# Number systems violating easy primes

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?

• If I recall correctly, an integral domain is a unique factorization domain if and only if those two definitions are equivalent. – Cameron Buie Jul 5 '18 at 1:23
• 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. – Cameron Buie Jul 5 '18 at 1:34
• @CameronBuie, see en.wikipedia.org/wiki/Unique_factorization_domain#Properties. We need ACCP, which is essentially the existence of factorization. – lhf Jul 5 '18 at 1:35
• 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"? – rschwieb Jul 5 '18 at 2:07
• @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. – Hans Lundmark Jul 17 '18 at 12:07

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.

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.