Let
$$\mathbb{Z}_{(p)}:=\left\{\frac{a}{b}\in \mathbb{Q}:a,b\in\mathbb{Z}, p\nmid b\right\}$$
Is $\mathbb{Z}_{(p)}[x] $ a Principal Ideal Domain (PID)? Use the following theorem:
Let $ f,q\in \mathbb{Z}_{(p)}[x] $ $$\deg(fq)=\deg(f)+\deg(q)$$
My idea to prove $\mathbb{Z}_{(p)}[x]$ is a PID is prove $\mathbb{Z
}_{(p)}$ is a field then it is a PID but I wouldn't be using the previous theorem
Can you help me please?
Best Answer
Consider the ideal $(p,x)$.
This ideal is not whole the ring after localization because does not intersect the set of elements you are inverting.
Moreover, it is not generated by only one element: let suppose that there exists a dipendence relation between $x$ and $p$, you can lift this relation to a dipendence relation in $\mathbb{Z}[x]$, which is absurd because of the $\mathbb{Z}[x]$-analogue of the relation between degrees that you stated in the question.