Discrete valuations on algebraic function fields: an additional condition that I don't understand

Question

Let $K$ an algebraic function field over an algebraically closed field $k$. A discrete valuation on $K$ is defined as a map $v:K^\ast\to\mathbb Z$ such that

1. $v(xy)= v(x)+v(y)$
2. $v(x+y)\ge \min\{v(x),v(y)\}$

3. $\;v(k^\ast)=0$

Why do we require condition 3? usually for discrete valuation fields there is no such condition!

Answer 1

Algebraic function fields are related to geometric objects $X$ like algebraic varieties or Riemann surfaces. These objects are defined over $k$, meaning that the equations used to define them possess coefficients in $k$. The condition $v(k^\ast)=0$ ensures that the valuation $v$ is related to the geometry of $X$ as well and is not reflecting properties of $k$ that are not interesting/relevant from a geometric point of view.

You can see this very clearly in the case of a projective irreducible algebraic curves $X$ defined over the field $k$ and having no singularities. In this case the discrete valuations of its function field $K(X)$ satisfying property 3 are in bijection with the points of $X$.