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The following definition for a residue map is given in "Algebraic Curves over a Finite Field" by Hirschfeld, Korchmáros and Torres (page 265):

"Let $\Sigma$ be a field of transcendence degree 1 over $K$. For every place $P$ of $\Sigma$, put $res_P(x) = \begin{cases} a& \text{if } ord_p(x-a)\ge 0 \\ \infty & \text{if } ord_p(x)<0 \end{cases}$ Then the residue map, $x \to res_p$, associated to $p$ is a $K$-morphism of $\Sigma \cup {\infty}$ onto $K \cup {\infty}$."

They go on and use properties of this map, for example linearity. But some things aren't so clear to me with this definition.

How is it well-defined? If $v_p(x-a) \ge 0$ for some $a \in K$, doesn't it follow that $v_p(x-b) \ge 0$ for any $b \in K$ because of the non-archimedian valuation? Shouldn't the $\ge$ be changed to a strict inequality, i.e. $>$?

If we change it to strict inequality, why is guaranteed that such $a$ exists? I agree that such $a$ exists, but not in $K$ - maybe in a larger set.

So, is there a mistake in the book or am I missing something (or maybe both)?

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The definition you quote seems to me completely nonsensical.

For example if you take $\Sigma=K(z)$ and for P the valuation at $0$, that is the one with valuation ring $K[z]_{(z)}$, then by everybody's definitiont the rational function $\frac {c}{z} \; (c\in K)$ , or better the differential form $\frac {c}{z}dz$, should have residue $c$ whereas the authors would have us believe that it is $\infty$.

Why not read the Master Himself, whose relevant pages 14 and following Google lets you read for free ?

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    I don't know the context, but the way I see it is that $x$ and $\sigma(x)$ are defined at $P$, and have the same value because $\sigma$ fixes $K$, so that $x-\sigma (x)$ vanishes at $P$ i.e. $v_P(x-\sigma(x))\gt 0$.2012-03-25