In exercise I.XXI of Geometry of Schemes by Eisenbud and Harris, one is asked to compute the ring of rational functions, firstly of an integral domain, then of a general Noetherian ring $R$,
$$\varinjlim_{U\in\mathscr U}\mathscr O_X(U),$$ where $(X,\mathscr O)$ is the affine scheme associated to $R$, and where $\mathscr U$ is the set of open dense subsets of $X$.
Thoughts:
For an integral domain $R$, I know that $X$ is irreducible and reduced; in particular, every open set is dense in $X$. Hence the limit in question is just the fraction field of $R$, right?
However, for an arbitrary Noetherian ring, I know almost nothing about this limit: there is no fraction field for $R$ at all, and we cannot think of a ring of sections as a ratio of elements in $R$!
Any help would be greatly appreciated, and thanks in advance.
Best Answer
The complement of a dense open is a closed set $V(I)$ which does not contain any irreducible component of $X$ -- equivalently $I$ is not contained in any minimal prime of $R$. If $\mathfrak{p}\in X - V(I)$, then by prime avoidance we can find $f$ such that $\mathfrak{p}\in D(f)$ and $f$ does not belong to any minimal prime of $R$ -- equivalently $D(f)$ is dense in $X$. Thus any dense open is covered by sets of the form $D(f)$ with $f$ not belonging to any minimal prime and you may compute the direct limit with respect to such $D(f)$.
From this, it is easy to see that the direct limit is the localization of $R$ at the multiplicative set of elements not contained in any minimal prime. In the case where $R$ is reduced, the set of elements not contained in any minimal prime is exactly the set of non-zero-divisors of $R$, so one obtains the total ring of fractions of $R$.