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Theorem 179 in the book Commutative Rings by Kaplansky states that:

Let $R$ be an integral domain. The following conditions are necessary and sufficient for $R$ to be a UFD.

(1) $R$ satisfies the ascending condition on principal ideals

(2) In the polynomial ring $R[x]$ all minimal prime ideals are finitely generated.

(3) For any prime ideal $P$ of grade one in $R$, $R_P$ is a UFD.

(4) In any localization of $R[x]$ all invertible ideals are principal.

I have a problem with his proof and I retype it here for more clearance.

proof

Sufficiency. We shall actually prove that $R[x]$ is a UFD; it is standard and easy that then $R$ is a UFD. We abbreviate $R[X]$ to $T$. It is also easy that the ascending chain condition on principal ideals is inherited by T.

Let $S$ be the set of all finite products of principals prime in $R$. By Theorem 177 (He means Nagata's theorem) it suffices to prove that $T_S=U$ is a UFD.

He then uses three conditions in Theorem 178 to prove $U$ is a UFD. Hence he reduces to the case prove $U_M$ is a UFD for all maximal idel $M$ in $U$.

Note that $M$ has the form $Q_S$ for suitable prime ideal in $T$ disjoint from $S$. Let $P=Q\cap R$. We claim that $P=0$ or has grade 1.

The problem is that he does not treat the case $P=0$. I think it's not trivial.

Here is my approach.

If $Q=0$ because $Q$ is maximal ideal such that disjoint from $S$ so every principal ideal $(a)$ does not disjoint from $S$ where $a$ is non-zero, non-unit, and since $S$ is saturated so $R[X]$ is a UFD . We therefore can assume $Q\neq 0 $. As $Q\cap R=0$ and from the fact that is we can not have a chain of three distinct prime ideals in $R[X]$ with the same contraction in $R$. Hence ht $Q=1$, from hypothesis (ii) $Q$ is finitely generated. Also we have $R[X]_Q\cong U_M$ so $U_M$ is a Noetherian domain with a unique non-zero prime ideal but from my previous question, it's still not enough to deduce $U_M$ is a UFD.

So I decide to post this question for the case $Q\cap R=0$.

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Anh_Rose 1210
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1 Answers1

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Suppose $P = Q \cap R = 0$, or in other words if we consider $R \setminus \{0\}$ (which is a multiplicatively closed system, as $R$ is an integral domain), then we have that $Q \cap (R \setminus \{0\}) = \varnothing$, so $R \setminus \{0\} \subset T \setminus Q$

This means if we consider $K= (R \setminus\{0\})^{-1} R$ the quotient field of $R$, we get that since $(R \setminus \{0\})^{-1}T=K[X]$ and $R \setminus \{0\} \subset T \setminus Q$, $T_Q$ (and hence also $U_M$) are localizations of $K[X]$. So since $K[X]$ is a PID, $T_Q$ and $U_M$ are PIDs, being localizations of a PID and thus they are UFDs.

Lukas Heger
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