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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?

malat
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Sebastien Palcoux
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    At least for integral domains the answer is no. – leoli1 Oct 09 '21 at 07:22
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    @leoli1 isn't it necessary that such an extension be an integral domain, since $0$ is a prime element of $\mathbb{Z}$? – Alex Wertheim Oct 09 '21 at 07:38
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    @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) – leoli1 Oct 09 '21 at 07:39
  • @leoli1 fair enough, I suppose I hadn't thought about these convention subtleties! And indeed, Wikipedia shows that your definition is standard, and what I had in mind is not. (Also, I'd be very interested to hear about your non-integral counterexample, if you wouldn't mind sharing it.) – Alex Wertheim Oct 09 '21 at 07:43
  • I don't have a counterexample. I just know for sure that the statement holds true if the extension ring is assumed to be an integral domain (because in an algebraic number field infinitely many primes split completely), but I don't know whether it is true without that assumption. – leoli1 Oct 09 '21 at 07:45
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    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 $b$. – user10354138 Oct 09 '21 at 08:57
  • @user10354138: Is your example an integral extension of $\mathbb{Z}$? – Sebastien Palcoux Oct 09 '21 at 12:00
  • @leoli1: Can you prove your statement in details (or provide a reference)? Next, is there an integral extension of $\mathbb{Z}$ which is not an integral domain? – Sebastien Palcoux Oct 09 '21 at 12:07
  • @SebastienPalcoux Every element $x\in\mathbb{Z}+\mathbb{Q}\epsilon$ is of the form $x=m+q\epsilon$ and satisfies $x^2-2mx+m^2=(x-m)^2=0$. – user10354138 Oct 09 '21 at 12:56
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    @SebastienPalcoux To see there are infinitely many primes that split (completely), recall Chebotarev's density theorem, see also this answer. – user10354138 Oct 09 '21 at 13:20
  • @user10354138: Thanks for your example! You may want to put it as an answer. – Sebastien Palcoux Oct 10 '21 at 05:34
  • If we allow the extension to be non-commutative (which is not usual for integral extension), we could also have $M_n(\mathbb{Z})$, for $n≥2$. See https://math.stackexchange.com/q/4273666/84284 – Sebastien Palcoux Oct 11 '21 at 15:30

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