Im stuck on something.. Why is $f(x)=2x$ irreducible in $\mathbb{Q}[x]$ but reducible in $\mathbb{Z}[x]$?
I know that $2$ is invertible in $\mathbb{Q}(x)$ but not in $\mathbb{Z}(x)$, the point of my question is:
Why are invertible elements not taken into consideration while decomposing something?
Any invertible element divides every element; if invertible elements “counted”, then nothing would be irreducible, and then notion becomes worthless. It would be as if, in the integers, we said that $p$ is prime if and only if it cannot be written as $p=ab$. If we used that definition, then no element would be prime, because we would always be able to write any integer $p$ as $p=1\times p$.
Similarly, if $u$ is invertible in a domain, then there exists $v$ such that $uv=1$. Then any element $f$ could be written as $f=u(vf)$, so nothing would be irreducible.