary example to show that there can not be any solution which only uses properties of the field $K$, without reference to $k$. Let $C$ and $D$ be two irreducible curves of different genuses. Let $K$ be the field of meromorphic functions on $C \times D$, let $k$ and $\ell$ be the fields of functions on $C$ and $D$. Then $K$ is a transcendence degree 1 extension of both $k$ and $\ell$, but has different genuses when considered in these two ways.

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Assuming you work over $\mathbb C$, you can see it in the structure of the Galois group. The points of the curve can be seen as non-Archimedean valuations on the field, and you can see whether a map of curves is ramified by looking at valuations (each prime in the smaller field should have valuation 1 or 0 for any valuation of the bigger field). Basically, this means that around any point in the domain and its target, the completed local rings look like $\mathbb{C}[[t]]$ and $\mathbb{C}[[s]]$, and the map should always be $s\mapsto a_1t+a_2t^2\cdots$, which $a_1\neq 0$ (so the map on completions is an isomorphism).

There's a maximal unramified field extension, and the Galois group of this extension is the profinite completion of $\pi_1(X)$ in the topological sense, so the genus can be obtained by looking at the abelianization.

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But one should be sure first that the field of constants is algebraically closed. –  JSE Oct 16 '09 at 3:25
Hmm, it's becoming more and more clear that in my world, there are only two kinds of fields: algebraically closed and finite. Oddly enough, not everyone seems to agree. –  Ben Webster Oct 17 '09 at 4:05

Danny Calegari addresses the question of how to see the genus of an affine curve. To obtain the genus of an algebraic curve from the function field, take two generic elements in the field (giving a map to ℂ2), and then take a minimal polynomial relation between them. Then compute the Newton polygon of this polynomial, and count the number of lattice points inside as described in Danny's blog entry. Maybe this is as unsatisfying an answer though as taking a map to ℂℙ1 and use Riemann-Hurwitz....

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The Riemann-Roch theorem (or even a fragment of it for a sequence of divisors of increasing degree) can be stated in terms of the field $K$, its constant field $k$ (as David Speyer pointed out this is very important) and the valuations of $K/k$. The genus is the unique number making the Riemann-Roch theorem true. More specifically, for divisors $D$ of sufficiently large degree, $g = \deg D - l(D) + 1$.

@shenghao If you use a similar definition for number fields, using $l(D)$ for the logarithm of $|L(D)|$ (instead of its dimension, which is not defined) you get that the logarithm of the absolute discriminant is the analogue of the genus.

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Chapter 3 in Neukirch's Algebraic Number Theory develops a geometric language for number fields which is analogous (this is THE analogy) to that which one has at one's disposal when studying function fields. This includes some notion of the genus of a number field. He proves a Riemann Roch type theorem and I think there is also some type of Hurwitz formula.

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Thanks for the pointer to Neukirch's book! That's useful. Have added a pointer to it on the nLab here: ncatlab.org/nlab/show/function+field+analogy#Neukirch92 –  Urs Schreiber Jul 19 at 14:24

See C. Chevalley, Introduction to the theory of Algebraic functions of one variable. That book develops the whole theory of algebraic curves using just its function field.

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Here is a review by A. Weil of this book: ams.org/journals/bull/2000-37-01/S0273-0979-99-00834-4/… –  Martin Brandenburg Apr 18 '11 at 11:01
the full review is here: ams.org/journals/bull/1951-57-05/S0002-9904-1951-09522-1/… –  roy smith Apr 22 '11 at 21:03

I saw a definition for the genus of a global field (like number field) somewhere, in the spirit of Hurwitz formula. The title of that book is something like "zeta functions and L functions"; I forget.

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