Here is a theorem (Theorem 4) in Kaplansky's Commutative Rings.
Let $R$ be an integral domain. Let $S$ be the set of all elements in $R$ expressible as a product of principal primes. Then $S$ is a saturated multiplicatively closed set.
The definition of a saturated multiplicatively closed set is as follows:
A multiplicatively closed set $S$ is saturated if $x\in S$ implies that $S$ contains all the divisors of $x$.
When the author was proving the theorem, he said that
To prove that $S$ is saturated we assume $ab\in S$ and have to prove that $a$ and $b$ lie in $S$.
My Question: I am wondering if $a=p$ is a prime and $b=1$, then $ab=p\cdot 1\in S$ implies that $a=p\in S$ and $b=1\in S$. But $1$ should not be in $S$. Maybe I misunderstand something or how to fix the problem.
Thanks a lot.