complete ring of quotients

Consider a commutative unitary ring $R$ and set

$$𝒮:=\{{\mathrm{Hom}}_R(I,R):I\text{ is dense in }R\}$$

(here ${\mathrm{Hom}}_R(I,R)$ is the set of $R$-module morphisms from $I$ to $R$) and define $A:={\bigcup }_{B\in 𝒮}B$.

Now we shall assign a ring structure to $A$ by defining its addition and multiplication. Given two dense ideals $I_1,I_2\subset R$ and two elements $f_i\in {\mathrm{Hom}}_R(I_i,R)$ for $i\in \{1,2\}$, one can easily check that $I_1\cap I_2$ and $f_2^{-1}(I_1)$ are nontrivial (i.e. they aren’t $\{0\}$) and in fact also dense ideals so we define

$f_1+f_2\in {\mathrm{Hom}}_R(I_1\cap I_2,R)$ by $(f_1+f_2)(x)=f_1(x)+f_2(x)$

$f_1*f_2\in {\mathrm{Hom}}_R(f_2^{-1}(I_1),R)$ by $(f_1*f_2)(x)=f_1(f_2(x))$

It is easy to check that $A$ is in fact a commutative ring with unity. The elements of $A$ are called .

There is also an equivalence relation that one can define on $A$. Given $f_i\in {\mathrm{Hom}}_R(I_i,R)$ for $i\in \{1,2\}$, we write

$$f_1\sim f_2\iff f_1|I_1\cap I_2=f_2|I_1\cap I_2$$

(i.e. $f_1$ and $f_2$ belong to the same equivalence class iff they agree on the intersection of the dense ideal where they are defined).

The factor ring $Q(R):=A/\sim$ is then called the complete ring of quotients.

Remark.

$R\subset T(R)\subset Q(R)$, where $T\mathit{}\mathrm{(}R\mathrm{)}$ is the total quotient ring. One can also in general define complete ring of quotients on noncommutative rings.

Title	complete ring of quotients
Canonical name	CompleteRingOfQuotients
Date of creation	2013-03-22 16:20:29
Last modified on	2013-03-22 16:20:29
Owner	jocaps (12118)
Last modified by	jocaps (12118)
Numerical id	17
Author	jocaps (12118)
Entry type	Definition
Classification	msc 13B30
Related topic	CompleteRingOfQuotientsOfReducedCommutativeRings
Related topic	EpimorphicHull
Defines	fraction of rings
Defines	complete ring of quotients