Mapping on Quadratic Integers over 2 to Conjugate is Automorphism

Theorem
Let $\Z \sqbrk {\sqrt 2}$ denote the set of quadratic integers over $2$:
 * $\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$

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

Then the mapping $\phi: \Z \sqbrk {\sqrt 2} \to \Z \sqbrk {\sqrt 2}$ defined as:
 * $\forall x = a + b \sqrt 2 \in \Z \sqbrk {\sqrt 2}: \map \phi {a + b \sqrt 2} = a - b \sqrt 2$

is a ring automorphism.

Proof
We have Quadratic Integers over 2 form Subdomain of Reals.

First we note that:
 * $\forall x \in \Z \sqbrk {\sqrt 2}: \map \phi x \in \Z \sqbrk {\sqrt 2}$

Proof of Bijection
Let $\map \phi {a + b \sqrt 2} = \map \phi {a' + b' \sqrt 2}$.

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 $y = c + d \sqrt 2 \in \Z \sqbrk {\sqrt 2}$.

We have that:

Hence:
 * $\forall y \in \Z \sqbrk {\sqrt 2}: \exists x \in \Z \sqbrk {\sqrt 2}: \map \phi x = y$

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.