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Many students are surprised to learn that the definition of prime is not generally “only divisible by 1 and itself” for general number systems.

What are some examples of numbers systems for which $p|ab$ implies either $p|a$ or $p|b$ is not equivalent to the definition $p$ only is divisible by 1 and itself? And explicit constructions if violating numbers in these systems?

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  • $\begingroup$ If I recall correctly, an integral domain is a unique factorization domain if and only if those two definitions are equivalent. $\endgroup$ – Cameron Buie Jul 5 '18 at 1:23
  • $\begingroup$ Nope. It seems that "irreducible$\implies$prime" also holds in GCD domains, which need not be UFDs. Perhaps "irreducible$\implies$prime" is equivalent to being a GCD domain, but that's beyond my knowledge. $\endgroup$ – Cameron Buie Jul 5 '18 at 1:34
  • $\begingroup$ @CameronBuie, see en.wikipedia.org/wiki/Unique_factorization_domain#Properties. We need ACCP, which is essentially the existence of factorization. $\endgroup$ – lhf Jul 5 '18 at 1:35
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    $\begingroup$ Even the definition "$p$ is only divisible by $1$ and itself" is too restrictive to be interesting. It does not even hold for $\mathbb Z$, but it does hold for $\mathbb N$. Others in the comments have already turned on to the obvious improvement: for which rings are "irreducible( which means $x=ab$ implies one of $a$ or $b$ is a unit) elements" the same as "prime elements"? $\endgroup$ – rschwieb Jul 5 '18 at 2:07
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    $\begingroup$ @CameronBuie: If I understand things correctly, integral domains where irreducible implies prime form a class of their own (which strictly includes the class of so-called Schreier domains, which in turn strictly includes the class of GCD domains). They are often called AP-domains, for “atoms (= irreducibles) are prime”, and also EL-domains, for “Euclid's Lemma”, in Pete Clark's commutative algebra notes. $\endgroup$ – Hans Lundmark Jul 17 '18 at 12:07
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In more general systems, these two notions are different and have different names:

  • "irreducible" means something that cannot be further decomposed.

  • "prime" means something that divides at least one factor of a product if it divides the product.

Consider the set $2\mathbb N$. Then $30$ is an irreducible element of $2\mathbb N$. Now, $30$ divides $60=6\cdot10$ but does not divide $6$ or $10$. Therefore, $30$ is not prime in $2\mathbb N$.

Another classical example is the Hilbert monoid $4\mathbb N+1$. Then $21$ is irreducible but not prime because $21$ divides $9 \cdot 49$ but does not divide $9$ or $49$.

In both systems, the numbers can always be decomposed into products of irreducibles but this decomposition is not always unique because not every irreducible is prime.

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If you want an example which is an integral domain, consider the ring $\mathbf{Z}[i \sqrt3]$ which consists of complex numbers of the form $$ z = a + b i \sqrt{3} ,\qquad a,b \in \mathbf{Z} . $$ Then $$ 4 = 2 \cdot 2 $$ and $$ 4 = (1+i \sqrt3)(1-i \sqrt3) $$ are two essentially different factorizations of $4$ into irreducible elements.

That the numbers $z=2$ and $z=1 \pm i \sqrt3$ are irreducible in this ring can be seen by contemplating the $\mathbf{Z}$-valued function $N(z)=|z|^2=a^2+3b^2$ which has the multiplicative property $N(z_1 z_2)=N(z_1)N(z_2)$. Since $N(z)=4$ in both cases, and $N=1$ only for the invertible elements $\pm 1$, a nontrivial factorization $z=w_1 w_2$ would have to have $N(w_1)=N(w_2)=2$, but the ring contains no elements with $N=2$.

On the other hands, those numbers cannot be prime in this ring, since a factorization of a ring element into primes is always essentially unique (i.e., unique up to reordering and multiplication by invertible elements), and we saw two different factorizations of $4$ above.

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