Prove no zero divisors of a ring with a radical in $\mathbb{Z}_7$

Question

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 $\mathbb{Z}_7$?

Answer 1

You have $a^2 \equiv 6b^2 \pmod{7}$. Assuming $b \not\equiv 0 \pmod{7}$ (consequently $b$ is invertible) we can rewrite the given congruence as $x^2 \equiv 6 \pmod{7}$. So the real question is does $6$ have a square root in $\Bbb{Z}_7$?

If you square the elements of $\Bbb{Z}_7$ you only get $\{0,1,2,4\}$. Thus $6$ is not a quadratic residue mod $7$, so the only way that the original equation has a solution is when both $a,b \equiv 0$. Thus no zero divisors.

Answer 2

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.