Showing a map from $\Bbb Q$ to the field of fractions of $\Bbb Z 1_{R}$ is well defined.

Question

Suppose that I have a field $R$ with characteristic $0$. Let $Q= \{uv^{-1} | u,v \in \Bbb Z 1_{R}, v \neq 0 \}$. I want to show that the map $f :\Bbb Q \rightarrow Q$ defined as $f(n/m) = (n1_{R})(m1_{R})^{-1}$ is well defined.

Suppose that $\frac {a}{b} = \frac{n}{m}$, then $am-nb = 0$ and $f(n/m)=(n1_{R})(m1_{R})^{-1}$ which I need to show is equal to $(a1_{R})(b1_{R})^{-1}$

Here I am not sure how to proceed. Hints appreciated.

Answer 1

To show that $f$ is well-defined, let $\frac{a}{b}=\frac{m}{n}$ where $b,n \neq 0$. Then $an-mb=0$. Since $\mathrm{char}R=0$, the homomorphism $\varphi:\mathbb{Z} \to R$ sending $1 \mapsto 1_R$ is injective.

In fact, you can show that for any field $F$, $\mathrm{char}F=m$ if and only if $\ker\{\varphi: \mathbb{Z} \to F\}=m\mathbb{Z}$. Sometimes this is taken as the definition of $\mathrm{char}F$.

Now consider $\varphi(an-mb)=0$. Since $f$ is a ring homomorphism, we can write

$$\varphi(a)\varphi(n)=\varphi(m)\varphi(b)$$ Injectivity of $\varphi$ implies that $\varphi(b),\varphi(n) \neq 0$. Hence, $$\frac{\varphi(a)}{\varphi(b)}=\frac{\varphi(m)}{\varphi(n)}$$

Using your notation, $\varphi(t)=t1_R$. Hence, the equality above is the same as

$$f(\frac{a}{b})=\frac{a1_R}{b1_R}=\frac{m1_R}{n1_R}=f(\frac{m}{n})$$ Hence, $f: \mathbb{Q} \to Q$ is well-defined.

More generally, suppose that $f: D_1 \to D_2$ is an injective ring homomorphism between two integral domains $D_1$ and $D_2$. Denote their fields of fractions by $K_1$ and $K_2$, respectively. Then $f$ extends to a ring homomorphism $\hat{f}: K_1 \to K_2$ given by $\frac{a}{b} \mapsto \frac{f(a)}{f(b)}$. The proof is exactly similar and I leave it to you. In your question, $D_1=\mathbb{Z}$ and $D_2=R$. Injectivity of $f$ comes from $\mathrm{char}R=0$. That's all.

Answer 2

$l:=(n1_R)(m1_R)^{-1}$

so

$(m1_R) l=n1_R$

then

$(b1_R) l= (n1_R)^{-1}(nb1_R) l= $

$(n1_R)^{-1}(am1_R)l=a1_R(n1_R)^{-1}(n1_R)l=a1_R$

To sum up

$l=a1_R(b1_R)^{-1}$

Answer 3

Characteristic $0$ is used to ensure that, for $n\ne0$, also $n1_R\ne0$, so $(n1_R)^{-1}$ makes sense.

In order to prove the map $f$ is well defined, you need to prove that, if $a/b=m/n$, then $$ (a1_R)(b1_r)^{-1}=(m1_R)(n1_R)^{-1} $$ Since $R$ is a field, in particular commutative, this is the same as proving that $$ (a1_R)(n1_R)=(b1_R)(m1_R) $$ Since $an=bm$ by assumption, this boils down to proving that, for integers $x$ and $y$, $$ (x1_R)(y1_R)=(xy)1_R $$ For $x\ge0$ this can be proved by induction: the case $x=0$ is clear. Suppose the statement holds for $x\ge0$; then $$ ((x+1)1_R)(y1_R)=(x1_R+1_R)(y1_R)=(x1_R)(y1_R)+1_R(y1_R)=(xy)1_R+y1_R=(xy+y)1_R $$ The final equality stems from properties of multiples: $xr+yr=(x+y)r$, for $x,y\in\mathbb{Z}$ and $r\in R$. Since $xy+y=(x+1)y$, we're done.

For $x<0$, just observe that $x1_R=-(-x)1_R$.

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Actually, the above is a proof that the map $x\mapsto x1_R$ is a ring homomorphism $\mathbb{Z}\to R$.

Once you have this, you can prove a more general result.

If $f\colon D\to F$ is an injective ring homomorphism, with $D$ an integral domain and $F$ a field, then there is a unique extension of $f$ to a ring homomorphism $g\colon Q(D)\to F$, where $Q(D)$ is the quotient field of $D$, defined by $$ g(x/y)=f(x)f(y)^{-1} $$

One essentially has to prove that $g$ is well defined, because the homomorphism property is an easy verification. Note that injectivity of $f$ guarantees that, for $y\ne0$, $f(y)\ne0$.

Suppose $a/b=x/y$ in $Q(D)$. This means that $ay=bx$. Hence $f(ay)=f(bx)$ and therefore $f(a)f(y)=f(b)f(x)$, so we can multiply both sides by $f(b)^{-1}f(y)^{-1}$, getting $$ f(a)f(b)^{-1}=f(x)f(y)^{-1} $$ as wished.