Division of an irreducible element in a PID

Question

I'm working through some ring theory and I'm stuck on a small thing in a proof. It's probably pretty easy, but I'm not seeing it.

Let $R$ be a principal ideal domain and $\pi \in R$ irreducible. Suppose that $a,b\in R$ and that $\pi | ab$. Then $\pi|a$ or $\pi|b$.

Consider the ideal $(a,\pi)$. Since $R$ is a principal ideal domain this ideal is generated by a single element, which we call $g$. Note that $g|\pi$. Hence either $g$ is a unit, or $g=\epsilon π$ with $\epsilon \in R^{∗}$. In the second case, since $g|a$ we also get $\pi|a$ and our assertion is proven.

I don't get why $g|\pi$. Can someone explain me this? Thanks!

Answer 1

You have $(\alpha, \pi) = \{x\alpha + y\pi : x,y \in R\} = (g) = \{rg: r \in R\}$ and take $x=0 , y=1$ then $\pi \in (g)$ so there exist $r' \in R$ such that $\pi =r'g$ hence $g| \pi.$

Now if $g$ is a unit, $(\alpha,\pi) = R$ so there exist $x,y \in R$ such that $$x\alpha+y\pi =1 $$

Multiple this by $b$ to get $$x\alpha b +y\pi b = b$$

Now since $\pi | ab$ you get that $\pi | b.$

Answer 2

$(a,\pi)=(g)$, now notice that $(a),(\pi)\subseteq(g)$, that is $g\mid a$ and $g\mid \pi$