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“Every prime element is irreducible,” – This is true in an integral domain, but not in every commutative semigroup with an identity. Consider the multiplicative group of integers modulo 6: 2 and 3 are prime but indefinitely reducible since $2=2\cdot4$ and $3=3^2$. Note that the explanation of irreducibility that follows – “i.e. is divisible only by divisors of unity or elements associated to it” – is redundant (an element associated to unity is obviously a divisor of unity) and wrong; if there is a unit, every irreducible element is also divisible by itself, after all, though it is not a product $ab$ where ''both'' $a$ and $b$ are nonunits. Am I confused? -- [[User:Ivan|Ivan]] ([[User talk:Ivan|talk]]) 10:13, 31 March 2017 (CEST)