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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.

twothreefive
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    Davenport's ad-hoc proof of unique factorization obscures the essence of the matter. We can view unique factorization as the combination of two basic properties: $(1)$ every element $\neq 0$ has a factorization into atoms (irreducibles), and $(2)$ every atom $,p,$ is prime (i.e. satsifies Euclid's Lemma: $,p\mid ab\Rightarrow p\mid a,$ or $,p\mid b).\ \ $ – Bill Dubuque Jul 01 '22 at 20:05
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    $\Bbb E$ satisfies $(1)$ but not $(2)$, e.g. $,2\mid 6\cdot 6,$ but $,2\nmid 6,,$ which yields the nonunique factorization $,6\cdot 6 = 2\cdot 18,$ into atoms of $\Bbb E.\ \ $ – Bill Dubuque Jul 01 '22 at 20:05
  • @BillDubuque I'm grateful for your insight, I will have a look more into atoms/irreducibles. Also I didn't know that prime means that it satisfies Euclid's Lemma. – twothreefive Jul 01 '22 at 20:42
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    In more general numbers systems the properties of being irreducible (an atom) and satisfying Euclid's Lemma are not equivalent. It is the latter property that is used to define primes Factorizations into atoms are unique (up to order and units) iff every atom is prime (satisfies Euclid's Lemma). This basic result is covered in a first course in abstract algebra. – Bill Dubuque Jul 01 '22 at 20:54
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    See also my final pedagogical remark here regarding such "direct" elementary proofs that eliminate higher level concepts (i.e. are written in mathematical "assembly language"). – Bill Dubuque Jul 01 '22 at 20:59

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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.

legionwhale
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    Thank you so much. It took me forever (until your help!) to figure why it won't work for the even numbers! – twothreefive Jul 01 '22 at 18:41
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    @twothreefive It was tough to spot -- I couldn't find it just by reading the reasoning. – legionwhale Jul 01 '22 at 18:42
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    what property then would you say that the even numbers is lacking (relative to the natural numbers) in this proof? If the system of numbers 4x + 1 is lacking addition/subtraction for this proof, then what do the even numbers lack in this case? That 1, i.e. the multiplicative identity, is missing in the even numbers? – twothreefive Jul 01 '22 at 18:50
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    @twothreefive $1$, it seems, but I meant that I had to work through the reasoning with a specific example (I chose $n = 36$) before I could see what goes wrong -- I couldn't find it just reading the steps. (I'm no expert on rings without unity, by the way; this would be my first proper encounter with one.) – legionwhale Jul 01 '22 at 18:56
  • Oh okay. Yeah I had a feeling it has to do with the multiplicative identity missing and was just trying to figure which part of the proof uses it. And you found it :) – twothreefive Jul 01 '22 at 18:58
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    @twothreefive I have thought about it a bit more and significantly modified the answer; I was not happy with the original answer actually. It is now correct. – legionwhale Jul 01 '22 at 19:13
  • Let us continue this discussion in chat. – twothreefive Jul 01 '22 at 20:35