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Every number has the factors of $1$, itself, $-1$, and the negative version of itself (itself multiplied by $-1$).

So let's take for example $5$, it has the factors:

$ 1$
$ 5$
$-1$
$-5$

Since the definition of a prime number is a number with the factors 1 and itself, should this not mean that all numbers are composite?

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3 Answers 3

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The morally correct definition of prime number is given by Euclid's lemma. If you have a ring that is an integral domain ($ab=0\implies a=0$ or $b=0$), that is, a set with sum, multiplication, all the known rules and a $0$ and a $1$, a non-unit non-zero element is said to be prime if $p\mid ab\implies p\mid a$ or $p\mid b$. Where $p\mid a$ means that $a=pq$ for some other $q$. If a number $p$ has this property and if $u$ is invertible, i.e. there is $v$ for which $uv=vu=1$, then $up$ has this property too. If for two numbers $a,b$ there is a unit $u$ for which $a=ub$, we say that $a$ and $b$ are associates. When we want to look at factorization of numbers, we thus take from the set of all primes of your domain, a set of representatives: that is, a subset of the primes such that every prime is associate to one of the primes in our representatives set, and such that no representatives are associates. In the domain $\Bbb Z$ of integers, the (positive) prime numers $2,3,\ldots$ are a set of representatives of all the primes of $\Bbb Z$, $\pm 2,\pm 3,\ldots$. The units of $\Bbb Z$ are $1,-1$, which is what you observed.

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That would lose the useful idea of a prime number. We could also say that $1/2$ is a factor of 5. So we restrict the possible factors to positive integers.

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  • $\begingroup$ I suppose this makes sense, but why limit it when other types of numbers systems do exist? $\endgroup$
    – warspyking
    Nov 8, 2014 at 14:41
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    $\begingroup$ If you are dealing with whole numbers, then the primes you know of make sense. If you are talking about polynomials, like $x^2+1$, then you can talk about prime polynomials with no factors of smaller degree. If you allow complex numbers, then there are no prime polynomials. So 'you pays your money and you takes your choice'. $\endgroup$
    – Empy2
    Nov 8, 2014 at 14:50
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    $\begingroup$ @Michael There are prime polynomials in $\Bbb C[X]$! Namely, $x-\alpha$ for $\alpha\in\Bbb C$. $\endgroup$
    – Pedro ♦
    Nov 8, 2014 at 15:03
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    $\begingroup$ Fair enough.... $\endgroup$
    – Empy2
    Nov 8, 2014 at 15:05
  • $\begingroup$ @Michael Not fair enough. True. $\endgroup$
    – Pedro ♦
    Nov 8, 2014 at 23:51
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You're essentially asking for how to generalize the notion of prime from the positive integers to the (nonzero) integers. This is complicated by the fact that the positive integers have only one unit, $1$, while the integers have two, $1$ and $-1$. So where you can insist on the prime numbers in two factorization being equal in the positive integers, this statement only holds because they are special in having only a single unit. The general case is that two factorizations must be identical up to units, which in this case means that $n$ and $-n$ are identified for the purpose of factorization.

You might consider what happens in rings which have more than two units.

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