TITLE: Prove no zero divisors of a ring with a radical in $\mathbb{Z}_7$
QUESTION [6 upvotes]: When having the ring
$$R:= \{a+b\sqrt{6}:\, a,b\in \mathbb{Z}_7\}$$
I have to prove the ring has no zero divisors. I want to proceed to prove with contradiction.
So I know you can rewrite this to $N(xy) = N(x) N(y)$ so we can rewrite it to:
$$N(a+b\sqrt{6}) = (a+b\sqrt{6})(a-b\sqrt{6})$$
First, let's assume $xy = 0$ for some $x\neq 0$ and $y\in \mathbb{Z}_7$. So $N(xy)=N(x)N(y)=0$. 
So either $N(x)=0$ or $N(y)=0$. If we assume $N(x)=0$, then we get: $a^2 -6b^2=0$ for some $a,b\in \mathbb{Z}_7$. 
Now I'm stuck at the contradiction part, why can't there be any zero divisors in $\mathbb{Z}_7$?

REPLY [2 votes]: The ring has no zero divisors because it embeds into a ring with no zero divisors, as the map
$$R\ \longrightarrow\ \Bbb{F}_7[x]/(x^2-6):\ \sqrt{6}\ \longmapsto\ x,$$
shows (you should verify that this is a ring homomorphism). This quotient ring has no zero divisors because the ideal $(x^2-6)\subset\Bbb{F}_7[x]$ is prime, because $x^2-6\in\Bbb{F}_7[x]$ is irreducible, because $6$ is not a square in $\Bbb{F}_7$. The latter is also the reason that $a^2-6b^2=0$ is impossible for $a,b\in\Bbb{F}_7$, which gives you your contradiction.