Zero Divisors in $\mathbb Z_8$

Detailed Solution

By definition, a zero divisor in this question must be a nonzero element $x\in\mathbb Z_8$ for which there exists a nonzero $y\in\mathbb Z_8$ such that $$ xy\equiv0\pmod 8. $$ We examine the nonzero elements $1,2,3,4,5,6,7$. The elements $2$, $4$, and $6$ are zero divisors because $$ 2\cdot4=8\equiv0\pmod8, $$ so $2$ is a zero divisor; $$ 4\cdot2=8\equiv0\pmod8, $$ so $4$ is a zero divisor; and $$ 6\cdot4=24\equiv0\pmod8, $$ so $6$ is a zero divisor. Now consider $1,3,5,7$. Each of these is relatively prime to $8$: $$ \gcd(1,8)=\gcd(3,8)=\gcd(5,8)=\gcd(7,8)=1. $$ Therefore, each of $1,3,5,7$ is a unit in $\mathbb Z_8$. A unit cannot be a zero divisor: if $xy\equiv0\pmod8$ and $x$ has a multiplicative inverse, multiplying by $x^{-1}$ gives $$ y\equiv0\pmod8, $$ contradicting the requirement that $y\neq0$. Hence the nonzero zero divisors are exactly $$ \{2,4,6\}. $$ Notice that $0$ is not counted because the definition explicitly requires $x\neq0$. Therefore, there are exactly $3$ zero divisors. Option B is incorrect because it includes $0$, which is excluded by the definition given in the question. Option C is incorrect because not every nonzero element of $\mathbb Z_8$ is a zero divisor. In particular, $1,3,5,$ and $7$ are units and hence are not zero divisors. Option D is incorrect because $2$, $4$, and $6$ explicitly satisfy the zero-divisor condition. Thus the correct option is $\boxed{\mathrm{A}}$.