ag.algebraic geometry - How do you see the genus of a curve, just looking at its function field?

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.