I am writing a proof of the following theorem:
In a domain in which factorization into irreducible is possible:
Factorization is unique $\iff$ Every irreducible is prime.
I essentially have two questions about the method:
- For the converse, in the book that I study (Algebraic number Theory and Fermat's last theorem, by David Tall and Stewart Ian), . I have an issue with that, as i don't really understand the structure of that induction. To me, it feels like a "backward induction". Is my way of writing it wrong?
- is there a point in $\implies$ which isn't correct?
Proof from THE BOOK (Pardon my Eötvös reference) :
https://i.stack.imgur.com/PeFC0.png
My Version (converse):
$\impliedby$: Suppose there are two Factorizations into (prime) irreducibles of an element $$u_1\prod_i^mp_i=x=u_2\prod_j^n q_j$$ We will show that $n = m$ and that there is a permutation $\pi \in S_n$ such that $p_i$ and $q_{\pi(i)}$ are associates.
Suppose $m \geq 1$, by assumption $p_m|(u_2 \prod_j^n q_j) \implies p_m|u_2 \lor p_m|q_j$, the first case would imply that $p_m$ is a unit, which can't be: The second case must hold for some $j$, which we renumber such that $j = n$. Now, both must be associates, so we have
$$u_1 p_m \prod_i^{m-1}p_i =u_2 (u_mp_m) \prod_j^{n-1} q_j \implies u_1 \prod_i^{m-1}p_i = (u_2u_m) \prod_j^{n-1} q_j $$
We can repeat this step until we reach $m = 0 \lor n = 0$ and for $m \neq n$, a contradiction arises.
$\implies$: Suppose that factorization into irreducibles is unique and let $p$ be an irreducible that divides $x = ab$. We have $a = u_1 \prod_i a_i$ and $b = u_2 \prod_j b_j$. Since $p|x$, we also have $x = cp \implies ab = cp $ with $c = u_3 \prod_k c_k$. We have $u_1 u_2 = u_3$ and we are left with $p \prod_k c_k = \prod_i a_i \prod_j b_j$ with every term irreducible and by uniqueness, $p$ must divide either some $a_i$ or some $b_j$ and is therefore prime.
Best Answer
The proof of $\impliedby$ is by standard induction, though you rephrased it slightly probably giving rise to your confusion.
The induction hypothesis here is that any factorisation into $m-1$ irreducible elements is unique from which we then deduce that a factorisation into $m$ element has to be unique as well. For this we start with a factorisation into $m$ elements and argue that we can "get rid of" (see Bill Dubuque's comment) one of the irreducible factors. This reduces us to a factorisation into $m-1$ elements for which we know that the result holds.
The base case for the induction is the emtpy factorisation, i.e. a unit. As the a product of ring elements is a unit iff all its constitutes are units this is clear.
Your proof of $\implies$ is correct.