# Is $0$ a composite number and $-1$ a prime number?

If in the set of natural numbers, all prime numbers $p$ have only two divisors, $1$ and $p$, and all composite numbers have at least three divisors, then can we also use these definitions for the set of integers to say that all negative numbers are composite numbers except $-1$, and $0$ is also a composite number?

Does it have any practical purpose?

• $-1$ is a unit, not a prime. I think $0$ is usually excluded from the set of composite numbers (a non-zero,non-unit, non-irreducible number), but I'm not sure about that. – Daniel Fischer Oct 29 '13 at 12:52
• A quick google on negative primes gave me this result (maybe it can help you a bit): primes.utm.edu/notes/faq/negative_primes.html – Arthur Oct 29 '13 at 12:54

In a commutative ring $R$ with $1$, we say $p\in R\setminus\{0\}$ is prime if the following conditions are satisfied:

1. $p$ is not a unit
2. If $p$ divides $ab$, then $p$ divides $a$ or $p$ divides $b$.

In $\Bbb Z$, $(-1)(-1) = 1$, so that $-1$ is a unit, and is therefore not prime. $0$ is not prime (using the ring-theoretic definition of prime) as we exclude it by definition, but it does behave similarly to primes: $0$ only divides $0$, and no two non-zero elements of $\Bbb Z$ will multiply to give $0$, so if $0\mid ab$, either $a = 0$ or $b = 0$, and hence $0$ will divide whichever one is $0$ (remember that by "$a$ divides $b$," we mean "there exists $k\in\Bbb Z$ such that $ak = b$," so it is fine to say "$0$ divides $0$," even though division by $0$ is undefined).

Using the above definition, we can extend the notion of primeness to not only the negative integers, but also any commutative ring! In fact, you'll find that a negative integer $p$ is prime if and only if $-p$ is prime. These primes differ by a unit $(-1)$, and in general such primes are called "associates."

A similar (but not equivalent) notion is that of irreducibility. A nonzero nonunit $r\in R$ is irreducible if it is not the product of two nonunits. In $\Bbb Z$, $p$ is prime if and only if $p$ is irreducible, but in general this is not true: in the ring $\Bbb Z[\sqrt{-5}] = \{a + b\sqrt{-5}\mid a,b\in\Bbb Z\}$, there are irreducible elements that are not prime! For example, it can be shown that $3$ is irreducible, but $3\mid 9 = (2 + \sqrt{-5})(2 - \sqrt{-5})$ and does not divide either factor!

The differences between these notions is an important aspect of number theory, and has led to many discoveries/creations, such as the notion of an "ideal," and the study of the question "in which rings $R$ does unique prime factorization hold?" One result states that for any number field (a "small" field containing the rational numbers $\Bbb Q$), in its ring of integers (the generalization of $\Bbb Z$ to fields other than $\Bbb Q$) every nonzero proper ideal will factor uniquely into the product of prime ideals, even though elements may not always factor uniquely into products of prime elements! As you may imagine, such questions are of interest to many mathematicians, and this type of generalization is very practical in mathematics - take something you know (like primeness) and extend it to apply to objects you don't understand quite as well as your original. If the proper generalization is used, one often finds surprising (and useful) analogies between the original object and the new, less-understood objects!

The generalisation of prime numbers to any integral domain is the notion of irreducible elements. Let $R$ be an integral domain, an element $p$ is called irreducible if for all non unit $q \in R$, and $u \in R$, $$p = uq \implies u \in R^\times .$$

In the integral domain $\mathbb Z$, $(-1)$ is a unit, so can't be irreducible.

The generalisation of composite numbers can be made in unique factorization domains, i.e. integral domain $R$ such that every non unit $q \in R \setminus \{0\}$ admits a factorization of the form $$p_1 \dots p_r,$$ the $p_i$ being irreducible, unique up to permutation of terms and the multiplication by a unit. A composite element is then an element with such a factorization and $r \geq 2$.

In the UFD $\mathbb Z$, $0$ is not a non unit of $\mathbb Z \setminus \{0\}$, so can't be composite.

• Well, not quite. The proper generalization of prime is that an element $p\in R\setminus R^{\times}$ is prime if when $p$ divides a product $ab$, then $p$ divides $a$ or $p$ divides $b$. The notion of irreducibility is actually not the same as this notion of prime, as there are integral domains with irreducible elements that are not prime. For example, $3$ is irreducible in $\Bbb Z[\sqrt{-5}]$, but $3$ is not prime ($3\mid (2 + \sqrt{-5})(2 - \sqrt{-5})$, but $3$ does not divide either factor). However, in $p\in\Bbb Z$ is prime iff $p$ is irreducible. – Stahl Oct 29 '13 at 16:37

A prime number by definition will have only two divisors and must be positive integers.

So by definition they start from 2 and are all positive numbers.

A composite number is also greater than 1.

No, $-1$ is a unit, not a prime number and certainly not a composite number.

Consider the function $f(n) = |m^n|$ for $n \geq 0$. If $m = 2$, this forms an increasing sequence. If $m = -2$, this also forms an increasing sequence (because the absolute value of $-8$ is 8, the absolute value of $-32$ is 32, etc.). But if $m = 1$, we just get a bunch of 1s. And if $m = -1$, we again get a bunch of 1s.

Now, I can certainly understand why you would think 0 is a composite number, it kinda makes sense because 0 is divisible by any integer. In fact, you can divide 0 by any real, imaginary or complex number other than 0 itself. But you can't divide any number by 0. Furthermore, if a nonzero algebraic integer is "reducible," even in a domain that is not a unique factorization domain, it may have multiple distinct factorizations but those distinct factorizations form a finite set. 0 is an algebraic integer but you could say it has infinitely many factorizations in any domain of algebraic integers.