Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $b$?

share|improve this question
    
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
add comment

4 Answers

up vote 0 down vote accepted

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 $-1$.

share|improve this answer
    
And why the downvote? –  Arturo Magidin Apr 11 '12 at 6:05
add comment

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.

share|improve this answer
add comment

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

share|improve this answer
add comment

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 $\pm1$).

share|improve this answer
    
@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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.