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Let $A$ be a noetherian ring and $X=\operatorname {Spec}A$ the corresponding affine scheme.
There are three rings which might reasonably be called the ring of rational functions on $X$.
a) The total ring of fractions $S^{-1}A$ obtained by inverting the monoid $S$ of regular elements of $A$ (=non zero divisors): $$\mathcal K (X)=\operatorname {Tot} A$$ b) The product of the localizations of $A$ at the finitely many minimal prime ideals of $A$: $$\mathcal R(X)=\prod_{\mathfrak p \in \operatorname {Specmin}A} A_\mathfrak p$$ c) The product of the localizations of $A$ at the finitely many associated prime ideals of $A$: $$\mathcal A(X)=\prod_{\mathfrak P \in \operatorname {Ass}A} A_\mathfrak P$$
Questions

  1. What are the relations between these rings?
  2. Is there a name or use for these rings?
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1 Answer 1

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The ring $\mathcal K(X)$ is called the ring of meromorphic functions on $X$ in EGA, the Stacks project or by Kleiman and many others.
The terminology is disastrous if one considers schemes over $\mathbb C$, since these schemes have a holomorphic structure for which "meromorphic" has a completely different meaning.
Be that as it may, there is a canonical ring morphism $$u=(u_\mathfrak P):\mathcal K (X)=\operatorname {Tot} A \to \mathcal A(X)=\prod_{\mathfrak P \in \operatorname {Ass}A} A_\mathfrak P$$ Indeed $S=A\setminus \cup_{\mathfrak P \in \operatorname {Ass}(A)}\mathfrak P\subset S_\mathfrak P =A\setminus \mathfrak P$ so that we have ring morphisms $$u_\mathfrak P:\operatorname {Tot} A=S^{-1}A\to A_\mathfrak P=S^{-1}_\mathfrak P A$$ which taken together give the morphism $u$.
We also have a natural projection $v:\mathcal A(X)\to \mathcal R(X)$, as well as the composition of those natural morphisms $w=v \circ u:\mathcal K (X)\to \mathcal R(X)$.
The ring $\mathcal R(X)$ is called the ring of rational functions in EGA 1, the Stacks project and many other places; this is an excellent terminology generalizing the standard case of an integral scheme.
On the other hand I have never seen any allusion to $\mathcal A(X) $ in the literature.

An example
Let $k$ be a field and consider the ring $A= k[X,Y]/\langle Y^2,XY\rangle=k[x,y]$ ($y^2=xy=0$), for which the associated prime ideals are $\mathfrak p=\langle y\rangle\subset \mathfrak P=\langle x,y\rangle$.
The canonical morphism $w=v \circ u:\mathcal K (X)\to \mathcal R(X)$ is thus the morphism $w:k[x]_{\langle x\rangle}[y]\to k(x)$, in which $w(x)=x$ and $w(y)=0$.
That morphism is not injective since the nonzero nilpotent meromorphic function $y$ is sent to zero, nor surjective since the rational function $\frac 1x$ is not the image of a meromorphic function on $X$.

Some remarks
As already mentioned the ring $\mathcal A(X)= k[x]_{\langle x\rangle}[y]\times k(x)$ doesn't seem very useful.
To tell the truth I don't quite understand the reason for the introduction of $\mathcal K(X)$ either, but I hope that some of the extremely competent algebraic geometers on this site will tell me how useful meromorphic functions are in certain situations.
Recall also that the definition of the sheaf $\mathcal K$ on a general non-affine scheme is not trivial at all. Kleiman showed in his celebrated article, mischievously called Misconceptions about $\mathcal K_X$, that Grothendieck, Hartshorne and Kleiman himself had given false definitions for that sheaf!

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    $\begingroup$ The sheaf $\mathcal K$ is used in studying line bundles. It is needed to extend the useful characterization "line bundles are (Cartier) divisors modulo rational functions" to a reducible scheme. $\endgroup$
    – Will Sawin
    Commented Mar 28, 2022 at 23:47
  • $\begingroup$ I think the map $u$ is injective: For this it is sufficient to show that $A\to\mathcal A(X)$ is injective, so suppose $a \in A$ is in the kernel. This means for each associated prime $\mathfrak p$ there exists $s_{\mathfrak p}\in A \setminus \mathfrak p$ with $a s_{\mathfrak p}=0$. Hence $s = \prod s_{\mathfrak p}$ is not contained in any associated prime, so that $s \neq 0$ and $s$ is not a zero divisor. However $as=0$ and so $a=0$. I always was under the impression that, at least for reduced $X$, the map $u$ is an isomorphism. But I'm not that sure right now. $\endgroup$ Commented Mar 29, 2022 at 9:29
  • $\begingroup$ Has someone considered the ring of functions with values in $\mathbf{P}^1$? $\endgroup$
    – Leo Alonso
    Commented Mar 29, 2022 at 11:37
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    $\begingroup$ @Leo is this a ring? $\endgroup$ Commented Mar 29, 2022 at 11:40
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    $\begingroup$ @LeoAlonso I believe the complaint is that maps to $\mathbf{P}^1$ does not form a ring. For example, there is no additive inverse for the constant map valued at $\infty$. $\endgroup$
    – Brian Shin
    Commented Mar 29, 2022 at 15:41

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