Do we have negative prime numbers? Do we have negative prime numbers?


$..., -7, -5, -3, -2, ...$
I don't know why this question has a down vote, because it identifies a subtle point about arithmetic which becomes particularly significant when the notion of "prime" is extended to other contexts, and is relevant so far as the integers are concerned when looking at issues like unique factorisation.

When we first encounter prime numbers, we do so in the context of the positive integers. The significant point about this context is that $+1$ is the only unit (the only positive integer with a multiplicative inverse). So the question here does not really arise. Often, when the main focus of work is the positive integers, the word prime will be used to imply a positive integer.

As soon as we start to extend this to the integers, and in particular, to consider the integers as having the structure of a ring, we add in a second unit $-1$ with $(-1)^2=1$. Even in this context it is possible to define the prime numbers as positive integers without too much inconvenience.

But if we extend further and add $i$ with $i^2=-1$ as another unit - note that $i\cdot -i=1$, we are in a different world. For example, $2=(1+i)(1-i)$ and $(1+i)=i(1-i)$ so that $2=i(1-i)^2$. Are these factorisations of $2$ to be taken as the same or different?

So very soon, in the context of ring theory and the theory of algebraic integers, we start talking about prime ideals (initially thought of as all the multiples of some prime $p$ - but extended beyond that idea too - an ideal which consists of all the multiples of a single element is called principal). And it is somewhat natural, if the ideal is principal, to call the generator a prime element of the ring. However, the primes are then only identified up to multiplication by units - $1+i$ generates the same ideal as $1-i$. One of the reasons for using ideals is that the uniqueness of factorisation can be maintained in this larger context. In $\mathbb Z$ both $2$ and $-2$ generate the same ideal.
Of course there are negative prime numbers. For a number $p$ in a given domain to be prime, it needs to satisfy $p|ab$ always means $p|a$ or $p|b$ (maybe both). For example, if $p = -3$, $a = -2$, $b = 30$, we see the condition is satisfied with $p|b$. But if $p = -4$, neither condition is satisfied. This means that $-3$ is a negative prime number and $-4$ is a negative composite number.

It also needs to be said that there is a negative unit, $-1$. A unit divides every number in the domain. If $p$ is prime and $p = ab$, either $a$ or $b$ (but not both) must be a unit. For example, $-3 = -1 \times 3 = 1 \times -3$.

This is no problem whatsoever for unique factorization (if it exists in the domain at hand). We simply say that factorization is unique regardless of ordering (e.g., $5 \times 3$ is not a distinct factorization of $15$) and regardless of multiplication by units.

However, for the prime counting function $\pi(x)$, we're generally only concerned with positive primes. After all, $\pi(7) = \infty$ would be kind of useless. Or we can just say that $\pi(x)$ counts how many primes there are between $0$ and $x$. Then it turns out that $\pi(x) = \pi(-x)$, e.g., $\pi(-10) = 4$ just as $\pi(10) = 4$.