I am stuck in understanding the backward proof in Euclid's lemma, namely:
If $m\ge 2$ is an integer such that $m|ab$ always implies $m|a$ or $m|b$, then $m$ is a prime.
The usual proof uses the contrapositive, and in the book words is
We prove the contrapositive: if m is composite, then there is a product ab divisible by m, yet neither factor is divisible by m. Since m is composite, m = ab, where a < m and b < m. Thus, m divides ab, but m divides neither factor (if m | a, then m ≤ a).
My problem is in the following example, if $a=1$ and $b$ is not prime, then $m$ still divides the product $ab$ and divides $b$, which cannot happen since we pick that $m$ is composite. In other words, the proof states that if $m$ is composite then $m$ neither divides $a$, nor $b$.
Just for reference, I give the definition that the book has on composite numbers:
An integer is called composite if it is not a prime.
Where I screwed it?