On the way to defining Cartier divisors on a scheme $X$, one sheafifies a presheaf base-presheaf of rings $\mathcal{K}'(U)=Frac(\mathcal{O}(U))$ on open affines $U$ to get a sheaf $\mathcal{K}$ of "meromorphic functions".1

(ETA: See Georges Elencwajg's answer for Kleiman's article on why $Frac(\mathcal{O}(U))$ doesn't define an actual presheaf. The correct base-free way to make a presheaf $\mathcal{K}'$ is to let $S(U)$ be the elements of
$\mathcal{O}(U)$ which are "stalk-wise regular", i.e. non-zerodivisors in $\mathcal{O}_p$ for every $p\in U$, and define
$\mathcal{K}'(U)=\mathcal{O}(U)[S(U)^{-1}]$. This agrees with the base-presheaf above on affines.)

Have you ever wondered what this sheaf does on affine opens? That's how I usually grasp what a sheaf "really is", but Hartshorne's Algebraic Geometry (Definition 6.11-, p. 141) doesn't tell us. The answer is non-trivial, but turns out to be nice for lots of nice rings. Q. Liu's Algebraic Geometry and Arithmetic Curves shows that:

  1. If $A$ is Noetherian, or reduced with finitely many mimimal primes (e.g. a domain), then
    $\mathcal{K}(Spec(A))=Frac(A)$. (Follows from Ch.7 Remark 1.14.)
  2. If $A$ is any ring, then $Frac(A)$ is a subring of $\mathcal{K}(Spec(A))$. (Follows from Ch.7 Lemma 1.12b.)

So for $A$ non-Noetherian, we could be getting some extra elements, and presumably, they could be units. In other words, we could have principal Cartier divisors that don't come from $Frac(A)$.

Is there an example where this happens?

Follow-up: Thanks to BCnrd's proof below, the answer is "no": even though $\mathcal{K}(Spec(A))$ can be strictly larger than $Frac(A)$, it can't contain additional units, so there are no such extra principal divisors!


1 Here "Frac" means inverting the non-zero divisors of the ring; I'm not assuming anything is a domain.

  • 8
    $\begingroup$ The definition of $\cal K'$ in my book edition 2002 is not correct (though I knew that EGA IV.20.1.3 is wrong and wrote an example where $Frac(O_{X,x})$ is different from $\cal K_x$!). The correct definition can be found in my 2006's paperback edition (see errata on my webpage). I learned the paper of Kleiman mentionned by Georges only last year from Gabber. :( $\endgroup$
    – Qing Liu
    Jun 17, 2010 at 22:49

2 Answers 2


In the setup in the question, it should really say "we could have invertible meromorphic functions on Spec($A$) that don't come Frac($A)^{\times}$", since those are what give rise to "extra principal Cartier divisors". This is what I will prove cannot happen. The argument is a correction on an earlier attempt which had a bone-headed error. [Kleiman's construction from Georges' answer is not invertible, so no inconsistency. Kleiman makes some unfortunate typos -- his $\oplus k(Q)$ should be $\prod k(Q)$, and more seriously the $t$ at the end of his construction should be $\tau$, for example -- but not a big nuisance.]

For anyone curious about general background on meromorphic functions on arbitrary schemes, see EGA IV$_4$, sec. 20, esp. 20.1.3, 20.1.4. (There is a little subtle error: in (20.1.3), $\Gamma(U,\mathcal{S})$ should consist of locally regular sections of $O_X$; this is the issue in the Kleiman reference mentioned by Georges. The content of EGA works just fine upon making that little correction. There are more hilarious errors elsewhere in IV$_4$, all correctable, such as fractions with infinite numerator and denominator, but that's a story for another day.) Also, 20.2.12 there is the result cited from Qing Liu's book in the setup for the question.

The first step in the proof is the observation that for any scheme $X$, the ring $M(X)$ of meromorphic functions is naturally identified with the direct limit of the modules Hom($J, O_X)$ as $J$ varies through quasi-coherent ideals which contain a regular section of $O_X$ Zariski-locally on $X$. Basically, such $J$ are precisely the quasi-coherent "ideals of denominators" of global meromorphic functions. This description of $M(X)$ is left to the reader as an exercise, or see section 2 of the paper "Moishezon spaces in rigid-analytic geometry" on my webpage for the solution, given there in the rigid-analytic case but by methods which are perfectly general.

Now working on Spec($A$), a global meromorphic function "is" an $A$-linear map $f:J \rightarrow A$ for an ideal $J$ that contains a non-zero-divisor Zariski-locally on $A$.
Assume $f$ is an invertible meromorphic function: there are finitely many $s_i \in J$ and a finite open cover {$U_i$} of Spec($A$) (yes, same index set) so that $s_i$ and $f(s_i)$ are non-zero-divisors on $U_i$; we may and do assume each $U_i$ is quasi-compact. Let $S$ be the non-zero-divisors in $A$. Hypotheses are preserved by $S$-localizing, and it suffices to solve after such localization (exercise). So without loss of generality each element of $A$ is either a zero-divisor or a unit. If $J=A$ then $f(x)=ax$ for some $a \in A$, so $a s_i=f(s_i)$ on each $U_i$, so all $a|_{U_i}$ are regular, so $a$ is not a zero-divisor in $A$, so $a$ is a unit in $A$ (due to the special properties we have arranged for $A$). Hence, it suffices to show $J=A$.

