# Is 1 a prime number?

Is 1 classified as a prime number?

And if so, why? If not, why not?

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Given how often "let p be an odd prime" shows up in theorems, sometimes I wonder if we'd be better off defining 2 as non-prime too ;) – user7530 Oct 4 '11 at 20:12
So, is there something about why two is prime? – Gustavo Bandeira Apr 3 at 16:17

## 8 Answers

One of the whole "points" of defining primes is to be able to uniquely and finitely prime factorize every natural number.

If 1 was prime, then this would be more or less impossible.

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In the same way, it's often convenient for a variety of reasons not to consider hte unit ideal as prime (e.g. for unique factorization in Dedekind domains). – Akhil Mathew Jul 20 '10 at 21:15

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That's a good way of looking at it, but I wouldn't call it a proof. We simply define 1 as a prime, there is no proof of that fact; it's a definition. – Edan Maor Jul 20 '10 at 22:48

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Akhil's reason is the real reason... – user126 Jul 21 '10 at 8:51

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@Stacked let's say I wanted to prime factorize 6. It is 2*3, but it's also 2*3*1. It is also 2*3*1*1, and 2*3*1*1*1; that is, 6 no longer has a unique prime factorization – Justin L. Aug 10 '10 at 23:06

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@Stacked: “Isn’t it now impossible to factorize 1?” Good question… it’s not impossible; it takes lateral thinking. 1 is the product of the empty list of factors (just as 0 is the sum of an empty list). See the answers to math.stackexchange.com/questions/6832/… for discussion of this! – Peter LeFanu Lumsdaine Nov 21 '10 at 8:57
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The main point of talking about prime numbers is Euclid's theorem that every positive integer can be written uniquely as a product of primes. As Justin remarks, this would break horribly if $1$ were considered prime, for example we could factor $2$ as $2\times1\times1\times1\times1\times1$. Instead we say that $1$ is not a prime, but it is the product of zero primes (see Why is $x^0 = 1$ except when $x = 0$? to understand why any prime multiplied by itself $0$ times is $1$) so Euclid's theorem works out nicely!

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Euclid's theorem could easily be modified to work with 1 as a prime. The reason why 1 isn't a prime is either because it makes a whole lot of different results nicer, or an accident of history, not because of one specific theorem – Casebash Jul 20 '10 at 23:25

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– Doug Spoonwood Aug 22 '11 at 21:16

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Hence the sentence "Euclid's theorem that every positive integer can be written uniquely as a product of primes" and then referring back to it later. – Noah Snyder Aug 23 '11 at 0:33

It's important to understand that this is not something that can be proved: it's a definition. We choose not to regard 1 as a prime number, simply because it makes writing lots of theorems much easier.

Noah gives the best example in his answer: Euclid's theorem that every positive integer can be written uniquely as a product of primes. If 1 is defined to be a prime number, then we'd have to change that theorem to: "every positive integer can be written uniquely as a product of primes, except for infinite multiplications by 1". So we choose to go with the easier path of defining 1 to not be a prime.

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actually 1 was considered a prime number until the beginning of 20th century. Unique factorization was a driving force beneath its changing of status, since it's formulation is quickier if 1 is not considered a prime; but I think that group theory was the other force. Indeed I prefer to describe numbers as primes, composites and unities, that is numbers whose inverse exists (so if we take the set of integer numbers Z, we have that 1 and -1 are unities and we still have unique factorization up to unities).

We can always amend the defition of a prime number and say it is a number with exactly two divisors: in this way 1 is not a prime by definition :-)

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– Charles Oct 4 '11 at 14:44

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In the classical Greek system, 1 was not even called a "number", and "prime" was a subclassification of "odd", so both 1 and 2 were excluded then. – GEdgar Oct 4 '11 at 15:05
If you're talking about the integers, you're missing one. Every integer is prime, composite, a unit, or zero. – Tanner Swett Mar 7 '12 at 0:33

It's worth emphasizing that, in addition to unique factorization, there are also somewhat deeper structural reasons underlying the convention that $\:1\:$ is neither prime nor a prime ideal, but $\:0\:$ is. Below I discuss the motivations for these differing conventions.

One important motivation for including the zero ideal as prime is that this facilitates powerful reductions. For example, in many ring theoretic problems involving an ideal $\rm\; I\;$, one can reduce to the case $\rm\;I = P\;$ prime, and then reduce to $\rm\;R/P\;$, therefore reducing to the case when the ring is a domain. In this case one simply says that one can factor out $\rm\; P\;$, so wlog assume $\rm\; P = 0\;$ is prime, hence the ring is a domain. For example, at the end of this post is an excerpt from Kaplansky's classic textbook "Commutative Rings", section 1-3: G-Ideals, Hilbert Rings, and the Nullstellensatz, where I've explicitly highlighted a few prototypical examples of such reductions - cf. reduce to...

Thus we have solid evidence for the utility of the convention that the zero ideal is prime. So why don't we adopt the same convention for the unit ideal $1$ or, equivalently, why don't we permit the zero ring as a domain? There are a number of reasons. First, in domains and fields it often proves very convenient to assume that one has a nonzero element available. This permits proofs by contradiction to conclude by deducing $1 = 0\:$. More importantly, it implies that the unit group is nonempty, so unit groups always exist. It'd be very inconvenient to have to always add the proviso (except if $\;\rm R = 0)$ to the ubiquitous arguments involving units and unit groups. More generally it's worth emphasizing that the usual rules for equational logic are not complete for empty structures. That is why groups and other algebraic structures are always axiomatized to prevent nonempty structures (see this thread for further details).

