Please correct me if I'm wrong
I've been trying to prove Euclid's lemma without Bezout's lemma and this is what I've come up with so far:
Since Euclid's lemma is used only for proving the 'uniqueness' part of the Fundamental Theorem of Arithmetic, I think the lemma can be proved using just the 'existence' part of the fundamental theorem without using Bezout's Lemma. But, I'm not sure if this proof is correct or if I'm missing something. But the proof is as follows:
Let $p$ and $|ab|$ the smallest pair of a prime number and a product such that $p \mid |ab|$ and $p \nmid |a|$ and $p \nmid |b|$. Since $x \mid y \iff x \mid |y|$, this step doesn't eliminate any of the cases for negative integers, but allows us to have the smallest such numbers. Also note that $a,b \neq 0 $ because then all $p$ would divide them and $|ab| \neq p$ because $|ab| = p$ would either mean $p$ is composite or that one of $|a|$ or $|b|$ $= p$, both of which are contradictions.
That gives us 3 equations:
- $pn = |ab|$ for some $n\gt 1$(because $|ab|\gt 0$ and $ab\neq p$)
- $|a| = px + i, 0\lt i\lt p, x\ge0$
- $|b| = py + j, 0\lt j\lt p, y\ge0$
Now there are two possibilities, both of which are discussed below:
Case 1: Suppose $|a|,|b|\lt p$.
Then $|ab| \lt p^2$, and hence in equation 1, $n\lt p$. Considering arbitrary prime decompositions of $a$ and $b$, we have
$pn = |q_1 \cdot q_2 \cdot .... \cdot q_u||r_1 \cdot r_2 \cdot ....\cdot r_v|$
$p = \frac{|q_1 \cdot q_2 \cdot .... \cdot q_u||r_1 \cdot r_2 \cdot ....\cdot r_v|}{n}$
Also $n\gt 1$ because $n$ is positive as all the other terms are positive and $\neq 1$ because then $p$ would equal $ab$ meaning $ab$ is prime which would be a contradiction, since a prime number can't have factors($a$ and $b$ in this case) other than 1 and itself. Note that $a,b \neq p$ because $p|p$
Now suppose $n$ divides some sublist of the product $|q_1 \cdot q_2 \cdot .... \cdot q_u||r_1 \cdot r_2 \cdot ....\cdot r_v|$. Then this division would have to leave at least two terms, because if it left only one term, say $r_i$ or $q_j$ then $p$ would equal this $q_j$ or $r_i$, meaning $p$ is a prime factor of either $a$ or $b$ which is against our initial assumption, and if n divided the whole thing, that would mean $n = ab$ which would mean $p = 1$ which is also a contradiction since 1 is not a prime number. Also these terms would have to be a combination of the prime factors of $a$ and $b$ because if they consisted only of prime factors of one of them $p$ would be a factor of one of them. So after the division we would remain with
$p = |q_k \cdot q_l ... \cdot q_m||r_x \cdot r_y ... \cdot r_z|$
But, this would mean $p$ is a composite number, which is again a contradiction. So we assume that $n$ doesn't divide any sublist of the prime decomposition. Then, considering an arbitrary prime decomposition of $n$, equation 3 would become
$p\cdot |s_1\cdot s_2\cdot ...\cdots_z| = |ab|$
But since $n\lt p$, $s_1,s_2...s_z\lt p$, which mean there are prime numbers lesser than $p$ that divide $|ab|$ but neither $a$ nor $b$ which is again a contradiction since $p$ is the smallest such prime.(Note that even if some sublist of $n$ divides some sublist of the prime decomposition there would be at least one $s_q\lt p$ that doesn't because the whole of $n$ doesn't divide the prime decomposition.) So case 1 is a contradiction.
Case 2: Suppose $|a|$ or $|b|$ $\gt p$
Then, in equations 2 & 3 atleast one of $x$ and $y \ge 1$. So equation 1 would become
$pn = (px+i)(py+j)$
$pn = p^2xy + pxj + pyi + ij$
$pn = p(pxy + xj + pyi) + ij$
$p(n -pxy - xj - yi) = ij < pn = ab$ (Note that $n\gt [n-pxy - xj - yi]$ because $i$,$j$ and atleast one of $x$ and $y$ are positive.
$pm = ij \lt ab$
But, $i,j \lt p$, so $p \nmid i$ and $p \nmid j$. So the above equation says that there is a $ij \lt |ab|$ such that $p \mid ij$ and $p \nmid i,j$, which is a contradiction since $|ab|$ is the smallest such product. So, case 2 is also a contradiction.
Since, all possible cases yield contradictions, there can be no smallest $p$ and $|ab|$ such that $p \mid |ab|$ but $p \nmid |a|,|b|$, meaning if $p$ divides $|ab|$, $(p \mid a) \lor (p \mid b)\to$ QED
Is this proof correct or am I missing something here?
