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

$f\left(a+b\sqrt{2}\right)=a-b\sqrt{2}$

b. Use theorem 6.11 to show that is also injective and hence f is an isomorphism. [You may assume that$\sqrt{2}$ is irrational.]

Short answer

(a) and (b) are proved.

Step-by-step solution

Solution to Part (b)

Assume $a-b\sqrt{2}\in ker\left(f\right)$then, $f\left(a+b\sqrt{2}\right)=0$which is equal to $a-b\sqrt{2}=0$.

If $b\ne 0$then the equation $a-b\sqrt{2}=0$becomes $\sqrt{2}=\frac{a}{b}\in \mathrm{\mathbb{Q}}$which is a contradiction being $\sqrt{2}$irrational. Thus,b=0 which also implies that a=0

By using theorem 6.11 which states that assume $f:R\to S$be a homomorphism of rings with kernelK . Then, $K=\left(0_R\right)$if and only if f is injective.

Therefore, the kernel of f is trivial then by applying theorem 6.11 is injective.

Hence, the statement is proved.