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.