# What's the point of removing $1$ from the list of primes and having an empty product in the fundamental theorem of arithmetic?

I was reading about how we stopped considering the number $$1$$ as prime since this would mean the fundamental theorem of arithmetic wouldn't apply any more (since we can repeat the number $$1$$ over and over again and then the factorization wouldn't be unique) however, this got me wondering about the unique factorization of $$1$$ itself and I found out that we call it the "empty product" but what is the point of this? If the point from removing $$1$$ from the list of the prime numbers was to not make it an exception while stating this theorem and some others but this added another definition to make up for it so it feels like there was no point in doing it in the first place. Is there any other reason why we removed $$1$$ from the list of primes and made this?

• The empty product is defined more generally than just for this theorem and is useful. In other words: empty products are useful and natural in general. Dec 14, 2020 at 12:34
• would you mind elaborating more please? like giving an example and such Dec 14, 2020 at 12:37
• I don't know if this will help. In any case, the empty product in a category C, when it exists, is a terminal object of that category. The positive integers are a partially ordered set via $a<b$ iff $b$ divides $a$, and thus as a category. Since $1$ divides everybody it is the terminal object of that category and thus the empty product. More generally GCDs are categorical colimits and LCMs are categorical limits. Dec 14, 2020 at 12:45
• Dec 14, 2020 at 13:30
• It seems part of your puzzlement is due to your belief that allowing $1$ as a (empty) product (of primes) is done to "make up for" excluding $1$ as a prime number. Why do you believe that (and what precisely do you mean by "make up for it")? Dec 14, 2020 at 14:27

There is actually a more fundamental reason why we exclude $$1$$. Primes are usually defined (in any integral domain $$R$$) as $$p \in R$$ such that whenever $$p | ab$$ then $$p | a$$ or $$p | b$$.

However, this definition always excludes units. Since units divide everything, every unit would be trivially a prime. This is because we generally want to aim for unique prime factorization. That is, for any $$n \in R$$ we want a unique list of distinct primes $$p_1,..., p_k$$ and positive integers $$\alpha_1,..., \alpha_k$$ such that

$$n = \prod_{i=1}^k p_i ^{\alpha_i}$$

If $$1$$ was a prime, then we would have $$1 = 1^2 = 1^3 = ...$$ so we lose uniqueness. Unique prime factorization can be useful for many arguments and if we count primes as units we would always have infinitely many factorizations.

It is generally useful to define empty products (and sums, intersections, etc). If we have any associative and commutative binary operation on a set $$S$$ such that there is a neutral element with respect to this binary operation, we define the empty operation to return the neutral element. For instance:

• $$\sum_{x\in\emptyset}x=0$$, the empty sum is $$0$$.
• $$\prod_{x\in\emptyset}x=1$$, the empty product is $$1$$.
• $$\bigcup_{A\in\emptyset}A=\emptyset$$, the empty union is the empty set.
• $$\bigotimes_{V\in\emptyset}V=\mathbb F$$, the empty tensor product is the base field.

The last one may be a bit abstract, but I wanted to show that there are more than just the three "obvious" operations where we do this. The reason why we do this is that this way, if $$A,B$$ are disjoint subsets of $$S$$, then we have the simple rule that $$\sum_{x\in{A\cup B}}x=\sum_{x\in A}x+\sum_{x\in B}x.$$ The same goes for all other types of commutative, associative binary operations. This is important in different fields of mathematics. For instance, we need the empty tensor product to define a useful tensor algebra. So defining an empty operation is not something new, done specifically to salvage the unique prime factorization theorem (though that would be worthwile all by itself). It's just a natural definition to choose, and it's useful elsewhere as well.