The unique factorisation theorem for positive integers states that every positive integer can be uniquely expressed as a product of primes.

What does "uniqueness" mean here? Let $n\in \mathbb{Z}^{+}$ and $n = {p_1} {p_2} \ldots {p_r}= {q_1}{q_2}\ldots {q_s}$ be any two factorisations of $n$ as product of primes. Then the uniqueness here simply means that the multisets $\left[p_1, p_2, \ldots, p_r \right]= \left[q_1,q_2, \ldots , q_s\right]$

So far so good. Now, I was wondering if there's a way to write the unique factorisation for any non-zero integer $n$. I guess you may write the prime factorisation of $n<0$ as the prime factorisation of $|n|$ with a "minus sign" but I'm not sure what exactly would remain unique about it then. Because $-1 = (-1)^3=(-1)^5 = \ldots$. (The last part is similar to prime factorisation of $n≠0$ discussed in Chapter 1, Theorem 1 of A Classical Introduction to Modern Number Theory by Ireland and Rosen).

For the record, unlike Ireland & Rosen where it takes that: If $p$ is a prime then $-p$ is also prime, I define primes to be positive integers having two distinct positive divisors, $1$ and $p$ to avoid running into problems like $6= 2 \cdot 3 =(-2)(-3)$ for example. But is there a more natural generalisation of unique factorisation theorem to every non-zero integer? I guess I should also mention that I have not studied Ring Theory yet, just in case. So layman language would be very appreciated.

  • $\begingroup$ Beware that the introductory exposition in Ireland & Rosen on this topic is a bit shoddy and may lead to confusion, e.g. see this answer. $\endgroup$ Dec 14, 2021 at 8:36

1 Answer 1


If you go on to study "abstract algebra" you will encounter the natural generalization you are looking for. Here's a brief look ahead.

A ring is a mathematical structure where addition and multiplication make sense and follow the usual rules. So the set $\mathbb{Z}$ of integers is ring, and so is the set $\mathbb{C}[z]$ of polynomials with complex coefficients.

A unit in a ring is an element whose reciprocal is also in the ring. The units in $\mathbb{Z}$ are $\pm 1$; the units in $\mathbb{C}[z]$ are the nonzero constant polynomials.

In an integral domain a nonunit $p$ is irreducible if whenever $p=rs$, one of $r$ or $s$ must be a unit. It's prime if when $p$ divides $ab$ it divides $a$ or $b$.

In the integers and a polynomial ring over a field these are equivalent, which is one of the key properties that makes unique factorization work:

The fundamental theorem of arithmetic says that every nonzero nonunit in the integers is uniquely a product of primes, where uniqueness means "up to order and to multiplication of the factors by units".

The fundamental theorem of algebra says that every complex polynomial of degree $n > 0$ has $n$ roots when you count them with the right multiplicities. Since roots correspond to linear factors, that says every nonconstant polynomial is "uniquely a product of primes" in the same sense.

You will also encounter interesting rings where the "fundamental theorem" fails because irreducible elements might not be prime.

  • $\begingroup$ "FTA says that every non-zero non-unit..." Why did you exclude units? For example, $1$ can be uniquely expressed as product of no primes (empty product). No? $\endgroup$
    – William
    Dec 13, 2021 at 23:01
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    $\begingroup$ I exclude the units because you don't have unique factorization for them even allowing an empty set of primes. Consider $1 = 2 \times (1/2) \times 3 \times (1/3)$ in the polynomial ring. If you want to allow the empty product of primes you have to work harder to say precisely what "unique" means. So choose the definition that makes life easiest. $\endgroup$ Dec 13, 2021 at 23:05
  • $\begingroup$ I'm sure you're right. I was referring to what I learned here on this site, a few hours ago see here. I suppose you're saying that including units is painful, in general UFDs even though it may not be as painful in $\mathbb{Z}$. $\endgroup$
    – William
    Dec 13, 2021 at 23:13
  • $\begingroup$ You can get away without defining units when you state the fundamental theorem of arithmetic. You just have to deal in some ad hoc way with $\pm 1$. In general rings it's hard to discuss unique factorization without first defining units. $\endgroup$ Dec 13, 2021 at 23:30
  • $\begingroup$ Your definition of a "prime" is incorrect. What you defined is called an irreducible (or atom). Primes are nonunits satisfying Euclid's Lemma $\,p\mid ab\Rightarrow p\mid a\,$ or $\,p\mid b\,$ or, equivalently, $\,R/p\,$ is a domain. $\endgroup$ Dec 14, 2021 at 8:32

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