Let $R$ be a ring (possibly noncommutative 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](https://en.wikipedia.org/wiki/Chebotarev%27s_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](https://math.stackexchange.com/questions/4271735/is-there-an-integral-extension-of-mathbbz-keeping-its-prime-elements#comment8888446_4271735)).


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).


In [this answer](https://math.stackexchange.com/a/4277145/84284) Keith Conrad provides an example with finitely many new irreducible elements (up to unit):



> 
> 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.
> 
> 
> 


A *finite extension* of $\mathbb{Z}$ is a ring which is an extension of $\mathbb{Z}$ and a $\mathbb{Z}$-module of finite type. The example above is not.


**Question**: Is there a *finite* extension $R$ of $\mathbb{Z}$ keeping its irreducible elements, and also with new irreducible elements (up to units), but only finitely many (up to units)?


As Keith Conrad pointed out in [this comment](https://math.stackexchange.com/questions/4275346/existence-of-a-finite-extension-of-the-primes#comment8904743_4277145), if $R$ is an integral domain, then we are in the situation of Chebotarev's density theorem mentioned above. So $R$ must be noncommutative and/or with zero-divisors.




---


"up to units" means that if $r$ is an irreducible element and $u,v$ are units, then $r$ and $urv$ count for one.



Existence of a finite extension of ℤ providing a finite extension of the primes
This post started as some minor observations, but I believe it now contains a full proof that there is no such ring. Throughout, we let $R$ be an example of the kind wanted.


**Observation 1**: $R$ is indecomposable.


*Proof of observation 1*: Write $R=S\_1\times S\_2$, with $S\_1,S\_2\neq 0$. Suppose, by way of contradiction, that $S\_1$ has finite characteristic $m>1$. Let $p$ be any integer prime divisor of $m$. Then $p=(p,p)$ is not irreducible (since $(1,0)(p,p)=0$), contradicting our assumption that the primes of $\mathbb{Z}$ stay irreducible in $R$ (and using your definition of irreducible, which includes not being a zero-divisor).


Thus, each $S\_i$ has characteristic $0$. Moreover, the argument above shows that each integer prime cannot be a zero-divisor in any $S\_i$.


If an integer prime $p$ is a unit in $S\_i$, then $S\_i$ contains a copy of $\mathbb{Z}[1/p]$, which is not finitely generated as a $\mathbb{Z}$-module, contradicting the fact that submodules of finitely generated $\mathbb{Z}$-modules are finitely generated.


Also, if $p$ is not irreducible in $S\_1$ (or, by symmetry $S\_2$), say $p=ab$ where $a,b$ are nonunits, then $(p,p)=(a,p)(b,1)$ is a product into nonunits, contradicting the irreducibility of the primes of $\mathbb{Z}$.


The elements $\{(p,1)\, :\, p>1\text{ prime}\}$ are not units, not zero-divisors, and are easily shown to be irreducible in $R$. They are not associate to each other, nor to the integer primes $(p,p)\in R$. Thus, this would contradict our assumption of having only finitely many new primes (up to associates).


**Observation 2**: If $r\in R$, then the subring $\mathbb{Z}[r]\subseteq R$ is isomorphic to $\mathbb{Z}[x]/I$ where $I$ is a principal ideal generated by a monic polynomial $f$.


*Proof of observation 2*: Since $\mathbb{Z}[r]$ is a $\mathbb{Z}$-submodule of $R$, it must be finitely generated. Hence, $r$ satisfies some monic polynomial. Let $f\in \mathbb{Z}[x]$ be the monic polynomial of smallest degree satisfied by $r$.


Now, $\mathbb{Z}[r]\cong \mathbb{Z}[x]/I$ where $I$ is the ideal of polynomials satisfied by $r$. It thus suffices to show that $I=f\mathbb{Z}[x]$. Supposing otherwise, let $g\in I-f\mathbb{Z}[x]$.


Working over $\mathbb{Q}[x]$ for a moment, take $d=\gcd(f,g)$, a monic polynomial. Since $f$ is monic, we know that $d\in \mathbb{Z}[x]$. By the extended Euclidean algorithm, we can write $d$ as a $\mathbb{Q}[x]$-linear combination of $f$ and $g$. Hence $cd\in I$ for some minimal $c\in \mathbb{Z}\_{>0}$. We know $c\neq 1$ by the minimality of the degree of $f$ (since $d$ is monic).


Let $p$ be any prime dividing $c$. Then $p\cdot (c/p)d(r)=0$, and hence $p$ is a zero-divisor in $R$, which is a contradiction.


**Observation 3**: The monic polynomial $f$ is a power of an irreducible polynomial $q\in\mathbb{Z}[x]$.


*Proof of observation 3*: Assume $f$ is not a power of an irreducible, so $f=gh$ over $\mathbb{Z}[x]$, where $\gcd(g,h)=1$ and $\deg(g),\deg(h)\geq 1$.


By an argument used previously, we can find some constant $c\in \mathbb{Z}\_{>0}$ that is a $\mathbb{Z}[x]$-linear combination of $g$ and $h$, say $c=gg'+hh'$.


