Let $k$ be an algebraically closed field. Let $\pi : C \to Spec(k)$ be a smooth curve (i.e. a separated, smooth morphism of finite presentation and relative dimension $1$) I am trying to understand a few assertions about what an effective Cartier divisor on $C$ looks like that I read in Katz-Mazur's book Arithmetic of elliptic curves (freely available here : https://web.math.princeton.edu/~nmk/katz-mazur.djvu)

Here are a few assertions that I am having trouble to understand :

1) Any section $\sigma : Spec(k) \to C$ of $\pi$ defines an effective cartier divisor (a section of a separated morphism is a closed immersion). That would algebraically translate, if i'm not mistaken, to the fact that if $A$ is a dimension one, regular finite type algebra over $k$, and $\mathfrak{m}$ is a maximal ideal that is the kernel of a morphism $A \to k$ then $m$ is a locally free $A$-module of rank one.

2)If $D \subset C$ is a closed subscheme that is finite over $Spec(k)$ then $D$ is an effective cartier divisor. Using the same notation as before if would translate as : if $I$ is an ideal of $A$ such that $A/I$ is a finite dimensional $k$-vector space then $I$ is locally free of rank one.

3)An effective Cartier divisor in $C$ is a finite $k$-scheme : if $I$ is a locally free $A$-ideal of rank $1$ then $A/I$ is a finite dimensional vector space.

So basically I think if we put it all together it comes down to the following statement :

Let $k$ be an algebraically closed field, $A$ be regular $k$ algebra of finite type and dimension $1$ and $I$ be an ideal of $A$. Then $I$ is invertible (i.e. locally free of rank $1$) if and only if $A/I$ is a finite dimensional $k$ vector space.

Is this true ? And if yes why ?

  • $\begingroup$ An $A$ like the one you describe is a Dedekind domain (i.e. locally it is a dvr) and your assertions will follow. $\endgroup$
    – Mohan
    Sep 13, 2015 at 13:43
  • $\begingroup$ Does it have to be a domain ? Would you mind elaborating a bit ? Because even in the case of dedekind domain i'm not sure how this follows ? $\endgroup$
    – J. Doe
    Sep 13, 2015 at 13:44
  • $\begingroup$ Since the curve is smooth, if it irreducible, it will be a domain. If not, it will be disjoint union of curves and you can still argue the same for each component. $\endgroup$
    – Mohan
    Sep 13, 2015 at 13:46
  • $\begingroup$ Ah yes I see. because smooth => reduced. thank you very much $\endgroup$
    – J. Doe
    Sep 13, 2015 at 14:42
  • $\begingroup$ Isn't J.Doe the generic point in the scheme of American names ? :-) $\endgroup$ Sep 13, 2015 at 18:10

1 Answer 1


Yes, your final statement is true.
This is because if moreover $A$ is a domain then it is a Dedekind ring: actually every non-zero ideal $0\subsetneq I\subset A$ is invertible, since by the fundamental result on Dedekind rings we can write $I=\mathfrak m_1^{a_1}\dots \mathfrak m_r^{a_r} $, where the $ \mathfrak m_i$'s are maximal ideals, necessary invertible with $A/\mathfrak m_i=k$, and $a_i\gt0$.
Notice that Dedekindness follows from the regularity hypothesis, which is equivalent to all localizations $A_\mathfrak m$ being discrete valuation rings.

If the ring $A$ satisfies your hypotheses but is not a domain it can be written canonically as a product $A=A_1\dots \times A_s$ where the $A_i$'s are Dedekind domains to which the above applies.
More precisely every ideal $I\subset A$ is a product of ideals $I=I_1\times \dots\times I_s \quad (I_i\subset A_i)$ and that product is invertible if and only if all $I_i$'s are non zero, which is equivalent to $A/I$ (or all $A_i/I_i$'s) being finite-dimensional over $k$.
The geometric picture is blindingly clear: the scheme $$\operatorname {Spec}(A)=\coprod _{i=1}^s\operatorname {Spec}(A_i) $$ is the disjoint union of its irreducible components, which coincide with its connected components, and of course a sheaf is invertible on $\operatorname {Spec}(A)$ if and only if its restrictions to these components $\operatorname {Spec}(A_i)$ are all invertible.

  • $\begingroup$ Thanks a lot ! I can't upvote your answer because I don't have enough points. $\endgroup$
    – J. Doe
    Sep 13, 2015 at 14:41
  • $\begingroup$ Just to make sure : We don't need $k$ to be algebraically closed ? and in fact what you are saying says that every closed subscheme of C is an effective cartier divisor ? $\endgroup$
    – J. Doe
    Sep 13, 2015 at 14:46
  • $\begingroup$ Just nitpicking. Every {\bf proper closed} subscheme is an effective Cartier divisor. $\endgroup$
    – Mohan
    Sep 13, 2015 at 16:23
  • 1
    $\begingroup$ Ahah yes of course you are right :) thanks $\endgroup$
    – J. Doe
    Sep 13, 2015 at 16:28

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