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From my previous question, I know that$$\zeta_X(s) = {{P(u)}\over{(1-u)(1-qu)}}$$for some polynomial $P(u)$ of degree $2g$, where

  • $X$ is the set of all places of $F$, a function field in one variable over a total constant field $k$,
  • $u := q^{-s}$,
  • $g$ is the genus of $F$.

My question now is, must the limit$$\lim_{s \to 1} (1 - q^{1-s})\zeta_X(s)$$necessarily converge to$${h\over{(q-1)q^{g-1}}}?$$Here, $h$ denotes the number of divisor classes of $F$ of degree $0$.

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    $\begingroup$ In your previous question I suggested you look at Roquette's paper rzuser.uni-heidelberg.de/~ci3/rv.pdf, and here too the result you ask is presented there. It is equation (42), up to simple scaling factors (your $1-q^{1-s}$ is replaced there with $s-1$, and their ratio as $s \rightarrow 1$ is $\log q$). Are you reading any of the references being suggested to you?? Another book-length treatment you might like is Mike Rosen's Number Theory in Function Fields. $\endgroup$
    – KConrad
    Commented May 29, 2015 at 4:24
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    $\begingroup$ @KConrad: I agree that the question is basic and can be answered after minimal self-study. On the other hand, I would like to defend the OP, who is probably a student, in that only a few hours passed since she/he received from you the link to Roquette's paper, and the quoted formula is on page 49. $\endgroup$
    – GH from MO
    Commented May 29, 2015 at 4:55
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    $\begingroup$ @GHfromMO, sure, page 49 is deep in Roquette's paper, but I had specifically suggested in my comment in the OP's previous question to look at section 5.2.2. If the OP had started reading section 5 from the start then page 49 would have come up pretty soon, and the displayed formula is hard to miss. $\endgroup$
    – KConrad
    Commented May 29, 2015 at 5:20
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    $\begingroup$ @KConrad: Thank you. I missed your note "esp. Section 5.2.2". BTW I did not know about Roquette's paper until your comment today, and now that I have read parts of it, I find it highly interesting and entertaining. $\endgroup$
    – GH from MO
    Commented May 29, 2015 at 5:24

1 Answer 1

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Yes. From the functional equation (Theorem 4 in Chapter VII-6 of Weil: Basic number theory) we know that $P(u)=q^gu^{2g}P((qu)^{-1})$. We also know that (see same theorem) that $P(1)=h$. Hence the limit in question equals $$ \frac{P(q^{-1})}{1-q^{-1}}=\frac{q^{-g}P(1)}{1-q^{-1}}=\frac{P(1)}{q^{g-1}(q-1)}=\frac{h}{q^{g-1}(q-1)}.$$

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