prove that $({a + b \sqrt{2},+,*)}$ is not a field How can I prove that $({a + b \sqrt{2},+,*)}$ is not a field for $a,b$ in $Z$ ? I know it's a Integral domain and I just need to prove that it does not have inverse to prove that it's not a field.
$$(a + b \sqrt{2}) *k =1 $$ and $$k = \frac{1}{a + b \sqrt{2}}$$
but I get $k = 1 $ for $a=1,b=0$ which is inverse for a particular case but not any other case. Is this enough to prove it's a field?
 A: It is enough to show a non-zero element that doesn't have an inverse in the ring. The simplest example is $2$.
Suppose $2(a + b \sqrt{2})=1$. Then $2b\sqrt{2}=1-2a$. If $b=0$, we get $a=\frac12 \not\in\mathbb Z$. If $b\ne0$, we get $\sqrt{2}=\frac{1-2a}{2b}\in\mathbb Q$, a contradiction, since $\sqrt2$ is irrational.
A: Note that we have the map $\|\cdot\|\colon k\to \Bbb Z$, $a+b\sqrt 2\mapsto a^2-2b^2$. This has the property that $\|\alpha*\beta\|=\|\alpha\|\cdot\|\beta\|$. As $\|1\|=1$, we conclude $\|\alpha^{-1}\|=\frac1{\|\alpha\|}$, and this fails to be an integer whenever $\|\alpha\|>1$ or $<-1$, e.g., for $\alpha=\sqrt 2$.
Alternatively, you may verify directly that $(a+b\sqrt 2)(0+1\sqrt 2)=1$ has no integer solutions.
A: Hint $\ $ Every $\,\alpha = a+b\sqrt 2\in\Bbb Z[\sqrt 2]\,$ is a root of a monic quadratic with integer coefficients $$\,f(x)\, =\, (x\!-\!\alpha)(x\!-\!\bar\alpha)\, =\, x^2-(\overbrace{\alpha+\bar\alpha}^{\large 2a})\,x + \!\!\!\!\overbrace{\alpha\bar \alpha}^{\large\ \ a^2-2b^2}\!\!\!\in\Bbb Z[x]$$
By the Rational Root Test any rational root of $\,f\,$ is an integer, therefore the only rationals in $\,\Bbb Z[\sqrt2]\,$ are integers, so it cannot be a field (e.g. $\,1/2\not\in \Bbb Z[\sqrt 2])$
Remark $\ $ Unlike some direct arguments, this argument generalizes widely, e.g.
$\qquad$ If $\ D \subset E\,$ is an integral extension of domains then $\,D\,$ is a field $\iff E\,$ is a field.
