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If the ring is $\mathbb{Z}$,we all know the answer is no because $p_1p_2...p_n+1$ is also a prime if we only have finite primes.I'm not sure if the idea also works in other rings.I think it's true if the ring is a UFD,and maybe we can consider some special situations such as a domain or a commutative ring.

jdhejw
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  • See https://mathoverflow.net/questions/99606/infinite-domain-with-finite-number-of-prime-idealselements. – kipf Oct 05 '23 at 07:55
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    Do you mean "finitely many"? $\mathbb{Z}$ has finite primes e.g. $2$. Pedantic - yes but this is mathematics. – badjohn Oct 05 '23 at 08:26

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The ring of $p$-adic integers $\Bbb Z_p$ is a local ring. Hence the only prime element is the element $p$. Clearly $\Bbb Z_p$ is an infinite ring.

Note that in $\Bbb Z$ it is not true, that $p_1p_2\cdots p_n+1$ is again a prime - see Prove or disprove $p_1p_2\cdots p_n+1$ is prime for all $ n\geq 1$

We have $$ 2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13+1=59\cdot 509. $$

Dietrich Burde
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  • Thank you but you maybe misunderstand me.I mean that if we only have finite primes in $\mathbb{Z}$,then $p_1...p_n+1$ must be a prime so it contradicts. – jdhejw Oct 05 '23 at 08:43
  • @jdhejw Yes, I am sorry, you are right. I was thinking that we cannot have finitely many primes then, so that this need not be a prime. – Dietrich Burde Oct 05 '23 at 08:51