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Consider a commutative unitary ring $R$ and set $$\mathcal S:=\{ \mathrm{Hom}_R(I,R) : I \textrm{ is dense in } R \}$$ (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 fractions.
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 1 $R\subset T(R)\subset Q(R)$ , where $T(R)$ is the total quotient ring. One can also in general define complete ring of quotients on noncommutative rings.
- Huckaba
- J.A. Huckaba, "Commutative rings with zero divisors", Marcel Dekker 1988
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