Obvious there is no unique factorization in the even numbers, e.g. $180$ can be factorized in several ways by "$\mathbb{E}$-primes" e.g. $180 = 6\times 30 = 10\times180 = 2\times90$, but where does the following proof for FTA fail for even numbers?
(The proof was originally for natural numbers in Davenport Chapter 1 Section 4, but substitute natural number for even number, prime number for $\mathbb{E}$-prime and so on.)
Preliminary remark. If the factorization of a particular number $m$ into primes is unique, each prime factor of $m$ must occur in that factorization. For if $p$ is any prime which divides $m$, we have $m = pm'$, where $m'$ is some other number, and if we now factorize $m'$ into primes we obtain a factorization of $m$ into primes by simply putting on the additional factor $p$. Since there is supposed to be only one factorization of $m$ into primes, $p$ must occur in it.
Proof. We prove the uniqueness of factorization by induction. This requires us to prove it for any number $n$, on the assumption that it is already established for all numbers less than $n$. If $n$ is itself a prime, there is nothing to prove. Suppose, then, that $n$ is composite, and has different representations as products of primes, say
$$n = pqr… = p'q'r'…$$
where $p, q, r, …$ and $p', q', r', …$ are all primes. The same prime cannot occur in both representations, for if it did we could cancel it and get two different representations of a smaller number, which is contrary to the inductive hypothesis.
We can suppose without loss of generality that $p$ is the least of the primes occuring in the first factorization. Since $n$ is composite, there is at least one prime besides $p$ in the factorization, and therefore $n ≥ p^2$. Similarly, $n ≥ p'^2$. Since $p$ and $p'$ are not the same, at least one of these two inequalities must be true with strict inequality, and it follows that $pp' < n$. Now consider the number $n − pp'$. This is a natural number less than $n$. Since $p$ divides $n$ it also divides $n − pp'$, and therefore by the preliminary remark it must occur in the factorization of $n − pp'$. Similarly $p'$ must occur. Hence, the factorization of $n − pp'$ into primes must take the form
$$n − pp' = pp'QR…$$
where $Q, R, …$ are primes. This implies that the number $pp'$ is a factor of $n$. But $n = pqr…$, so it follows on canceling $p$ that $p'$ is a factor of $qr…$. This is impossible by the preliminary remark, because $qr…$ is a number less than $n$, and $p'$ is not one of the primes $q, r, …$ occurring in its factorization. This contradiction proves that $n$ has only one factorization into primes.
(End of proof.)
I ask this because the author (Davenport) said that the uniqueness of factorization proof doesn't apply in number systems like numbers of the form $4x+1$ because they do not admit addition or subtraction, because adding or subtracting in this system $4x+1$ will go out of the system. The proof above uses subtraction in $n − pp'$ and so it won't apply in number systems that do not admit subtraction/addition. What makes so special about natural numbers then is that it admits addition/subtraction.
But then how about even numbers? Even numbers admit addition and substraction. So wouldn't the proof above work for even numbers? What even numbers don't admit is the existence of a multiplicative identity, but I don't see how the multiplicative identity is used in the above proof.
Thanks in advance.
Best Answer
Let $0 \neq d \in \mathbb{E}$. Note that we do not, in fact, have that $d \vert d$, since there is no $c \in \mathbb{E}$ such that $d=cd$.
Thus, in particular $pp' \nmid pp'$ and, consequently, we do not necessarily have that $pp' \vert n$.
The proof above would work if we had a multiplicative identity (such that $d = 1 \cdot d \implies d | d$), but note that this is only a sufficient condition, not a necessary one.