Question

Let $f\hspace{0.17em}:R\to R_1$ be an isomorphism of integral domains. Let F be the field of quotients of R and ${\mathit{F}}_{\mathbf{1}}$the field of quotients of ${\mathit{R}}_{\mathbf{1}}$. Prove that the map$f*:F\to F_1$ given by$\mathit{f}\mathbf{*}\left(a/b\right)\mathbf{=}\mathit{f}\left(a\right)\mathbf{/}\mathit{f}\left(b\right)$ is an isomorphism.

Short answer

It is proved that f* is an isomorphism.

Step-by-step solution

Define a map f

Let $f:R\to R_1\Rightarrow f\left(a\right)=a$.

Now, clearly f is an isomorphism.

Showing that f* is well-defined

Define the map $f*:F\to F_1$given by $f*\left(a/b\right)=f\left(a\right)/f\left(b\right)$.

First, we have to show it is well-defined as:

$\begin{array}{rcl}a/b& =& c/d\\ ad& =& bc\\ f\left(a\right)f\left(d\right)& =& f\left(b\right)f\left(c\right)\left[\because f\left(a\right)=a\right]\\ f\left(a\right)/f\left(b\right)& =& f\left(c\right)/f\left(d\right)\end{array}$

This implies, $f*\left(a/b\right)=f*\left(c/d\right)$.

Therefore, mapping is well-defined.

Showing that f* is a homomorphism

To prove: f* is a homomorphism, we have to prove:

(i) $f*\left(a/b+c/d\right)=f*\left(a/b\right)+f*\left(c/d\right)$

(ii) $f*\left(a/b*c/d\right)=f*\left(a/b\right)f*\left(c/d\right)$

Find $f*\left(a/b+c/d\right)$as:

$\begin{array}{rcl}f*\left(a/b+c/d\right)& =& f*\left(\frac{ad+bc}{bd}\right)\\ & =& \frac{f\left(ad+bc\right)}{f\left(bd\right)}\\ & =& \frac{ad+bc}{bd}\\ & =& \frac{a}{b}+\frac{c}{d}\end{array}$

Simplify the equation as:

$\begin{array}{rcl}f*\left(a/b+c/d\right)& =& \frac{f\left(a\right)}{f\left(b\right)}+\frac{f\left(c\right)}{f\left(d\right)}\\ & =& f*\left(a/b\right)+f*\left(c/d\right)\end{array}$

Find $f*\left(a/b*c/d\right)$as:

$\begin{array}{rcl}f*\left(a/b*c/d\right)& =& f*\left(\frac{ac}{bd}\right)\\ & =& \frac{f\left(ac\right)}{f\left(bd\right)}\\ & =& \frac{ac}{bd}\\ & =& \frac{a}{b}*\frac{c}{d}\end{array}$

Simplify the equation as:

$\begin{array}{rcl}f*\left(a/b*c/d\right)& =& \frac{f\left(a\right)f\left(c\right)}{f\left(b\right)f\left(d\right)}\\ & =& f*\left(a/b\right)f*\left(c/d\right)\end{array}$

Hence, f* is a homomorphism.

Showing that f* is surjective

Find surjective as:

$\begin{array}{rcl}f*\left(a/b\right)& =& f*\left(c/d\right)\\ \frac{f\left(a\right)}{f\left(b\right)}& =& \frac{f\left(c\right)}{f\left(d\right)}\\ \frac{a}{b}& =& \frac{c}{d}\left[\because f\left(a\right)=a\right]\end{array}$

Hence, $f*\left(a/b\right)=f*\left(c/d\right)\Rightarrow a/b=c/d$.

Therefore, f* is a well-defined surjective Homomorphism.

Thus, f* is an isomorphism.