Let $J=g\mathbb{Z}[x]$ and $K=h\mathbb{Z}[x]$. Let $S\_1=\mathbb{Z}[x]/J$ and $S\_2=\mathbb{Z}[x]/K$. Consider the map $\varphi\colon \mathbb{Z}[x]/I\to S\_1\times S\_2$ where $a+I\mapsto (a+J,a+K)$. This is an injective ring homomorphism. Thus, we can view $\mathbb{Z}[r]$ as a subring of this direct product. Moreover the image of $\mathbb{Z}[r]$ contains the elements $(c+J,0+K)$ and $(0+J,c+K)$. (Indeed, $h(r)h'(r)\mapsto (hh'+J,hh'+K) = (c+J,0+K)$.) Since the image of $\mathbb{Z}[r]$ contains $(1+J,1+K)$, we see (by a trivial use of Dirichlet's theorem) that the image of $\mathbb{Z}[r]$ contains elements of the form $(p+J,1+K)$ and $(1+J,p+K)$, with $p$ an integer prime. Thus (suppressing the coset notation) we have $(p,p)=(p,1)(1,p)$ in this image. Back inside $\mathbb{Z}[r]$, write this factorization as $p=ab$. Without loss of generality, $a$ is a unit in $R$, with inverse $a^{-1}$.


The ring $\mathbb{Z}[r,a^{-1}]$ is a finitely generated $\mathbb{Z}$-module (being a $\mathbb{Z}$-submodule of $R$), and it naturally maps onto $(\mathbb{Z}[x]/J)[1/p]$, which is not a finitely generated $\mathbb{Z}$-module, which is a contradiction.


**Observation 4**: $\deg(q)=1$.


*Proof of observation 4*: If $\deg(q)>1$, then there are infinitely many primes $p$ that factor nontrivially in $\mathbb{Z}[x]/(q)$ (by an argument supplied by John Voight) and such a factorization lifts to $\mathbb{Z}[x]/(q^n)\cong \mathbb{Z}[r]$; see [this link](https://mathoverflow.net/questions/406752/hensels-lemma-bezouts-identity-and-the-integers) for the full argument.


Now, if such a prime $p$ is to be irreducible in $R$, one of those factors must become a unit $u$ in $R$. But then $\mathbb{Z}[r,u^{-1}]$ is a commutative ring, mapping to $(\mathbb{Z}[x]/(q))[a^{-1}]$ (where $a$ is that corresponding nontrivial factor, but modulo $q$ rather than $q^n$), which is not a finitely generated $\mathbb{Z}$-module, giving us a contradiction, as before.


**Observation 5**: If $J$ is the set of nilpotent elements in $R$, then $J$ is a (nilpotent) ideal.


*Proof of observation 5*: An arbitrary element $r\in R$ satisfies $q^n$, with $q(x)=x-k\in \mathbb{Z}[x]$, and hence $r=k-t$ where $t$ is nilpotent of index $n$. Therefore, $r(k^{n-1}+k^{n-2}t + \cdots + t^{n-1})=k^n$. If $k\neq 0$, then $k^n$ is not a zero-divisor (since all the primes of $\mathbb{Z}$ are not zero divisors), and hence $r$ is not a zero-divisor. Thus, the zero-divisors are exactly the nilpotent elements.


Next, if $t\in R$ is nilpotent, say of index $n\geq 1$, and $r\in R$ is arbitrary then $t^{n-1}(tr)=0$ with $t^{n-1}\neq 0$. Hence $tr$ is a zero-divisor, hence nilpotent. Thus $J$ is closed by right multiplication from $R$; and by left multiplication by a symmetric argument.


In particular $J$ is closed under multiplication. By the paper [Rings in which nilpotents form a subring](https://www.jstor.org/stable/44000112) by Janez Ster ([arXiv version here](https://arxiv.org/abs/1510.07523)), we know that $J$ is also closed under addition, since $R$ satisfies Koethe's conjecture. (Quick argument: The Jacobson radical $J(R)$ is nilpotent, since tensoring up to $\mathbb{Q}$ keeps it in a finite-dimensional $\mathbb{Q}$-algebra.) Thus, $J$ is an ideal.


**Observation 6**: The type of ring we want cannot exist.


*Proof of observation 6*: Fix a $\mathbb{Z}$-basis for $J$, say $t\_1,\ldots, t\_k$. Then a $\mathbb{Z}$-basis for $R$ is $1,t\_1,\ldots, t\_k$ (since every element of $R$ is an integer shift from a nilpotent, by observation 4).


The units of $R$ are exactly $\pm 1+t$ for some $t\in J$. Thus, the associates of a prime $p\in \mathbb{Z}$ are of the form $\pm p+t'$ where $t'\in J$ is divisible by $p$. Thus $p+t\_1$ is not associate to any of the integer primes. A similar argument shows that $p+t\_1$ and $p'+t\_1$ are not associate, for distinct primes $p,p'$. Further, $p+t\_1$ is not a unit, and not a nilpotent (hence not a zero-divisor). It thus suffices to show that $p+t\_1$ is irreducible, and then we'll have infinitely many "new" nonassociate irreducibles.


If $p+t\_1$ factors as $(a+t)(b+t')$ with $a,b\in \mathbb{Z}$ and $t,t'\in J$, then $ab=p$. Hence, without loss of generality, $a=\pm 1$. Thus, $a+t$ is a unit, so every factorization is trivial.


