I will be taking a Rings and Fields course in the Fall, so I figured I would read ahead in the textbook (A First Course in Abstract Algebra, by Anderson and Feil) to prepare. Recall the following definitions:
- An integer $p$ (other than $\pm 1$) is irreducible iff $p=ab \implies a = \pm 1$ or $b = \pm 1$.
- An integer $p$ (other than $0$ and $\pm 1$) is prime iff $p \mid ab \implies p \mid a$ or $p \mid b$.
While trying to follow a proof in my textbook that every irreducible integer is also prime, the authors state without proof that:
If $p$ is irreducible and $p \not \mid a$, then $\text{gcd}(p,a)= \pm 1$.
I don't really see why this is obvious, so I tried to fill in the details by proving it for myself:
My Proof: Suppose instead that $p$ is irreducible and $p \not \mid a$, but $\text{gcd}(p,a) \neq \pm 1$. Then we will derive a contradiction by showing that $p \mid a$. Let $g=\text{gcd}(p,a)$. Then we know in particular that $g \mid p$ and $g \mid a$. Since $g \mid p$, we know that $\exists x \in \Bbb Z$ such that $p=gx$. Thus, since $p$ is irreducible, it follows that $g=\pm 1$ or $x=\pm 1$. But since $g\neq \pm 1$, we know that $x = \pm 1$. Substituting this back into $p=gx$ yields $p=\pm g \iff g=\pm p$. Now recall that $g \mid a$. Hence, we know that $\exists y \in \Bbb Z$ such that $a=gy$. By using substitution and rearranging this last equation, we obtain $a=(\pm p)y=p(\pm y)$. Hence, since $\pm y \in \Bbb Z$, it follows that $p \mid a$, as desired.
Is my proof correct? I'm fairly sure it is, but it feels unnecessarily verbose. I feel like I'm missing some obvious shortcut that will make the authors' claim trivially true. Does a more elegant proof exist?
