# How to compute the ring of rational functions over a Noetherian ring?

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.

• Maybe this is too simple to even spend some time to explain. In that case, a reference is also welcomed. Aug 18 '13 at 11:37
• You may find this helpful: en.wikipedia.org/wiki/Total_quotient_ring Aug 18 '13 at 17:02
• @Andrew Indeed it is related. But my problem is: I do not know what the open dense subsets of $Spec R$ are, thus I cannot compute the direct limit. Also, I think that, in some cases, this limit is just the total quotient ring of that Nötherian ring? Aug 18 '13 at 17:56
• Dear @awllower, I think the second part of this exercise assumes a little too much from beginners. For example, Vakil's online book dedicates about six pages of explanations and exercises on associated points of locally noetherian schemes before defining "rational function". You might benefit from reading this section (5.5 of the 2013-06-11 draft). Aug 18 '13 at 19:45
• @Andrew Many thanks: Now I see that this is not as simple as supposed to be. Also I cannot find the note you referred to; maybe you can tell me where to find it? Thanks again. Aug 19 '13 at 3:39

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 $R$.
• Thanks: in this case the limit would become the total ring of fractions of $R$, if I get it right from your arguments. But I am wondering about the first sentence: why can the complement of a dense open not contain any irreducible component of $X$? This is the only remaining point that troubles me now. Thanks for your effort! Nov 9 '13 at 10:35
• If subset $U\subseteq X$ doesn't meet an irreducible component, then it is contained in the union of the (finitely many) remaining irreducible components, so the closure of $U$ must also be contained in this union and $U$ cannot be dense. Nov 9 '13 at 22:21