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.
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5$[3]=[3]\cdot[3]$ too. – hmakholm left over Monica Oct 17 '18 at 03:48
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1You always have $[1]\cdot [n]=[n]$. What matters is if that is the only representation as a product. – JMoravitz Oct 17 '18 at 04:04
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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.

Lubin
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2For the classes to be disjoint, don't you need to explicitly require that a reducible element (or its factors) is not a zero divisor? – hmakholm left over Monica Oct 17 '18 at 04:24
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Thank you for such nice explanation. but I don't understand the last line, that neither reducibles nor irreducible. Actually I don't know am I wrong or right that R, commutative ring with 1,is called reducible if p is not irreducible. – sumi Oct 17 '18 at 08:10
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2@sumi Although regarded as ordinary English words, you would expect everything to be either reducible or irreducible, it is not usual to consider it that way. It is usual to distinguish the units as neither. Consider the rational numbers, which are reducible and which are irreducible? I have not seen Lubin's categorisation before but I like it. He separates out the zero-divisors as well. I expect that I have not seen because I have only seen reducibility discussed in integral domains. – badjohn Oct 17 '18 at 09:18
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@sumi Worth emphasis is that there is no standard terminology for factorization theory in rings with zero-divisors, where basic notions such as "associate" and "irreducible" bifurcate into at least a few inequivalent notions, e.g. see the papers cited here. – Bill Dubuque Mar 09 '19 at 02:55