# $-1$ as the only negative prime.

I was recently thinking about prime numbers, and at the time I didn't know that they had to be greater than $1$. This got me thinking about negative prime numbers though, and I soon realized that, for example, $-3$ could not be prime because $3 \cdot (-1) = -3$. In some sense $-1$ could be though because its only factors that are integers are $-1$ and $1$, and this is allowed for primes. Is there some way, by this logic, that $-1$ can be considered a prime then?

• $-1$ is a unit, so it is ignored just as $1$ is in determining primality. $-1$ is not prime because it is a unit, but $-3$ is. Indeed $-p$ is prime for any prime $p$. – Matt Samuel Feb 7 '16 at 21:31
• I keep looking at this post and thinking "this is a good post, why all $-1$s?" And then I remember. – PyRulez Feb 8 '16 at 0:30
• People needed a short name for positive non-invertible indecomposable integer. They chose the word "prime". – DanielWainfleet Feb 8 '16 at 2:02
• Conway sometimes treats -1 as prime e.g. here mathforum.org/kb/message.jspa?messageID=1093178 – Oscar Cunningham Feb 8 '16 at 4:18
• @user254665: That kind of historic reconstruction to explain the present is usually extremely inexact, and it is certainly in this case. I don't want to go into linguistic issues (the English language did not even exist when this first came up), but the notion of prime number arose many many centuries before negative numbers were conceived. And the terms invertible and decomposable are of much more recent coinage. – Marc van Leeuwen Feb 8 '16 at 6:40

## 8 Answers

If we define prime so that $-1$ is prime, unique factorization into primes fails, since $6=3\cdot 2=3\cdot 2\cdot (-1)^2$. So it is not useful to define $-1$ as a prime.

When we get to higher math, we find that when we talk about "primes" in other systems, we are required to treat any pair of numbers that divide each other as "equivalent." That is, if $a$ divided $b$ and $b$ divides $a$, then we treat $a$ and $b$ as "equivalent" for the purposes of primeness and factorization.

In particular, any number that divides $1$ is equivalent to $1$, since $1$ divides everything. The numbers that are equivalent to $1$ are called "units."

In the integers, the only units are $+1$ and $-1$, so we can just avoid this complication by only talking about the positive integers. In other rings, we are not so lucky.

• This is an interesting answer. I feel it could be improved somewhat by a specific example or two of the notion of primality in systems other than the integers. (Or pointers towards such examples.) That would help a lot in making the issue intuitively clear. – Nathaniel Feb 8 '16 at 2:34
• Given the level of the question, I'm not convinced going into any specific example is going to help the OP, nor give the OP an "intuitive' feel for things. @Nathaniel – Thomas Andrews Feb 8 '16 at 2:41
• If we accept $-3$ as prime, unique factorization fails as well: $(-3)^2=3^2$. That's why "we" use prime ideals instead of primes ... – Hagen von Eitzen Feb 8 '16 at 3:06
• @HagenvonEitzen Did you even read what I wrote? :) $-3$ and $3$ are equivalent, so those are not different factorizations. Talk of ideals are, for this OP, likely premature. – Thomas Andrews Feb 8 '16 at 3:08
• @ThomasAndrews answers here aren't just for the OP though - they're also meant to be a long-term store of useful information for other readers. (Not that I'm trying to bug you to improve an already good answer, I'm just explaining why I made that recommendation in the first place. Hagen von Eitzen's mention of the term "prime ideal" was enough for me to look up more information about it.) – Nathaniel Feb 8 '16 at 8:17

Your logic argument for $-1$ being the only negative prime:

• $-3$, for example, could not be prime because it is equal to $3\cdot(-1)$
• $-1$, however, is prime because it is divisible only by itself and by $1$

According to this logic, $+1$ is the only positive prime, since:

• $3$, for example, could not be prime because it is equal to $(-3)\cdot(-1)$

Instead of primes, consider the set $S = \{p^{2^n} \space\vert\space n,p \in \mathbb{N}, p \space \text{prime}\}$. These are the primes as well as the squares of primes, fourth powers, eighth powers, etc. Every positive integer can be represented uniquely as a product of distinct elements of $S$, in other words multiplication is a bijection between $\mathbb{N} \setminus \{0\}$ and the set of finite subsets of $S$, this can be seen by writing a prime factorization with each exponent in base $2$. Clearly, multiplication is also a bijection between $\mathbb{Z} \setminus \{0\}$ and the set of finite subsets of $S \cup \{-1\}$. So while $-1$ isn't quite like a prime it seems fair to say that it is like a prime to the power of a power of $2$.

If you think negative numbers are weird, try imaginary numbers.

There's not really a good way to think of positive numbers. You could say an imaginary number is positive if its real part is positive, but then you still get results like $(1+2i)(1+2i) = -3+4i$ so you immediately lose the fact that the product of two positive numbers is positive. That means once you start thinking about unique factorization you can escape the "positive" numbers very easily, and you're back to the problem you started with.

Unique factorization seems to fail, because $17 = (4+i)(4-i) = (-1+4i)(-1-4i)$. But does that really count? Play with more complicated examples, you'll see the pattern that the factorizations you write down seem suspiciously like each other.

I would say the simplest reason $6 = 2\cdot3 = -2\cdot-3$ doesn't violate prime factorization is we are interested in finding structure in numbers, and if that means our unique factorization theorem must protect itself from "noise" that the negative numbers introduce. It should be clear that the negative signs are obfuscating structure, not true counterexamples.

