Proving (for integers) that primality implies irreducibility Let $Z = \{n \in \mathbb{Z} \; | \; |n| > 1\}$. Let $p\in Z$.
$p$ is irreducible if for some $a,b \in \mathbb{Z}$: 
$$p = ab \implies |a| = 1 \vee |b| = 1$$
$p$ is prime if for $a,b \in \mathbb{Z}$: 
$$p \mid ab \implies p\mid a \vee p\mid b$$
I would like to prove that $\textrm{prime}(p) \implies \textrm{irreducible}(p)$. 
Let $p$ be some prime number such that for $a,b \in \mathbb{Z}$, $p = ab$. 
Then, since every number is divisible by itself, we must have that $p \mid ab$. By definition of $|$, there is some $u \in \mathbb{Z}$ such that $up = ab$.
Since $p$ is prime, we know that $p\mid a$ (without loss of generality), so there is some $v \in \mathbb{Z}$ such that $vp = a$. 
Then, $up = vpb$, so $bv = u$. Well, great, but I'd like to show that $b = \pm 1$. I took a wrong turn somewhere, but where?

You started with $p=ab$, and then weakened this hypothesis to $p|ab$.  In fact, with the original, stronger, hypothesis, you have $u=1$, so $b$ is a unit since it divides $1$.
