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In a unique factorization ring with unity (I am not considering commutativity and zero divisors in definition of UFD) irreducible implies prime.

And it was proved in ring with unity without zero divisors (commutativity not necessary) prime implies irreducible in question Prime which is not irreducible in non-commutative ring with unity without zero divisors.

So question is:

Does in a ring with unity prime implies irreducible or not?

Definitions: $p$ is prime iff $p|ab$ implies that $p|a$ or $p|b$, and $x$ is irreducible iff $x = ab$ implies that either $a$ or $b$ is a unit.

glS
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Sushil
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    I'm not an expert on the topic, but my impression was that prime is equivalent to irreducible. Am I wrong? – Avi Feb 15 '15 at 14:10
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    @Laertes in UFD's the concepts are the same. But I know cases ($\mathbb Z[\sqrt{-5}]$ is one of them) in which not every irreducible is prime. The OP seems looking for cases in wich not every prime is irreducible. – drhab Feb 15 '15 at 14:30
  • @drhab I don't see the distinction; isn't the definition of a prime ideal one that is irreducible? – Avi Feb 15 '15 at 14:37
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    @Laertes See my answer here : http://math.stackexchange.com/questions/1076517/irreducible-and-prime-elements/1076574#1076574 – Olórin Feb 15 '15 at 14:40
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    @Laertes Have a look at the question that the OP is referring to. There you find definitions of prime and irreducible. They are not the same. – drhab Feb 15 '15 at 14:41
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    Consider $\mathbb{Z}/(6)$. The primes there are not irreducible. – Daniel Fischer Feb 15 '15 at 15:08
  • @DanielFischer Just a question: why in form of a comment and not an answer? Isn't this what the OP is asking for? – drhab Feb 15 '15 at 15:14
  • @drhab As usual, time. Writing a short comment is quicker than writing an answer (which I have now done). – Daniel Fischer Feb 15 '15 at 16:27
  • @drhab Thank you, sorry for the confusion. – Avi Feb 15 '15 at 18:39

2 Answers2

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In a ring with zero divisors, a prime element need not be irreducible. A simple example is $\mathbb{Z}/(6)$, where we see that the elements $\overline{2},\, \overline{3}$ and $\overline{4}$ are all prime - the primes $\overline{2}$ and $\overline{4}$ are associated, $\overline{4} = \overline{5}\cdot\overline{2}$ - as one verifies, but we have

$$\overline{2} = \overline{2}\cdot \overline{4},\quad \overline{4} = \overline{2}\cdot \overline{2}\quad\text{and}\quad \overline{3} = \overline{3}\cdot \overline{3},$$

so none of them is irreducible.

Daniel Fischer
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Using the definition of "irreducible" you wrote in the question, the answer is no, as illustrated in Daniel's answer. However, the generally accepted definition of "irreducible" by those who do research on factorization in commutative rings with zero divisors is: $a$ is irreducible if $a = bc \Rightarrow (a) = (b)$ or $(a) = (c)$. Using this definition, the answer is yes. The definition in the question is what factorization experts would call "very strongly irreducible". See "Factorization in commutative rings with zero divisors" by Anderson and Valdes-Leon for more information.

Jason Juett
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