If you play with the complex integers you will find that the "noise" that interferes with unique factorization are $1$ $-1$, $i$ and $-i$. These are the units: the numbers that divide $1$, and therefore divide anything however well they please.

The natural numbers in the integer get to cheat a little bit because they are closed under multiplication, so they have enough mathematical structure to be factored neatly and avoid the unit-noise without any extra work. But once we look at other number systems we have to think about units and watch out for how they interfere with factorizations.

Also, we do not say $-1$ is prime because we do not want primes to include numbers that divide every number. We need prime numbers to describe the multiplicative structure of the integers or of another ring, and units are the trivial part of this structure and need to be treated separately.

When talking about primes it can be useful to know that mathematicians divide the integers into four types. Following are the four types and their definitions.

First, there's zero. It's in a class by itself. It's special because it's the additive identity.

Second, there are units. A unit is any number that is a divisor of 1. (Why 1? Because it's the multiplicative identity.) The only integers which divide into 1 are 1 and -1. So those are the only two units.

Third, there are primes. A prime is an integer that satisfies the following two properties: First, it is neither zero nor a unit. Second, whenever it divides into a number ab, it also divides into either a or b. (Informally, p's ability to divide into ab can't be because "part" of it divides into a and "part" divides into b. The integer p must be, both figuratively and literally, indivisible.)

Fourth, there are composites. A composite is an integer that is neither zero nor a unit nor a prime.

According to these definitions, 13 and -13 are primes. 6 and -6 are composites. And zero, 1 and -1 are neither prime nor composite.

These definitions, of course, are arguably arbitrary. But they are the definitions that mathematicians use. The popular definition of a prime ("a number greater than 1 that can be divided only by itself and 1") is a simplification that isn't entirely accurate.

I think the answers so far don't give the OP's intuitions enough credit. There is, indeed, something quite special about the set $$X = \{-1,2,3,5,7,\ldots\},$$ and the OP deserves an answer that makes this explicit. I'll do my best to give one.

Firstly, observe the following.

• Every non-zero integer can be written as a product of the elements of $X.$

• If $Y$ is a proper subset of $X$, then it's not the case that every non-zero integer can be written as a product of the elements of $Y$.

In other words, $X$ is a minimal generating subset for the monoid $(\mathbb{Z}_{\neq 0}, \times,1).$

Of course, so too is the subset $$\{-1,-2,3,5,7,\ldots\}.$$ It follows that $X$, as defined above, isn't altogether that special. Nonetheless, the presence of $-1$ is essential:

Proposition. Every minimal generating subset of $\mathbb{Z}_{\neq 0}$ contains $-1$.

So $-1$ is actually pretty important.

At this point, it's worth observing something cool about $\mathbb{N}_{\neq 0} = \{1,2,3,4,\ldots\}.$ In particular, notice that $\mathbb{N}_{\neq 0}$ has a unique minimal generating subset, namely the (usual) prime numbers.

Finally, there's something else cool about $-1$ that makes it a bit primelike. If $p \in \mathbb{N}$ is a prime number in the usual sense and $n$ is an integer, we can ask: "what is the minimum number of copies of $p$ needed to build $n$?" This is usually denoted $\nu_p(n).$ Following this, it makes sense to make the following definition. Suppose $Z$ is a subset of $\mathbb{Z}_{\neq 0}$ that generates it multiplicatively. Suppose also that $q \in Z$. Then it makes sense to write $\nu_q^Z(n)$ for the least number of occurrences of $q$ in any factorization of $n$ as a product of elements of $Z$.

Now recall that we defined $$X = \{-1,2,3,5,7,\cdots\}.$$

It follows that given a non-zero integer $n$, the expression $\nu_{-1}^X(n)$ equals $0$ if $n$ is positive, and it equals $1$ if $n$ is negative. We can also define $$Y = \{-1,-2,-3,-5,-7,\cdots\}.$$ It follows that given an integer $n>0$, the expression $\nu_{-1}^Y(n)$ equals $0$ if the number of primes in the factorization of $n$ is even, and it equals $1$ if the number of primes in the factorization of $n$ is odd.

"A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself. A natural number greater than 1 that is not a prime number is called a composite number." - point is that prime numbers are defined as natural numbers so the question is not relevant!

Source: Prime Numbers

• From your Wikipedia link: "In the ring Z of integers, the set of prime elements equals the set of irreducible elements, which is {..., -11, -7, -5, -3, -2, 2, 3, 5, 7, 11, ...}." – Daniel R. Collins Feb 10 '16 at 19:35

You should be aware that some texts do in fact define primality for negative numbers (and not just $-1$). For example, the following development is from Hungerford's Abstract Algebra: An Introduction (sec. 1.3):

DEFINITION. An integer $p$ is said to be prime if $p \ne 0, \pm1$ and the only divisors of $p$ are $\pm1$ and $\pm p$.

EXAMPLE. $3, -5, 7, -11, 13,$ and $-17$ are prime, but $15$ is not (because $15$ has divisors other than $\pm1$ and $\pm15$, such as $3$ and $5$). The integer $4567$ is prime; to prove this from the definition requires a tedious check of all its possible divisors.

It is not difficult to show that there are infinitely many distinct primes (Exercise 25). Because an integer $p$ has the same divisors as $-p$, we see that

$p$ is prime if and only if $-p$ is prime.

In this case the Fundamental Theorem of Arithmetic is stated in terms of both positive and negative prime factors (Theorem 1.1), with the standard natural-number statement given thereafter as a corollary (Corollary 1.2).

Note that this is before rings (et. al.) appear in this text (although of course they do later), so the definition is not restricted to that domain.