Solution of $a.b+b=0.$ over the ring $\mathbb{Z}_8$

Question

Let $\mathbb{Z}_8$ be the ring containing elements integer modulo 8 with operation $+$ and $.$ being addition and multiplication modulo 8 resp. I want to find $a$ for every $0\neq b \in \mathbb{Z}_8$ such that $$a.b+b=0.$$

PS- I can write $a.b+b=(a+1).b=0$, which gives $a+1=0$ implies $a=-1=8-1=7$ over $\mathbb{Z}_8$. But when I check element wise, for $b=2,4,6$ we can have also $a=3$ as a solution. I didn’t expect 2 solutions when I am solving the equation $a.b+b=(a+1).b=0$. I see this is happening with non unit elements but Why is it so?

Answer 1

The property $xy=0\implies x=0 \text{ or }y=0$ is only valid in integral domains. If this is not true, we call $x$ and $y$ zero divisors.

A fairly elementary theorem in ring theory says that $\mathbf{Z}/p\mathbf{Z}$ is an integral domain if and only if $p$ is prime, and $8$ is not prime.

Answer 2

When you write $(a+1)b = 0$, then that's fine. To see why it doesn't follow that $a+1 = 0$, however, let's remember what that means in terms of regular integers: It means that for the $b$ you have chosen, we are looking for a number $a+1$ which, when multiplied by that $b$, gives us a number divisible by $8$.

Clearly, if $b$ is odd, then $a+1$ must be divisible by $8$. But if $b$ is even, then it's enough that $a+1$ is divisible by $4$, and if $b$ is divisible by $4$, then $a+1$ just has to be even. This is what gives you the additional solutions (two solutions in total for $b = 2$ or $6$, and four solutions for $b = 4$).