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!

Footnotes:

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

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!

Footnotes:

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

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

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 $Frac(A)$.

Does anyone know an example where this happens, so we can see what they're like?

Footnotes:

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

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!

Footnotes:

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