Let $R$ be a ring (possibly non-commutative with zero-divisors). A non-unit and non-zero-divisor element $r \in R$ will be called irreducible if for all $a,b \in R$ such that $r=ab$, then $a$ or $b$ is a unit.
Many extensions $R$ of $\mathbb{Z}$ do not keep its set of irreducible elements, for example $2 = (1+i)(1-i)$ in $\mathbb{Z}[i]$. More generally (Chebotarev's density theorem) if $R$ is the ring of integers of a Galois extension of $\mathbb{Q}$ of degree $n$ then the prime numbers that completely split in $R$ have density $1/n$ (so that there are infinitely many ones).
Here are examples of extensions of $\mathbb{Z}$ keeping its irreducible elements:
- $\mathbb{Z}[X]$: every prime number is an irreducible polynomial, but $\mathbb{Z}[X]$ contains also infinitely many other irreducible polynomials (up to units),
- $\mathbb{Z} + \mathbb{Q}\epsilon$ (with $\epsilon^2=0$): it keeps the irreducibles of $\mathbb{Z}$ but there is no new ones up to units (see this comment).
Observe that the extensions of $\mathbb{Z}$ mentioned above (keeping its irreducible elements) either have infinitely many new irreducible elements (up to units) or have none (up to units).
Question: Is there an extension of $\mathbb{Z}$ keeping its irreducible elements, and also with new irreducible elements (up to units), but only finitely many (up to units)?
"up to units" means that if $r$ is an irreducible element and $u,v$ are units, then $r$ and $urv$ count for one.
Best Answer
Extend $\mathbf Z[x]$ by inverting all nonconstant irreducible elements except for $x$. That gives you a UFD whose primes elements are the ordinary prime numbers and $x$, up to units.