I'm working on a homework problem and I'm completely stumped proving the following implication:

If $R$ is an integral domain such that every nonzero prime ideal contains a prime element, then every nonzero, nonunit element of $R$ is expressible as a product of primes.

Hints would be appreciated, nothing I've tried seems to be working. Thanks.


The result you want to prove is known as Kaplansky's criterion for unique factorization domains (UFD). To prove it you have to consider the set $S$ consisting of the units of $R$ together with (finite) products of all the prime elements of $R$.

Basically the idea is to show that $S=R\setminus \{0\}$. This can be done using the following result:

Lemma: $S$ is a saturated multiplicatively closed set.

Proof: If $x, y\in S$, then we can write both $x$ and $y$ as product of primes, so $xy$ can be also written as product of primes. This shows that $S$ is a multiplicatively closed set.

Now, to prove that $S$ is saturated we have to show that if $x\in S$, then every divisor of $x$ is in $S$ too. If we write $x=up_1\cdots p_n$, where $u$ is an unit and the $p_i$'s are prime, then it can be shown by induction on $n$ that every divisor of $x$ is in $S$. I let you to do this proof by induction.

Now, we argue by contradiction assuming that there is a nonzero element $a\in R$ such that $a\notin S$, so the ideal generated by $a$, $\langle a\rangle$, is disjoint from $S$, i.e., $S\cap \langle a\rangle=\emptyset$, because if there were some $ra\in S$, then $a$ would be in $S$ (because $a\mid ra$ and $S$ is saturated by the lemma above), contradicting our hypothesis that $a\notin S$.

Therefore the set $A=\{I\; \text{nonzero ideal of}\; R:I\cap S=\emptyset\}$ is non-empty and then by Zorn's Lemma $A$ has a maximal element $P$ such that $P$ is not only an ideal, but in fact a prime ideal. By our general hypothesis $P$ contains a prime element, let's say $p$, i.e., $p\in P$, but by the definition of $S$ is clear that $p\in S$, so $p\in P\cap S$, which contradicts $P\cap S=\emptyset$. This contradiction comes from our assumption that $a\notin S$.

Hence every nonzero $a\in R$ belongs to $S$, i.e., $S=R\setminus \{0\}$ and this means that every nonzero, nonunit element of $R$ is expressible as a product of primes.

  • 1
    $\begingroup$ Good, now we finally have a self-contained proof in an appropriate question to link to in the future. There was a complete proof at a different question, but nobody would have ever thought to find it at the question asked, and the presentation was so <s>elaborate</s> ahem insightful that it may have gone unappreciated. $\endgroup$ – rschwieb Feb 9 '17 at 5:11
  • $\begingroup$ @rschwieb thanks for the feedback. I appreciate your comment. $\endgroup$ – Xam Feb 9 '17 at 12:42

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