# nLab fundamental theorem of arithmetic

The Fundamental Theorem of Arithmetic

# The Fundamental Theorem of Arithmetic

## Idea

The so-called Fundamental Theorem of Arithmetic (sometimes FTA, although this can also mean the Fundamental Theorem of Algebra) is the statement that the ring of integers is a unique factorization domain (UFD). The theorem is much older than the abstract concepts of ring theory, and indeed the concept of UFD is an abstraction of this theorem. Inasumuch as ‘arithmetic’ can refer to number theory, this is a pretty basic theorem in that field.

## Statements

Let $\mathbb{N}_+$ be the positive natural numbers, let $\mathbb{N}_+ \times \mathbb{N}_+$ be the cartesian product of $\mathbb{N}_+$, with canonical product projection functions $\pi_1:\mathbb{N}_+ \times \mathbb{N}_+ \to \mathbb{N}_+$ and $\pi_2:\mathbb{N}_+ \times \mathbb{N}_+ \to \mathbb{N}_+$, and let $A^*$ be the free monoid on the set $A$. Let $p:\mathbb{N}_+ \to \mathbb{N}_+$ be the function which takes a positive natural number $n$ to the $n$-th prime number.

Since the positive natural numbers form a monoid with respect to multiplication, finite products exist, given by the monoid homomorphism

$\prod_{i = 0}^{\mathrm{len}(-) - 1}(-)(i):\mathbb{N}_+^* \to \mathbb{N}_+$

inductively defined on $\mathbb{N}_+^*$ by

$\prod_{i = 0}^{\mathrm{len}(\epsilon) - 1} \epsilon(i) = 1$
$\prod_{i = 0}^{\mathrm{len}(h(a)) - 1} (h(a))(i) = a$
$\left(\prod_{i = 0}^{\mathrm{len}(a) - 1} a(i)\right) \cdot \left(\prod_{i = 0}^{\mathrm{len}(b) - 1} b(i)\right) = \prod_{i = 0}^{\mathrm{len}(a b) - 1} (a b)(i)$
###### Theorem

There is a bijection $\mathrm{uf}:\mathbb{N}_+ \cong (\mathbb{N}_+ \times \mathbb{N}_+)^*$ such that for every positive natural number $n$,

$n = \prod_{i = 0}^{\mathrm{len}(\mathrm{uf}(n)) - 1} p(\pi_1(\mathrm{uf}(n)(i)))^{\pi_2(\mathrm{uf}(n)(i))}$

equivalently, for every list of pairs of positive natural numbers $a:(\mathbb{N} \times \mathbb{N})^*$

$\mathrm{uf}^{-1}(a) = \prod_{i = 0}^{\mathrm{len}(a) - 1} p(\pi_1(a(i)))^{\pi_2(a(i))}$

Equivalently, instead of using lists of pairs of positive natural numbers, one could just use finite multisets of positive natural numbers, which are lists of lists of positive natural numbers.

###### Theorem

There is a bijection $\mathrm{uf}:\mathbb{N}_+ \cong (\mathbb{N}_+^*)^*$ such that for every positive natural number $n$,

$n = \prod_{i = 0}^{\mathrm{len}(\mathrm{uf}(n))} \prod_{j = 0}^{\mathrm{len}(\mathrm{uf}(n)(i)) - 1} p(\mathrm{uf}(n)(i)(j))$

equivalently, for every finite multiset of positive natural numbers $a:(\mathbb{N}_+^*)^*$

$\mathrm{uf}^{-1}(a) = \prod_{i = 0}^{\mathrm{len}(a) - 1} \prod_{j = 0}^{\mathrm{len}(a(i)) - 1} p(a(i)(j))$
###### Corollary

Last revised on December 9, 2022 at 18:51:25. See the history of this page for a list of all contributions to it.