# Prove that a UFD $R$ is a PID if every nonzero prime ideal in $R$ is maximal.

Let $$R$$ be a UFD. Prove that $$R$$ is a PID if every nonzero prime ideal in $$R$$ is maximal.

The author gives the following hint: assume that every nonzero prime ideal of $$R$$ is maximal, and prove that every maximal ideal in $$R$$ is principal; then use Proposition 3.5 to relate arbitrary ideals to maximal ones, and prove that every ideal of $$R$$ is principal

Proposition 3.5. Let $$I ≠( 1 )$$ be a proper ideal of a commutative ring $$R$$. Then there exists a maximal ideal $$\frak{m}$$ of $$R$$ containing $$I$$.

I could prove every maximal ideals is principal, but then I got stuck. I know the question has been asked in MSE, but I couldn't find an answer that use the same method as the hint, so I really want to know how to use the method

• It might be worth pointing out that the book in question is Aluffi's Algebra: Chapter $\mathit0$. – mrtaurho Feb 14 at 18:06

You have already proved that every maximal ideal is principal.

Now take any proper ideal $I=(f_1, \dotsc, f_s) \subset R$ and let $I \subset (m)$ be a maximal ideal containing it.

In particular, for any $x \in I$, we have that $x$ admits the factor $m$. Thus you can consider the ideal $J=\left(\frac{f_1}{m}, \dotsc, \frac{f_s}{m} \right) \supset I$. It is easy to show that the $J$ is strictly larger than $I$, because we have $I=mJ$.

Hence by noetherian induction, we can assume that $J$ is principal, say $J=(f)$. Then $I=mJ=(mf)$.

Of course you need the noetherian property, but it is well known that one can test all ideals being finitely generated on prime ideals. And you have already checked this.

Ok, here is an argument without using the noetherian property, but only the ACCP property, which is a gimme for UFD's.

After having constructed $J=J_0$ with $I=m_0J_0$ (You can do this without assuming that $I$ is finitely generated), we have two cases:

1. $J_0=R$. Then $I=(m_0)$ and you are done.

2. $J_0$ is a proper ideal. Then we can repeat the argument and find $m_1$ with $J_0=m_1J_1$. Iterating this gives you $I=m_0m_1 \dotsb m_tJ_t$. Since $R$ is a UFD, this will eventually stop, since a non-zero element in $I$ has only finitely many irreducible factors. When this stops after $d$ steps, we have $J_d=R$ and thus $I=(m_0m_1 \dotsb m_d)$.