5
$\begingroup$

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?

$\endgroup$
12
  • 2
    $\begingroup$ At least for integral domains the answer is no. $\endgroup$
    – leoli1
    Oct 9, 2021 at 7:22
  • 1
    $\begingroup$ @leoli1 isn't it necessary that such an extension be an integral domain, since $0$ is a prime element of $\mathbb{Z}$? $\endgroup$
    – Alex Wertheim
    Oct 9, 2021 at 7:38
  • 1
    $\begingroup$ @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) $\endgroup$
    – leoli1
    Oct 9, 2021 at 7:39
  • 3
    $\begingroup$ 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 $b$. $\endgroup$
    – user10354138
    Oct 9, 2021 at 8:57
  • 1
    $\begingroup$ @SebastienPalcoux To see there are infinitely many primes that split (completely), recall Chebotarev's density theorem, see also this answer. $\endgroup$
    – user10354138
    Oct 9, 2021 at 13:20

0

Reset to default

You must log in to answer this question.