Below is the mentioned Kaplansky excerpt on reduction to domains by factoring out prime ideals.

Let $\rm\; I\;$ be any ideal in a ring $\rm\; R\;$. We write $\rm\: R^{*}\:$ for the quotient ring $\rm\; R/I\;$. In the polynomial ring $\rm\; R[x]\;$ there is a smallest extension $\rm\; IR[x]\;$ of $\rm\; I\;$. The quotient ring $\rm\; R[x]/IR[x]\;$ is in a natural way isomorphic to $\rm\; R^*[x].\;$ In treating many problems, we can in this way reduce to the case $\rm\; I = 0\;$, and we shall often do so.

THEOREM 28. $\;$ Let $\rm\; M\;$ be a maximal ideal in $\rm\; R[x]\;$ and suppose that the contraction $\rm\; M \cap R\ =\ N\;$ is maximal in $\rm\; R\;.\$ Then $\rm\; M\;$ can be generated by $\rm\; N\;$ and one more element $\rm\; f\:.\$ We can select $\rm\; f\;$ to be a monic polynomial which maps mod $\rm\; N\;$ into an irreducible polynomial over the field $\rm\; R/N\;$.

Proof. $\;$ We can reduce to the case $\rm\; N = 0,\;$ i. e., $\rm\; R\;$ a field, and then the statement is immediate.

THEOREM 31. $\;$ A commutative ring $\rm\; R\;\;$ is a Hilbert ring if and only if the polynomial ring $\rm\; R[x] \;\;$ is a Hilbert ring.

Proof. $\;$ If $\rm\; \;\rm R[x]\;$ is a Hilbert ring, so is its homomorphic image $\rm\; R\;$. Conversely, assume that $\rm\; R\;$ is a Hilbert ring. Take a G-ideal $\rm\: Q\:$ in $\rm\; R[x]\;$; we must prove that $\rm\: Q\:$ is maximal. Let $\rm\: P = Q \cap R\:$; we can reduce the problem to the case $\rm\ P = 0\:,\$ which, incidentally, makes $\rm\ R\$ a domain. Let $\rm\; u\;$ be the image of $\rm\; x\;$ in the natural homomorphism $\rm\ R[x] \ \to\ R[x]/Q\ .\ \$ Then $\rm\; R[u]\;$ is a G-domain. $\$ By Theorem 23, $\rm\ u\$ is algebraic over $\rm\ R\$ and $\rm\ R\$ is a G-domain. Since $\rm\:R\:$ is both a G-domain and a Hilbert ring, $\rm\ R\$ is a field. $\$ But this makes $\rm\ R[u] = R[x]/Q\$ a field, $\$ proving $\rm\ Q\$ to be maximal.

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Prime numbers are the multiplicative building blocks of the natural numbers in the sense that every natural number is either a prime or a product of primes (the empty product gives 1). Multiplicatively 1 does not contribute anything and so it is not a building block.

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As this question has been just bumped up, might as well mention this great article by Chris Caldwell (who maintains The Prime Pages) and Yeng Xiong:

• "What is the smallest prime?", J. Int. Seq., Vol. 15 (2012), Article 12.9.7,
arXiv:1209.2007 [math.HO]

The abstract:

What is the first prime? It seems that the number two should be the obvious answer, and today it is, but it was not always so. There were times when and mathematicians for whom the numbers one and three were acceptable answers. To find the first prime, we must also know what the first positive integer is. Surprisingly, with the definitions used at various times throughout history, one was often not the first positive integer (some started with two, and a few with three). In this article, we survey the history of the primality of one, from the ancient Greeks to modern times. We will discuss some of the reasons definitions changed, and provide several examples. We will also discuss the last significant mathematicians to list the number one as prime.

It shows that "There does not appear to be any period of time during which most mathematicians deemed one to be a prime" (to a large extent because "For much of history it did not even make sense to ask if the number one was a prime", as 1 was not considered a number). But just as a teaser, here are a couple more quotes (I've removed the references to make this readable):

Others who listed one as prime in this period are F. Wallis (1685), J. Prestet (1689), C. Goldbach (1742), J. H. Lambert (1770), A. Felkel (1776) and E. Waring (1782). [...]
Even after Gauss' pivotal text, many continued to write that unity was prime. Among these are: A. M. Legendre (1830), E. Hinkley (1853), M. Glaisher (1876), K. Weierstrass (1876), R. Frick & F. Klein (1897), A. Cayley (1890), L. Kronecker (1901), G. Chrystal (1904) and D. N. Lehmer (1914). [...]
G. H. Hardy listed one as a prime in several editions of his text A Course of Pure Mathematics.

Read the article; it's really interesting! They also have a "The history of the primality of one—a selection of sources".

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I don't think anyone's offered this: We know that $\sqrt{p}$ is irrational if $p \in \mathbb{Z}^+$ is prime. But 1 is its own square root, and is rational, thus 1 is not a prime.

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This is a consequence of our definition of what it means for $p$ to be prime - you might as well have said "1 is not prime, thus 1 is not prime"! – Clive Newstead Jul 28 '12 at 13:29

## protected by Nate EldredgeJul 28 '12 at 13:10

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