# Is there an integral extension of $\mathbb{Z}$ keeping its prime elements?

Let $$R$$ be a ring. An extension $$S$$ of $$R$$ is called prime-keeping if every element $$p$$ prime in $$R$$ is also prime in $$S$$.

Consider the ring $$\mathbb{Z}$$ and the following two extensions: $$\mathbb{Z}[X]$$ and $$\mathbb{Z}[i]$$. The first one is prime-keeping whereas the second is not, because $$2 = (1+i)(1-i)$$. Now the second one is an integral extension of $$\mathbb{Z}$$ whereas the first one is not.

Question: Is there a non-trivial integral extension of $$\mathbb{Z}$$ which is prime-keeping?

• At least for integral domains the answer is no. Oct 9, 2021 at 7:22
• @leoli1 isn't it necessary that such an extension be an integral domain, since $0$ is a prime element of $\mathbb{Z}$? Oct 9, 2021 at 7:38
• @AlexWertheim I guess that depends on the definition of prime element. In the definition I know a prime element has to be regular (or at least non-zero) Oct 9, 2021 at 7:39
• Unless I am missing something, isn't $\varinjlim_n\mathbb{Z}[\frac1n\epsilon]/(\epsilon^2)=\mathbb{Z}+\mathbb{Q}\epsilon$ an obvious example? $p$ in $\mathbb{Z}$ divides $(a+q\epsilon)(b+r\epsilon)=ab+(ar+bq)\epsilon$ iff $p$ divides $ab$ iff $p$ divides $a$ or $p$ divides $b$. Oct 9, 2021 at 8:57
• @SebastienPalcoux To see there are infinitely many primes that split (completely), recall Chebotarev's density theorem, see also this answer. Oct 9, 2021 at 13:20