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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,

  1. Can anyone give examples of many theorems involving primes would be complicated to state had $1$ been considered a prime?

  2. What are other reasons for not considering $1$ a prime apart from what Rotman said?

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  • $\begingroup$ 1) Take your favorite theorem from number theory, and where prime occurs, you will most likely have to replace it with "primes except for 1". 2) None - the definition just makes things cleaner. $\endgroup$ – Batman Dec 18 '14 at 15:04
  • $\begingroup$ 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... $\endgroup$ – ShallowBlue Dec 18 '14 at 15:11
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    $\begingroup$ Think about it: If 1 were prime, it would be the only prime. $\endgroup$ – Nick Dec 18 '14 at 15:12
  • $\begingroup$ @Nick that doesn't necessarily follow if their definition of prime is "divisible by only 1 and itself" $\endgroup$ – ShallowBlue Dec 18 '14 at 15:12
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    $\begingroup$ @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. $\endgroup$ – Pedro Tamaroff Dec 18 '14 at 22:40
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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.

  1. 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.
  2. $\mathbb F_p = \{0, \ldots, p-1\} \simeq \mathbb Z/p\mathbb Z$ 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)
  3. 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)
  4. 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.
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    $\begingroup$ Can you clarify how your answer answers my questions? $\endgroup$ – user170039 Dec 18 '14 at 15:07
  • $\begingroup$ @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. $\endgroup$ – AlexR Dec 18 '14 at 15:08
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    $\begingroup$ @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 $1$...". $\endgroup$ – drhab Dec 18 '14 at 15:18
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    $\begingroup$ The zero ring doesn't lack unity, but doesn't qualify as a field, because the usual definition for field requires $0\ne1$. $\endgroup$ – egreg Dec 19 '14 at 16:11
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    $\begingroup$ @AlexR That's the same idea, but your phrasing is “$\{0\}$ lacks multiplicative identity”, which is not precise. $\endgroup$ – egreg Dec 19 '14 at 16:48
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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.

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