Since the zero scheme $V({\rm{Ann}}(s_i))$ is disjoint from $U_i$ (as $s_i|_ {U_i}$ is a regular section), the closed sets $V({\rm{Ann}}(s_i))$ and $V({\rm{Ann}}(s_2))$ have intersection disjoint from $U_1 \cup U_2$. In other words, the quasi-coherent ideals ${\rm{Ann}}(s_1)$ and ${\rm{Ann}}(s_2)$ generate the unit ideal over $U_1 \cup U_2$. A quasi-coherent sheaf is generated by global sections over any quasi-affine scheme, such as $U_1 \cup U_2$ (a quasi-compact open in an affine scheme), so we get $a_1 \in {\rm{Ann}}(s_1)$ and $a_2 \in {\rm{Ann}}(s_2)$ such that $a_1 + a_2 = 1$ on $U_1 \cup U_2$. Multiplying both sides by $s_1 s_2$, we get that $s_1 s_2 = 0$ on $U_1 \cup U_2$. But $s_1$ is a regular section over $U_1$, so $s_2|_ {U_1} = 0$. But $s_2|_ {U_2}$ is a regular section, so we conclude that $U_1$ and $U_2$ are disjoint. This argument shows that the $U_i$ are pairwise disjoint.

Thus, {$U_i$} is a finite disjoint open cover of Spec($A$), so in fact each $U_i = {\rm{Spec}}(A_i)$ with $A = \prod A_i$. But recall that in $A$ every non-unit is a zero-divisor. It follows that the same holds for each $A_i$ (by inserting 1's in the other factor rings), so each regular section $s_i|_ {U_i} \in A_i$ is a unit. But the preceding argument likewise shows that $s_i|_ {U_j} = 0$ in $A_j$ for $j \ne i$, so each $s_i \in A$ has a unit component along the $i$th factor and vanishing component along the other factors. Hence, the $s_i$ generate 1, so $J = A$. QED

  • $\begingroup$ Yay! $ $ $\endgroup$ Jun 18, 2010 at 3:59
  • $\begingroup$ Wow, I'm glad now I was careful to write "presumably" before "they could be units"... Thank-you very much for the proof! $\endgroup$ Jun 18, 2010 at 17:37
  • $\begingroup$ Andrew, my pleasure. That was a fun one to figure out. $\endgroup$
    – BCnrd
    Jun 18, 2010 at 17:44

Dear Andrew, your prescription for $\mathcal K'(U)$ does NOT yield a presheaf: you cannot restrict an element of that ring to a smaller open subset $V \subset U$ because non zerodivisors do not restrict to non zerodivisors. Don't feel bad about this error, you are in good company: Grothendieck, Kleiman and Hartshorne (among others) made the same mistake.Kleiman saw the light and wrote an article aptly named

Misconceptions about $K_X \quad $ (L'Enseignement Mathématique, 25(1979), 203-206)

where he gives a correct definition. He addresses your question by constructing a beautifully geometric (but sophisticated) example of an affine scheme X=Spec(A) where $\Gamma(X,\mathcal K) $ (with the correct definition of $\mathcal K$ !) is strictly bigger than Frac(A).

A gift from L'Enseignement Mathématique Our generous Swiss friends allow us to freely download all issues of their journal from 1899 to 2004 [click on Tome 1(1899)-50(2004), in the green column on the left]:


Kleiman's article in particular is here [take the line correponding to page 203 and click on the white PDF logo on the left]:


  • $\begingroup$ Dear, hmm, BCnrd (see, I can keep a secret !): although I haven't checked your proof in detail I am impressed, as usual, by your spotting the error in EGA and correcting it independently of the paper I mention. I'm sure it will be a breeze for you to study Kleiman's example and to confirm that it is compatible with your proof. $\endgroup$ Jun 17, 2010 at 23:41
  • $\begingroup$ A $\mathit{mistake}$ in EGA? Unnonticed for, what, 20 years? I don't think oy vey is adequate response. Time to drink away the original sin. $\endgroup$ Jun 18, 2010 at 2:10
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    $\begingroup$ Victor, although each volume of EGA came up with (sometimes long!) errata lists for the previous volumes, by definition errors in the final EGA IV$_4$ never got an errata list. Anyway, the time lag between EGA IV$_4$ and 1979 is closer to 10 years than 20, but my guess is that experts knew this error right away (and it is easy to fix once one notices the issue), and probably Kleiman only got around to publishing something later when he found an actual counterexample. There are certainly other errors in the final volume (as I note in my answer), but all correctable in my experience. $\endgroup$
    – BCnrd
    Jun 18, 2010 at 2:22
  • $\begingroup$ Georges, thank-you for the fantastic reference! Looks like this topic was really a mess for some time... $\endgroup$ Jun 18, 2010 at 17:34

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