Unique factorisation theorem for $\mathbb{Z} \setminus \{0\}$ 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.
 A: 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.
