Let $R$ be a UFD. I'm trying to understand whether the following statements are equivalent:

1) $r\in R$ lies in some prime ideal $P$ but not in the ideal $P^2$.

2) $r$ is divisible by a prime element of $R$ but not by its square.

I think I can see why 2) implies 1). Suppose 2) holds, say $r$ is divisible by a prime element $p$ but not by $p^2$. Then $r$ lies in the ideal $P=(p)$, which is prime because $p$ is prime. Also $r$ does not lie in $P^2$, because otherwise it would be divisible by $p^2$. Is that right?

Suppose 1) holds. $P$ is an abstract prime ideal. Does it necessarily contain a prime element? If so, I would be able to conclude that $r$ is divisible by that prime element. (I think it must be finitely generated since $R$ is a UFD, do the generators have to be prime elements?) Does the condition $r\notin P^2$ imply that $r $ is not divisible by the square of that element?


1 Answer 1


Your reasoning that $(2) \implies (1)$ is correct.

As for your questions in the end:

It is not true that $P$ must be finitely generated. If every prime ideal of a ring is finitely generated, then in fact every ideal is finitely generated, i.e. the ring is Noetherian (this is sometimes called Cohen's theorem). In general, UFDs are not Noetherian. A classic example is $K[x_1, x_2, x_3, \ldots]$, the ring of polynomials in an infinite number of variables over a field.

However, this is not important for your problem.

In a UFD we can factor $r$ as $r = \prod p_i^{n_i}$ with the $p_i$ distinct primes.
Show from the definition of prime ideal that $r \in P$ prime implies $p_i \in P$ for some $i$. Remember, $P$ prime means that $ab \in P \implies a\in P \text{ or } b\in P$. (In general you've now shown that a prime ideal of a UFD always contains a prime element. In fact this behavior characterizes UFDs, and is sometimes called Kaplansky's theorem)

Finally, show from this and the assumption that $r \notin P^2$ that $p_i \mid r$ but $p_i^2 \nmid r$


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