Smart way to check the property $ab=0\implies a=0$ or $b=0$ to form an integral domain Is there a smart way to check if something satisfies the rule $ab=0\implies a=0$ or $b=0$ to show its an integral domain.
I am looking at $\{a+b\sqrt{2}:a,b\in \Bbb Z\}$ and $\{a+b\sqrt{2}:a,b\in \Bbb Q\}$ and trying to see if they are integral domains, and it occurred to me that I am bad at this, really bad.
So what I tried to do was take $(a+b\sqrt{2})(c+d\sqrt{2})=ac+bc\sqrt{2}+ad\sqrt{2}+2bd=0$
and I sit there and try to see how I can get this product to equal zero, without $(a+b\sqrt{2})$ or $(c+d\sqrt{2})$ to be equal to zero.
What is the smart mathematically experienced way to check that sort of thing?
 A: To see that $\{a+b\sqrt{2}\;|\; a,b \in \mathbb{Z}\}=\mathbb{Z}[\sqrt{2}]$ is a ring, you can check that it's a subring of $\mathbb{R}$ (show closure under subtraction and multiplication - also clearly $1$ belongs to this subring). 
An integral domain is a commutative ring with $1$, $1 \not= 0$, and no zero divisors.
Once we know that $\mathbb{Z}[\sqrt{2}]$ is a subring of $\mathbb{R}$, we know that $1 \not=0$ and it is commutative (since these things are true in $\mathbb{R}$). 
This only leaves us to check that there are no zero divisors. But this too is a property inherited from $\mathbb{R}$. Notice that a zero divisor in $\mathbb{Z}[\sqrt{2}]$ would be a zero divisor in $\mathbb{R}$. But $\mathbb{R}$ has no zero divisors (it's a field) so neither can your ring.
In general, a (non-zero) subring (with $1$) of an integral domain is itself an integral domain.
If you want to check for zero divisors directly, you can: $(a+b\sqrt{2})(c+d\sqrt{2}) = (ac+bd2)+(ad+bc)\sqrt{2}=0$ forces $ac+bd2=0$ and $ad+bc=0$. 
Without loss of generality, suppose $a+b\sqrt{2} \not=0$. 
Case 1: $a\not=0$ then $d=-bc/a$ so $0=ac+2bd=ac+2b(-bc/a)$ so that $a^2c=2b^2c$. This is impossible unless $c=0$ (use prime factorizations to see that the right hand side has even/odd power of $2$ then left hand side has odd/even power of $2$). Now $c=0$, so $0=ad+b(0)=ad$ so $d=0$ since $a\not=0$. Therefore, we must have $c+d\sqrt{2}=0$.
Case 2: $a=0$ and since $a+b\sqrt{2} \not=0$ we must have $b \not=0$. Thus $0c+2db=0$ and $0d+bc=0$. But $b \not=0$ so $d=0$ and $c=0$. Again $c+d\sqrt{2}=0$.
No zero divisors!
