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,

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?

Answer 1

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.

Answer 2

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.