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Let $C$ be a smooth curve over a field $k$, $\overline{C}$ the smooth compactification and $S=\overline{C} \setminus C$. We think of $S$ as a reduced divisor defined over $k$. Take the sheaf of logarithmic differentials $\Omega^1(\log S)$ and a non-zero rational section $\eta$. Then there is a notion of residue of $\eta$ at points in $S$.

Question: Let $s$ be a point in $S$. Is $\mathrm{Res}_s(\eta)$ an element of $k$ or of the residue field $k(s)$?

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  • $\begingroup$ Is $k(s)$ separable over $k$? If so, then the residue should be an element of $k$. $\endgroup$
    – Jason Starr
    Commented Jun 25, 2015 at 14:12
  • $\begingroup$ Yes, in the case I'm interested in $k$ has characteristic zero. Could you please explain why? Thanks! $\endgroup$
    – zyx
    Commented Jun 25, 2015 at 14:18
  • $\begingroup$ I believe Count Dracula is correct. I was assuming that the residue would be the trace of the residue obtained from a geometric point after base change. But that seems to be wrong. $\endgroup$
    – Jason Starr
    Commented Jun 25, 2015 at 14:42
  • $\begingroup$ I think the story is more complicated than it seems at first. Tate defines residues to be elements in $k$, even if the residue field extension is not separable. Presumably for any other definition of residue, if you apply the trace, you get back to Tate's definition. $\endgroup$
    – Jason Starr
    Commented Jun 25, 2015 at 15:24

1 Answer 1

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The residue of $\text{d}t/(t^2 + 1)$ at the point $(t^2 + 1)$ of $\mathbf{A}^1_\mathbf{Q}$ is the class of $-t/2$ in $\mathbf{Q}[t]/(t^2 + 1)$.

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  • $\begingroup$ Could you give more details about how you computed it? $\endgroup$
    – zyx
    Commented Jun 25, 2015 at 14:27
  • $\begingroup$ I used the definition of the residue. $\endgroup$
    – Count Dracula
    Commented Jun 25, 2015 at 14:31
  • $\begingroup$ I agree with your computation. But I am confused about how this jibes with the Residue Theorem as stated, for instance, on p. 15, Proposition 6 of Serre's "Algebraic Groups and Class Fields". To extend the Residue Theorem over non-closed fields, presumably at some point we need to take traces. $\endgroup$
    – Jason Starr
    Commented Jun 25, 2015 at 14:54
  • $\begingroup$ Hey ... wait a minute! According to Tate's "Residues of Differentials on Curves", the residue is always defined to be an element of $k$. I guess that Tate builds the trace into his definition of the residue. $\endgroup$
    – Jason Starr
    Commented Jun 25, 2015 at 15:22
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    $\begingroup$ If you apply the trace to the example given in the answer, you find 0, which is compatible with the residue theorem but not very interesting. $\endgroup$
    – abx
    Commented Jun 25, 2015 at 16:42

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