Let $X = \operatorname{Spec}(R)$ be an affine Noetherian scheme, and $U_i = D(f_i)$ (with $f_i \in R$) be (finitely many) principal open sets with $X = \bigcup_i U_i$.
We have the sheaf of meromorphic functions $\mathcal{K}_X$ on $X$. Let there be $x_i \in \mathcal{K}_X(U_i) = \operatorname{Frac}(R_{f_i})$ with ${x_i}_{|U_i \cap U_j} = {x_j}_{|U_i \cap U_j}$, i.e. $x_i = x_j$ in $\operatorname{Frac}(R_{f_i f_j})$.
Since $\mathcal{K}_X$ is a sheaf, there has to be some $x \in \mathcal{K}_X(X) = \operatorname{Frac}(R)$ with $x_{|U_i} = x_i$. My quenstion is how to calculate this $x$.
For example, if we forget the Frac, then Liu, Prop. 2.3.1 gives a formula to glue $x_i \in R_{f_i}$ to one $x \in R$. I tired to do what he did there in my case, but it didn't quite work. I got stuck on that I would need to find $y_i \in \operatorname{Frac}(R)$ with $x_i = y_i f_i^{-m}$ in $\operatorname{Frac}(R_{f_i})$.
Best Answer
With some work, I managed to find the answer myself. The trick was in the proof that $\mathcal{K}_X(U) = \operatorname{Frac}(\mathcal{O}_X(U))$ for $U \subseteq X$ affine, see Liu, Prop. 7.1.15.
We write $x_i = \dfrac{\frac{a_i}{f_i^{n_i}}}{\frac{b_i}{1}}$ (w.l.o.g. we can assume the denominator at the bottom to be $1$, since we can put everything there into $a_i$). We set $g_i := f_i b_i \in R$. Since $\frac{b_i}{1}$ is regular in $R_{f_i}$, we can identify $\operatorname{Frac}(R_{f_i}) = \operatorname{Frac}(R_{g_i})$. But now $x_i \in R_{g_i} \subseteq \operatorname{Frac}(R_{g_i}) = \operatorname{Frac}(R_{f_i})$.
We set $I \trianglelefteq R$ to be the ideal generated by the $g_i$. We want to show that there is some regular element in $I$. For every associated prime $\mathfrak{p}$ of $R$ there is some $f_i$ with $f_i \notin \mathfrak{p}$, since $X = \bigcup_i D(f_i)$. Then $\mathfrak{p}_{f_i}$ is an associated prime of $R_{f_i}$. If we had $\frac{g_i}{1} \in \mathfrak{p}_{f_i}$, then, since $f_i$ is a unit in $R_{f_i}$, we would have $b_i \in \mathfrak{p}_{f_i}$, a contradiction to $b_i$ regular. Therefore also $g_i \notin \mathfrak{p}$. Therefore $I \not\subseteq \mathfrak{p}$. Since $R$ is Noetherian, $R$ has only finitely many associated primes, and we get $I \setminus \bigcup_{\mathfrak{p} \in Ass(R)} \mathfrak{p} \neq \emptyset$ by prime avoidance (note that the proof of the prime avoidance lemma is constructive). Since we know that $\bigcup_{\mathfrak{p} \in Ass(R)} \mathfrak{p}$ is exactly the set of zero divisors, we can choose $a \in I$ regular.
We can now glue the $x_i$ (as elements of $R_{a g_i}$) to a $x \in R_{a}$ with the method from Liu, Prop. 2.3.1. We have $x \in R_{a} \subseteq \operatorname{Frac}(R)$, and it is easy to check that this is the $x$ we were looking for.