Why $1$ isn't a prime?

Question

I was wondering the reason behind defining the Prime Numbers in a manner of which $1$ isn't an example. I read in Rotman's A First Course in Abstract Algebra that one reason that $1$ is not called a prime is that many theorems involving primes would otherwise be more complicated to state.

So, here are my questions,

Can anyone give examples of many theorems involving primes would be complicated to state had $1$ been considered a prime?
What are other reasons for not considering $1$ a prime apart from what Rotman said?

I'm tempted to go to ring theory and use the definition of a prime or irreducible element, but they specifically say all non-zero non-unit elements that satisfy some property... — ShallowBlue, Dec 18 '14 at 15:11
Think about it: If 1 were prime, it would be the only prime. — Nick, Dec 18 '14 at 15:12
@Nick that doesn't necessarily follow if their definition of prime is "divisible by only 1 and itself" — ShallowBlue, Dec 18 '14 at 15:12
@afding: Touche' but what of factors of a number. $$72 = 2^3\times 3^2 \times 1^{\infty}$$ My point is that developing a rigorous definition of primes with 1 would be complicated even fundamentally. — Nick, Dec 18 '14 at 15:18
@Batman The correct answer in the general contexts of UFDs is that primes are non units, and 1 is a unit in the ring of integers. — Pedro, Dec 18 '14 at 22:40

AlexR · Answer 1 · 2014-12-18T15:18:25.073

7

Here is a small list of some theorems and facts that go nuts if $1\in\mathbb P$. This is by no means comprehensive but it should grant some insight as to why this "choice" is the most natural one.

Every natural number can be uniquely factorised into primes $$n = \prod_{i=1}^k p_i^{\alpha_i}$$ Where $p_i \in\mathbb P$ and $\alpha_i\in\mathbb N$. If $1\in\mathbb P$, you don't have $\alpha_1$ unique for $p_1 = 1$ anymore.
$\mathbb F_p = \{0, \ldots, p-1\} \simeq \mathbb Z/p\mathbb Z$ is a field for every prime $p\in\mathbb P$. If $p=1, \mathbb F_1 = \{0\}$ is no longer a field (it lacks multiplicative identity)
Note that $p\in\mathbb P \Leftrightarrow |\{n\in\mathbb N : n|p\}| = |\{1,p\}| = 2$; you can't use this as a definition if $1\in\mathbb P$ (this may or may not be considered a disadvantage)
The term "coprime" for "$a$ and $b$ don't have a common prime divisor" makes less sense if $1$ is considered a prime, because $a$ and $b$ would always have a common prime divisor.

edited Dec 18 '14 at 15:18

answered Dec 18 '14 at 15:05

AlexR

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2

Can you clarify how your answer answers my questions? – Dec 18 '14 at 15:07
@user170039 This is a list for both questions. Point 3. especially is an example for question 2., Points 1. and 2. are theorems answering question 1. – AlexR Dec 18 '14 at 15:08
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@user170039 If $1$ is "allowed" as prime then the factorization under 1. is not unique anymore. And very often you will be forced to say things as: "..all primes except $1$...". – drhab Dec 18 '14 at 15:18
@user170039 Since your comment got upvoted, I added a small comment to each point telling exactly what goes nuts. – AlexR Dec 18 '14 at 15:19
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The zero ring doesn't lack unity, but doesn't qualify as a field, because the usual definition for field requires $0\ne1$. – egreg Dec 19 '14 at 16:11
@egreg As far as I'm concerned, a field requires $\mathbb F\setminus {0}$ to be a group, but $\mathbb F_1 \setminus {0} = \emptyset$ and the $\emptyset$ can't be a group. – AlexR Dec 19 '14 at 16:37
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@AlexR That's the same idea, but your phrasing is “${0}$ lacks multiplicative identity”, which is not precise. – egreg Dec 19 '14 at 16:48
But for unique prime factorization in general UFDs (even just the integers for example), it is only unique up to multiplication of units and permutation. – user21820 Dec 25 '14 at 02:58
@user21820 In the context of UFDs, units are no primitive elements and the permutation is covered by the product in the case of integers. The exponents are unique as stated in the answer. – AlexR Dec 25 '14 at 11:30
@AlexR: I meant that I don't find the reason of uniqueness for natural numbers valid because it excludes prime factorization of integers. As you said, it is the uniqueness of the exponents that is the important point, which I think you should have made clearer. Many people who read your answer might not consider beyond natural numbers so the unique factorization in natural numbers can be misleading. As for the part about permutation, in my opinion I think we shouldn't say "uniquely factorized" (in your first line) without saying "up to permutation". – user21820 Dec 26 '14 at 05:54

score 0 · Answer 2 · answered Dec 19 '14 at 15:56

If we intuit prime (ideals) as "having nontrivial content", then units (like $\pm1$) "have no content", so shouldn't be considered prime.

Alternatively, note that when discussing prime numbers $p$, even more fundamental than FTA (unique factorization of $\mathbb{Z}$, or equivalently that the irreducible elements in $\mathbb{Z}$ coincide with the prime elements) is Euclid's lemma, the fact that any of the following equivalent conditions holds:

$p\mid xy$ if and only if $p\mid x$ or $p\mid y$;
If $x,y\in \mathbb{Z} \setminus p\mathbb{Z}$, then $xy\in \mathbb{Z} \setminus p\mathbb{Z}$;
$\mathbb{Z}\setminus p\mathbb{Z}$ is a multiplicative set (which in particular requires $\mathbb{Z}/\setminus p\mathbb{Z}$ to contain the "empty product $1$", thus excluding $p=1$);
$\mathbb{Z}/p\mathbb{Z}$ (ring of integers modulo $p$) is an integral domain (which are defined to be nonzero, thus excluding $p=1$).

The (equivalent) third and fourth conditions more generally define prime ideals in arbitrary (commutative) rings. See also Bjorn Poonen's somewhat related article on why rings should (be defined to) have identity elements.

Why $1$ isn't a prime?

2 Answers2