In an algebraic function field $F/k$ we have the degree zero divisor class group $\text{Cl}^0(F/k)$. Now since any such function field corresponds to the function field of a normal projective curve over $k$ (in particular regular, Noetherian), it seems to me quite reasonable to ask about the corresponding object in this geometric case. How does it look like, is it simply the same, only dealing with cartier divisors?

And further: Does anything change if we don't start with the function field but with a $\underline{\text{non-regular}}$ integral and complete (projective) curve over $k$?

  • $\begingroup$ What is your definition of $Cl^0(F/k)$? $\endgroup$ – Mohan Jul 25 '16 at 12:08
  • $\begingroup$ @Mohan It's $\text{Div}^0(F)/\text{Princ}(F)$ where $\text{Div}^0(F)$ denotes the group of degree zero divisors of $F $ and $\text{Princ}(F)$ denotes the subgroup of principal divisors. A divisor is a formal sum of places of $F/k $ with coefficients in $\mathbb{Z}$, almost all zero. The degree of a divisor is the sum over all degrees of the appearing places times the respective non-zero coefficient. This definitions are the standard ones of Stichtenoth, Algebraic Function fields and Codes. $\endgroup$ – windsheaf Jul 25 '16 at 12:22
  • $\begingroup$ What is $X$? A non-singular smooth projective model of $F$? $\endgroup$ – Mohan Jul 25 '16 at 12:25
  • $\begingroup$ @Mohan I'm sorry, I meant $F$ in the arguments of $\text{Div}^0(\cdot), \text{Princ}(\cdot)$ $\endgroup$ – windsheaf Jul 25 '16 at 12:29
  • $\begingroup$ These concepts are then the same, since the places of a function field are just closed points of the corresponding non-singular curve. A lot of things will go wrong if you do not take the non-singular model. For example (assume $k$ is algebraically closed), the group you have will always be an abelian variety, but for a singular curve, the corresponding group (coming from Cartier divisors) will not be. $\endgroup$ – Mohan Jul 25 '16 at 14:45

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