Question

Let$E$be the ring of even integers with the $\ast$multiplication defined in Exercise 23 of section . Show that the map$f:$$E\to \mathrm{\mathbb{Z}}$ given by$f\left(x\right)=x/2$ is an isomorphism.

Short answer

It is proved that the map$f:$localid="1659335795569" $E\to \mathrm{\mathbb{Z}}$is an isomorphism.

Step-by-step solution

Show that f is a surjective and injection

According to example 23 (section 23), Consider the set$E=\left\{2k\overline{)k\in \mathrm{\mathbb{Z}}}\right\}$with operation$\ast$as a commutative ring with identity.

Consider an arbitrary integer exists as an even integer$2a$, given function$f:E\to \mathrm{\mathbb{Z}}$defined by $f\left(a\right)=a/2$is a surjection.

Consider arbitrary even integers,$a$and $b$, then we have:

$\begin{array}{rcl}f\left(a\right)& =& f\left(b\right)\\ \frac{a}{2}& =& \frac{b}{2}\\ a& =& b\end{array}$

Hence, function $f$ is an injection.

Show that function is isomorphism

Prove that $f\left(a+b\right)=f\left(a\right)+f\left(b\right)$and$f\left(a\ast b\right)=f\left(a\right)f\left(b\right)$.

$\begin{array}{rcl}f\left(a+b\right)& =& \frac{a+b}{2}\\ & =& \frac{a}{2}+\frac{b}{2}\\ & =& f\left(a\right)+f\left(b\right)\end{array}$

$\begin{array}{rcl}f\left(a\ast b\right)& =& f\left(\frac{ab}{2}\right)\\ & =& \frac{ab}{4}\\ & =& \frac{a}{2}\frac{b}{2}\\ & =& f\left(a\right)f\left(b\right)\end{array}$

Hence, it is proved that$f:E\to \mathrm{\mathbb{Z}}$$f$ is an isomorphism.