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Is it possible to have a number, $x$, divisible by some prime, such that that prime does not appear in the unique prime factorization of $x$?

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hey suppose $p \mid x$, then $x= p \cdot (K)$ so $p$ appears –  user9413 May 21 '11 at 12:46

4 Answers 4

up vote 8 down vote accepted

If we are working in a unique factorisation domain, then this is impossible: if $p$ divides $x$, then $x = p y$ for some $y$, and so $p$ (or an associate) must appear in the prime factorisation of $x$, which is unique by hypothesis.

In fact, if we go by the strict definition of prime, then if $p$ is prime and divides $x$, it (or an associate) must appear in any factorisation of $x$ into irreducibles, regardless of whether the factorisation is unique.

However, there are rings which are not unique factorisation domains. For example, in the ring $\mathbb{Z}[\sqrt{-5}]$, $(1 - \sqrt{-5})(1 + \sqrt{-5}) = 6 = 2 \cdot 3$, but $2$, $3$, and $1 \pm \sqrt{-5}$ are all irreducible and non-associate. (Of course, this also shows that none of them are prime.)

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+1: this is the best answer by far. –  Pete L. Clark May 21 '11 at 23:24
    
Thanks, but I'm still somewhat confused. So if $x \in \mathbb{N}$, then it is not possible, or is it? –  oadams May 22 '11 at 3:38
    
@Oadams: no, it is not possible in the integers, because the integers famously form a unique factorization domain. What makes Zhen Lin's answer so good (in my opinion) is that he draws attention to this as being the essential point and gives examples where the result you are asking about fails. But to be sure, these are for more general rings than just the usual integers! –  Pete L. Clark May 22 '11 at 6:09

let $p$ divide $x$, where $x = p_1p_2\cdots p_n$ so we have $\frac {x}{p} = \frac {p_1p_2\cdots p_n}{p}=k , k \in \mathbb N$ but if $p \neq p_1 , p \neq p_2 \cdots p \neq p_n $ that means one of the primes $p_1p_2\cdots p_n$ was divisble by p ! contradicting the fact that it was a prime. So no it is not possible.

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If $p|x$, then write the prime factorization of $\frac{x}{p} = p_1p_2....p_n$. But then $x=pp_1p_2...p_n$ is a prime factorization of $x$, and unique factorization shows it is the only prime factorization of $x$.

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This cannot happen in the natural numbers. You said your self that factorizations are unique. If prime $p$ divides $x$, but $p$ doesn't appear in the unique factorization of $x$, then write $x = kp$. Then factor $k$.

Now you have a second factorization of $x$ that does include $p$ violating your uniqueness assumption.

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