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 $b$?

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Yes and that follows by writing out another factorization for the number that divides $b = p_1 \cdots p_s$ and using unique factorization to dedude that the factors must be some of the primes in the factorization of $b$. –  Adrián Barquero Apr 7 '12 at 15:34
More explicitly, if $b = p_{1}^{e_1} \cdots p_{s}^{e_s}$ is the prime factorization of $b$ where the $p_i$'s are different primes, then you can prove that any divisor of $b$ is a product of the form $p_{1}^{a_1} \cdots p_{s}^{a_s}$ where the exponents $a_i$ satisfy that $0 \leq a_i \leq e_i$ for $i = 1, \dots , s$. –  Adrián Barquero Apr 7 '12 at 15:38

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 $-1$.

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And why the downvote? –  Arturo Magidin Apr 11 '12 at 6:05

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.

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Yes, you can, let $b=p_1^{a_1}p_2^{a_2}\cdots p_s^{a_s}$, and let number $a=r_1^{b_1}r_2^{b_2}\cdots r_t^{b_t}$ divides $b$. If $a$ is a product of primes of number's $b$ factorization, then $\dfrac{b}{a}=\dfrac{p_1^{a_1}p_2^{a_2}\cdots p_s^{a_s}}{r_1^{b_1}r_2^{b_2}\cdots r_t^{b_t}}=p_1^{c_1}p_2^{c_2}\cdots p_s^{c_s},\, c_i\leq a_i$. If $a=p_1^{d_1}p_2^{d_2}\cdots p_s^{d_s}p,\, d_i\leq a_i,\,p\neq p_i,d_i\geq 0,\,i=1\dots s,p\geq 2,\,p\in\mathbb{P}$, i.e. there is exactly one prime number in the factorization of $a$, which isn't in the factorization of $b$, then $\dfrac{b}{a}=\dfrac{p_1^{a_1}p_2^{a_2}\cdots p_s^{a_s}}{p_1^{d_1}p_2^{d_2}\cdots p_s^{d_s}p}=p_1^{e_1}p_2^{e_2}\cdots p_s^{e_s}\frac{1}{p},\,e_i\geq 0$, but $p\geq 2$ so $a$ doesn't divide $b$. Of course, the same holds if we add more than one "new" prime number

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Hint $\rm\ p\ |\ d\ |\ p_1\cdots p_n\ \Rightarrow\ p\ |\ p_j\:$ 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 $\rm\{p_1,\ldots,p_n\}.$

Generally $\rm\:d\ |\ b_1\cdots b_n\ \Rightarrow\ d = d_1\cdots d_n,\ \ d_j\ |\ b_j,\:$ 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 $\rm\:p_j\:$ (modulo a unit factor $\pm1$).

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@Downvoter If something is not clear, please feel free to ask questions and I will be happy to elaborate. –  Bill Dubuque Apr 27 '12 at 19:13