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](https://mathoverflow.net/questions/28553/extra-principal-cartier-divisors-on-non-noetherian-rings/28566#28566) 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.)

- 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? 




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**Follow-up:** Thanks to [BCnrd's proof below](https://mathoverflow.net/questions/28553/extra-principal-cartier-divisors-on-non-noetherian-rings/28591#28591), 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!




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Footnotes:


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



Extra principal Cartier divisors on non-Noetherian rings? (answered: no!)
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


