Existence of a finite extension of ℤ providing a finite extension of the primes

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) 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).

In this answer 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, 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.

• Of course you already implicitly reference it by linking to a comment and an answer, but it may be worth mentioning that this is a refinement of an MSE question, and specifying exactly what's different from that question. Oct 18, 2021 at 12:33
• @LSpice: You are right, I will edit that later. The refinement is that it asks for a finite extension of $\mathbb{Z}$. Now I wonder whether the last part of the question (i.e. "but only finitely many (up to units)") is automatic for a finite extension of $\mathbb{Z}$... Oct 18, 2021 at 12:42
• @WlodAA: yes, elements in $\mathbb{Z}[X]$. Oct 19, 2021 at 4:55
• I think the $M_n(\mathbb{Z})$ example is flawed. The unique ring map from $\mathbb{Z} \to M_n(\mathbb{Z})$ takes $p$ to a determinant $p^n$ matrix ($p$ times the unit matrix), so the linked answer doesn't apply. And indeed, $5$ splits in $M_2(\mathbb{Z})$, since $M_2$ contains a copy of $\mathbb{Z}[i]$. Oct 19, 2021 at 10:30
• If I were you I would remove the bit about $M_n(\mathbb{Z})$, now it is really confusing to tell what you're looking for. If you're not after ring extensions and happy with some injective multiplicative monoid map from $\mathbb{Z}$ to some other ring $R$ which takes primes to irreducibles and hits all but finitely many irreducibles, you can take basically any $R$ with countably many irreducibles (for example $\mathbb{Z}$ itself). Oct 19, 2021 at 19:00

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 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 by Janez Ster (arXiv version here), 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.

• Dear Pace, it seems that your Observation 4 fully settles the commutative case in the negative. Doesn't it? I have two minor remarks. In the proof of Observation 1, it's not necessary to assume that an irreducible element is not a zero divisor (reasoning with $(p,1)$ and $(1,p)$ one is led to a product of fields, which contains no irreducible elements). In the proof of Observation 2, it helped me to note that $d$ is either $1$ or $f$. Oct 23, 2021 at 11:34
• @LucGuyot I figured out how to do the full case, and it is now written up. Oct 23, 2021 at 14:53
• About the proof of Observation 6: it looks as if you implicitly assumed that the additive group of $R$ is torsion-free. Can we make this assumption? A potential candidate is the Nagata idealization $\mathbb{Z} (+) \mathbb{Z}/2\mathbb{Z}$ of the $\mathbb{Z}$-module $\mathbb{Z}/2\mathbb{Z}$. It fails to match the requirements only because $(2^n, 1 + 2 \mathbb{Z})$ is irreducible for every $n > 0$. (By the way, isn't $d$ necessarily equal to $1$ in the proof of Observation 2?) Oct 23, 2021 at 20:51
• @LucGuyot The primes of $\mathbb{Z}$ remain irreducible, and according to the given definition irreducibles must not be zero-divisors. Thus, the additive group of $R$ is torsion-free. Oct 24, 2021 at 1:09
• @PaceNielsen: Yes sure, I was confused by the example $\mathbb{Z} + \mathbb{Q}\epsilon$ (with $\epsilon^2=0$) which keeps the prime numbers as prime elements, whereas $\mathbb{Z} + \mathbb{Z}\epsilon$ keeps them as irreducible elements only (not prime, because every $p$ divides $\epsilon^2 = 0$ but does not divide $\epsilon$). Oct 26, 2021 at 3:03