In the ring $\Bbb Z_6$ commutative ring with $1$,why $[3]$ is not irreducible?

Question

Though I was thinking in that way. $[3]=[1]*[3]$, where $[1]$ is unit of $\Bbb Z_6$ and $[3]$ is nonzero nonunit, so why not $[3]$ is irreducible in $\Bbb Z_6$. Please explain.

Answer 1

When I was teaching this stuff, I would separate elements of a commutative ring into five disjoint sets:

• $\{0\}$;
• Nonzero zero-divisors, i.e. $z$ for which there was $y\ne0$ with $zy=0$;
• Units, i.e. elements $z$ for which there was $y$ with $zy=1$;
• Reducible elements, namely those that were products of two nonzero nonunits;
• Everything else, and these are the irreducibles (primes, in $\Bbb Z$).

If you accept this classification, then in $\Bbb Z_6$, the element $3$ is a zero-divisor, since $3\cdot2=0$. And in this ring, all elements fall into the first three of my categories: there are neither reducibles nor irreducibles.