Simple question related to the Fundamental theorem of arithmetic Given a number $b$, which we write as a product of prime numbers:
$$b = p_1 \cdots p_s$$
Can I then deduce that a number, which divides $b$ then has to be a product of the above primes in the factorization of $b$?
 A: Essentially yes: If $p_1,\ldots,p_s$ are primes (possibly repeated), then every divisor of
$$b = p_1p_2\cdots p_s$$
must be a (possibly empty) product of some (possibly all) of the $p_i$, times $1$ or $-1$.
A: A divisor of $b$ has to be a product of some of the prime factors of $b$, but "some" may mean $0$ of them (if the divisor is $1$).  The divisor can have no prime factors that are not prime factors if $b$.
If the divisor in question is $c$, then for some number $d$ we have $b=cd$.  If a prime number $p$ divides $b$, then $p$ must divide either $c$ or $d$ (or both).  The foregoing sentence is Euclid's lemma.
A: Hint $\rm\ p\mid d\mid  p_1\cdots p_n \Rightarrow\, p\ |\ p_j\:$ for some $\rm\:j\:$ by the Prime Divisor Property. Cancelling $\rm\:p\:$ from $\rm\:d\:$ and $\rm\:p_1\cdots p_n\:$ and inducting yields that the prime factors of $\rm\:d\:$ are a sub-multiset of $\rm\{p_1,\ldots,p_n\}.$
Further $\rm\:d\mid b_1\cdots b_n \Rightarrow\, d = d_1\cdots d_n, \ d_j\mid  b_j\,$ i.e. a divisor of a product is a product of divisors. This is a useful generalization of the Prime Divisor Property from atoms to composite numbers. Yours is the special case when all $\rm\:b_j = p_j\:$ prime, so $\rm\:d_j = 1\:$ or $\rm\:p_j\:$ (up to sign).
