Mapping on Quadratic Integers over 2 to Conjugate is Automorphism

Theorem
Let $\Z \left[{\sqrt 2}\right]$ denote the set:
 * $\Z \left[{\sqrt 2}\right] := \left\{{a + b \sqrt 2: a, b \in \Z}\right\}$

... that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are integers.

Then the mapping $\phi: \Z \left[{\sqrt 2}\right] \to \Z \left[{\sqrt 2}\right]$ defined as:
 * $\forall x = a + b \sqrt 2 \in \Z \left[{\sqrt 2}\right]: \phi \left({a + b \sqrt 2}\right) = a - b \sqrt 2$

is a ring automorphism.

Proof
We have Numbers of Type Integer a plus b root 2 Form an Integral Domain.

First we note that $\forall x \in \Z \left[{\sqrt 2}\right]: \phi \left({x}\right) \in \Z \left[{\sqrt 2}\right]$.

Proof of Bijection
Let $\phi \left({a + b \sqrt 2}\right) = \phi \left({a' + b' \sqrt 2}\right)$.

Then $a - b \sqrt 2 = a' - b' \sqrt 2$ and so $a + b \sqrt 2 = a' + b' \sqrt 2$.

So $\phi$ is injective.

Now let $y = c + d \sqrt 2 \in \Z \left[{\sqrt 2}\right]$.

We have that:
 * $\phi \left({c + \left({-d}\right) \sqrt 2}\right) = c - \left({-d}\right) \sqrt 2 = c + d \sqrt 2$

Hence:
 * $\forall y \in \Z \left[{\sqrt 2}\right]: \exists x \in \Z \left[{\sqrt 2}\right]: \phi \left({x}\right) = y$

and so $\phi$ is surjective.

So $\phi$ is a bijection.

Proof of Morphism
Now consider $x = a + b \sqrt 2, y = c + d \sqrt 2$.

Then:

So $\phi$ is a ring homomorphism which is also a bijection, whose image equals its domain.

Hence the result by definition of ring automorphism.