# complete ring of quotients

 $\mathcal{S}:=\{\mathrm{Hom}_{R}(I,R):I\textrm{ is dense in }R\}$

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

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

 $f_{1}\sim f_{2}\Leftrightarrow f_{1}|I_{1}\cap I_{2}=f_{2}|I_{1}\cap I_{2}$

(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 $T(R)$ is the total quotient ring. One can also in general define complete ring of quotients on noncommutative rings.

## References

Title complete ring of quotients CompleteRingOfQuotients 2013-03-22 16:20:29 2013-03-22 16:20:29 jocaps (12118) jocaps (12118) 17 jocaps (12118) Definition msc 13B30 CompleteRingOfQuotientsOfReducedCommutativeRings EpimorphicHull fraction of rings complete ring of quotients