Generalised Fractions

Let $R$ be a commutative ring. Fix a nonempty subset

$D \subseteq R^* =: R\backslash \{0\}$

that is closed under multiplication.

For any $u \in D$ and $v \in R^*$ , $uv \neq 0$ and $vu \neq 0$ .
$D$ is closed under multiplication.

For instance $D = \mathrm{U}(R) \subseteq R^*$ works. For a concrete example,

$D = \{-1,1\} = \mathrm{U}(\mathbb Z) \subseteq \mathbb Z^*$

works.

Problem 1. Show that the relation $\sim$ on $R \times D$ defined by

$(r_1,d_1) \sim (r_2,d_2) \quad \iff \quad r_1 d_2 = r_2 d_1.$

In this case, we denote $r/d \equiv [(r, d)]$ .

(Click for Solution)

Solution. Reflexivity and symmetry are obvious. For transitivity, suppose

$\begin{aligned} (r_1,d_1) \sim (r_2,d_2) \quad &\iff \quad r_1 d_2 = r_2 d_1, \\ (r_2,d_2) \sim (r_3,d_3) \quad &\iff \quad r_2 d_3 = r_3 d_2. \end{aligned}$

Then

$\begin{aligned} r_1 d_3 d_2 &= r_1 d_2 d_3 \\ &= r_2 d_1 d_3 \\ &= r_2 d_3 d_1 \\ &= r_3 d_2 d_1 = r_3 d_1 d_2, \end{aligned}$

so that $(r_1 d_3 - r_3 d_1) d_2 = 0$ . Since $d_2 \in D$ , we must have

$r_1 d_3 - r_3 d_1 = 0 \quad \Rightarrow \quad (r_1,d_1) \sim (r_3,d_3).$

Problem 2. Show that $(R \times D)/{\sim}$ can be equipped with a ring structure that turns $(R \times D)/{\sim}$ into a commutative ring with a multiplicative identity, and contains $R$ as a sub-ring. Furthermore, show that if $R$ is an integral domain, $Q(R) := (R \times R^*)/{\sim}$ forms a field.

(Click for Solution)

Solution. Intuitively, we would like elements of the form $r/d$ to match the addition and multiplication of rational numbers. Hence, define

$\begin{aligned} (r_1/d_1) + (r_2/d_2) &:= (r_1 d_2 + r_2 d_1)/(d_1 d_2), \\ (r_1/ d_1) \cdot (r_2 / d_2) &:= (r_1 r_2)/(d_1 d_2), \end{aligned}$

both of which can be checked to be well-defined since $d_1 d_2 \neq 0$ . It is not hard to check that the ring axioms hold with additive identity $0/d$ and multiplicative identity $d/d =: 1$ . To justify this claim, we observe that for any $r/d \in (R \times D)/{\sim}$ ,

$r/d \cdot d/d = (rd)/(d \cdot d).$

Since $r \cdot (d \cdot d) = (r \cdot d) \cdot d$ , we have $(r,d) \sim (rd, d\cdot d)$ , so that

$r/d \cdot d/d = (rd)/(d \cdot d) = r/d,$

as required.

Furthermore, define the map

$f : R \to (R \times D)/{\sim},\quad f(r) := rd/d$

for any $d \in D$ . We remark that the choice of $d$ does not affect the result since for any $d_1, d_2 \in D$ ,

$\begin{aligned} (rd_1)d_2 = (rd_2)d_1 \quad &\Rightarrow \quad (rd_1, d_1) \sim (rd_2, d_2) \\ &\Rightarrow \quad [(rd_1, d_1)] = [(rd_2, d_2)] \\ &\Rightarrow \quad rd_1/d_1 = rd_2/d_2. \end{aligned}$

Hence, it remains to check that $f$ is an injective ring homomorphism, which is not difficult, since

$\begin{aligned} f(r+s) &= (r+s)d/d \\ &= (rd+sd)/d \\ &= ((rd+sd)d)/(d\cdot d) \\ &= (rd \cdot d+sd \cdot d)/(d\cdot d) \\ &= (rd)/d + (sd)/d \\ &= f(r) + f(s) \end{aligned}$

and

$\begin{aligned} f(r \cdot s) &= (r \cdot s)d/d \\ &= ((r \cdot s)d \cdot d)/(d \cdot d) \\ &= ((rd) \cdot (sd))/(d \cdot d) \\ &= (rd)/d \cdot (sd)/d \\ &= f(r) \cdot f(s) \end{aligned}$

and

$f(r) = 0 \quad \Rightarrow \quad rd/d = 0 \quad \Rightarrow \quad r = 0.$

To justify the last implication, use $0 = 0/d$ so that

$\begin{aligned} rd/d = 0/d \quad &\Rightarrow \quad (rd, d) \sim (0, d) \\ &\Rightarrow \quad (rd) \cdot d = 0 \cdot d \\ &\Rightarrow \quad r \cdot (d \cdot d) = 0. \end{aligned}$

Since $d\cdot d \neq 0$ , we must have $r = 0$ , as required.

Finally, suppose $R$ is an integral domain. Fix any nonzero $r/d \in (R \times R^*)/{\sim}$ . If $r = 0$ , then $r/d = 0/d = 0$ , a contradiction. Therefore, $r \neq 0$ so that $r \in R^*$ . Then $d/r = (r/d)^{-1} \in R \times R^*$ , since