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My proof is different from the one my professor gave me, so I'm just checking that it's valid.

Let $R$ be an integral domain and suppose $p\in R$ is prime. Then, $P$ is a prime ideal.

Suppose, to the contrary, that $p$ is a unit. Then $\exists p^{-1}\in R$ such that $pp^{-1}=1$. Since $P$ is an ideal, $pp^{-1}=1\in P$. This implies that $P = R$, contradicting the fact that this cannot happen for a prime ideal. So $p$ is not a unit.

Now suppose that $p=ab$, for some $a,b \in R$. Then $ab\in P$. Since $P$ is a prime ideal, either $a\in P$ or $b\in P$. WLOG, suppose that $a\in P$. Then, $ab-a=0\in P$. This implies that $ab=a$, and by the cancellation property of integral domains, $b=1$, a unit. Hence, $p$ is irreducible QED

Dallas Pinkerton
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  • I don´t understand who is $P$, and why $ab-a = 0$? – ZAF Apr 04 '24 at 02:09
  • Sorry, $P$ is the ideal generated by $p$. I believe I see the error in the proof. $ab-a=0\in P$ iff ab=a, which we have not deduced. – Dallas Pinkerton Apr 04 '24 at 02:14
  • For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque Apr 04 '24 at 02:26

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