Question

$\mathrm{\mathbb{Z}}\left[\sqrt{2}\right]$is a ring by Exercise 13 of Section 3.1. Let$f:\mathrm{\mathbb{Z}}\left[\sqrt{2}\right]\to \mathrm{\mathbb{Z}}\left[\sqrt{2}\right]$ be the function defined by$f\left(a+b\sqrt{2}\right)=a-b\sqrt{2}$ .

1. Show that is a surjective homomorphism of rings.

Short answer

(a) and (b) are proved

Step-by-step solution

Solution to Part (a)

To show that f is a surjective homomorphism of rings compute as follows:

$\begin{array}{rcl}f\left(\left(a+b\sqrt{2}\right)\right)+\left(c+d\sqrt{2}\right)& =& f\left(\left(a+c\right)+\left(b+d\right)\sqrt{2}\right)\\ & =& \left(a+c\right)-\left(b+d\right)\sqrt{2}\\ & =& \left(a-b\sqrt{2}\right)+\left(c-d\sqrt{2}\right)\\ & =& f\left(a+b\sqrt{2}\right)+f\left(c+d\sqrt{2}\right)\\ & & \end{array}$

On further solving, we get,

$\begin{array}{rcl}f\left(\left(a+b\sqrt{2}\right)\right)+\left(c+d\sqrt{2}\right)& =& f\left(\left(ac+2bd\right)+\left(ad+bc\right)\sqrt{2}\right)\\ & =& \left(ac+2bd\right)-\left(ad+bc\right)\sqrt{2}\\ & =& \left(a-b\sqrt{2}\right)\left(c-d\sqrt{2}\right)\\ & =& f\left(a+b\sqrt{2}\right)f\left(c+d\sqrt{2}\right)\\ & & \end{array}$

Thus,f is a homomorphism. It is surjective since for all $a+\sqrt{2}\in \mathrm{\mathbb{Z}}\left[\sqrt{2}\right]$then we have$f\left(a-b\sqrt{2}\right)=a+\sqrt{2}$

Hence, the statement isproved.