Quadratic Integers over 2 are Not a Field

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 algebraic structure:
 * $\left({\Z \left[{\sqrt 2}\right], +, \times}\right)$

where $+$ and $\times$ are conventional addition and multiplication on real numbers, is not a field.

Proof
We start from the result Numbers of Type Integer a plus b root 2 Form an Integral Domain.

We have from that result that $1 + 0 \sqrt 2$ is the unity.

Now consider a representative element $a + b \sqrt 2 \in \Z \left[{\sqrt 2}\right]$.

From Difference of Two Squares we have:
 * $\left({a + b \sqrt 2}\right) \left({a - b \sqrt 2}\right) = a^2 - 2b^2$

which leads to:
 * $\left({a + b \sqrt 2}\right) \left({\dfrac {a - b \sqrt 2} {a^2 - 2b^2}}\right) = 1$

so demonstrating that the product inverse of $\left({a + b \sqrt 2}\right)$ is $\dfrac a {a^2 - 2b^2} + \dfrac {b \sqrt 2} {a^2 - 2b^2}$.

But for any two given integers $a$ and $b$, the values $\dfrac a {a^2 - 2b^2}$ and $\dfrac b {a^2 - 2b^2}$ are not in general integers.

For instance, when $a = 3$ and $b = 4$ we get:
 * $\dfrac a {a^2 - 2b^2} = \dfrac 3 {23}$ and $\dfrac b {a^2 - 2b^2} = \dfrac 4 {23}$

(unless I've got my sums wrong).

So not every element of $\left({\Z \left[{\sqrt 2}\right], +, \times}\right)$ has a product inverse, demonstrating that $\left({\Z \left[{\sqrt 2}\right], +, \times}\right)$ is not a field.