Number systems violating easy primes

Solution 1:

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.

Solution 2:

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.