Consider a function field $K(t)$ where $K$ in any field, and define the degree valuation as:

$$v_\infty\left(\frac{f}{g}\right):=\operatorname{deg }(g)-\operatorname{deg }(f)$$

Is $v_\infty$ a complete valuation?

Basically this is the valuation attached to the point at infinity of a smooth projective curve over $K$. For each other finite point $v_x$ of the curve the evaluation is in general not complete, thus we have to switch from $K(t)$ to $K((t))$.


No, it's not complete. If you write $x = 1/t,$ then this is just the usual valuation at $x = 0$, and the completion is $K((x))$, as you know.


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