Proof of Euclid Lemma explanation

Question

Proof:

Lemma (Euclid). Let $p$ be a prime, and let $a, b$ be integers. If $p | ab$ then $p | a$ or $p | b$. Assume $p$ is the smallest prime for which this assertion fails, and let $a$ and $b$ be such that $p | ab$ and $p ∤ a$ and $p ∤ b$. By replacing $a$ and $b$ with their remainders when dividing by $p$, we may assume that $1 ≤ a < p$ and $1≤ b < p$. Then $kp = ab$; clearly, $1 ≤ k < p$. We have $k ≠ 1$ since $p$ is a prime. Let $q$ be a prime divisor of $k$. Then $q | ab$, and so, by the minimality assumption on $p$, we have $q | a$ or $q | b$. Then dividing $q$ into $k$ and into one of $a$ or $b$, we obtain an equation $k ′p = a ′ b ′$ , where $1 ≤ k ′ < k$, $1 ≤ a ′ < p$, and $1 ≤ b ′ < p$. Repeating this step as long as necessary, we arrive at an equation $k ′′p = a ′′b ′′$ with $k ′′ = 1$, $1 ≤ a ′′ < p$, and $1≤ b ′′ < p$. This equation contradicts the primality of $p$, completing the proof.

I have bolded the part that I do not understand. The proof starts with the assumption that $p$ is the smallest prime number for which the lemma does not hold. My question is does that mean then that $q > p$ in this specific proof? Since the prime $q$ can divide $a$ or $b$.

Answer 1

Update: In this simpler rework of our initial argument, the method of infinite descent isn't used. But getting the 'step-down' counterexample is still the main idea.

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The OP's tightly packed proof (copied from a website) has many moving parts, so here we want to organize it so that the main ideas shine through.

If Euclid's lemma isn't true, it will fail for a least $p$ (the 'minimal criminal'), and, for that $p$, a choice of integers $a_0$ and $b_0$ with the smallest product $a_0 b_0$, giving us our 'conditioned counterexample',

$\quad p \mid a_0 b_0 \text{ and } p \nmid a_0 \text{ and } p \nmid b_0$

It must be true that $a_0 \lt p$ (if not replacing $a_0$ with $a_0 - p$ gives a smaller product $(a_0 - p) b_0$ and a 'better' counterexample), and for the same reason, $b_0 \lt p$.

Using simple logic to sharpen the inequalities, the counterexample satisfies

There exists $p, a_0, b_0 \in \Bbb N$ with $p$ a prime number such that
$\tag 1 (\forall \text{ prime } z) \; \big[\text{IF } z \lt p \text{ THEN } (\forall x,y \in \Bbb N)\;z \mid xy \, \text{ implies }\, z \mid x \text{ or } z \mid y\;\big]$ $\quad \text{ AND } \quad \; \,[\;p \mid a_0 b_0 \text { and } 1 \lt a_0 \lt p \text{ and } 1 \lt b_0 \lt p\;] $

Using the second conjunct of $\text{(1)}$ we can write write $kp = a_0 b_0$ for some $k \ge 2$. Let $q$ be any prime factor of $k$. Since $q \lt p$, by the first conjunct of $\text{(1)}$ it must divide into either $a_0$ or $b_0$, allowing us, after factoring it out of one of the multiplicands, to write as true

$$\tag 2 p \mid a_1 b_1 \text { and } 1 \le a_1 \lt p \text{ and } 1 \le b_1 \lt p $$

where $a_1 \lt a_0 \text{ or } b_1 \lt b_0$ (a 'step down' on the natural numbers $a_0$ and $b_0$) giving $a_1 b_1 \lt a_0 b_0$. Since $\text{(2)}$ also provides a counterexample, the starting counterexample $\text{(1)}$ must be rejected.

By employing reductio ad absurdum we have demonstrated Euclid's lemma.

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From Cut-The-Knot, What Is Infinite Descent?:

Let $P$ be a property that integers may or may not possess. If an assumption that a positive integer $n_0$ has property $P$ leads to the existence of a smaller positive integer $n_1 \lt n_0$ that also satisfies $P$, then no positive integer has that property.