In the book Rings, Fields and Groups by R B J T Allenby I stumbled upon a "proof" that a counterexample exists to a previously proven theorem in the following exercise:
Let $\mathbb{Q}_{\mathbb{Z}}[x]=\{f:f\in \mathbb{Q}[x]\ \text{and f has integer constant term}\}$ . Given $f,g\in \mathbb{Q}_{\mathbb{Z}}[x]$ let $d$ be a common divisor of maximum possible degree in $\mathbb{Q}_{\mathbb{Z}}[x]$. Show that $f_0=f/d$ and $g_0=g/d$ (both in $\mathbb{Q}_{\mathbb{Z}}[x]$) have gcd 1 in $\mathbb{Q}[x]$. Deduce that $z=mf_0+ng_0$ for suitable $z\in \mathbb{Z}$ and $m$ and $n$ in $\mathbb{Q}_{\mathbb{Z}}[x]$. Write $f_0=f_1+\alpha$, $g_0=g_1+\beta$ where $\alpha , \beta$ are the constant terms in $f̣_0$ and $g_0$. Show that $\alpha=f_0 - (\frac{f̣_1}{z})z$ lies in the ideal $[f_0,g_0]$ in $\mathbb{Q}_{\mathbb{Z}}[x]$ and similarly for $\beta$. Set $c=(\alpha,\beta)$ in $\mathbb{Z}$. Deduce $[f_0,g_0]=[c]$ and hence that $[f,g]=[cd]$, a principal ideal in $\mathbb{Q}_{\mathbb{Z}}[x]$. Deduce from $x=\frac{1}{2}x\cdot 2=\frac{1}{4}x\cdot 2^2=...$ that $x$ cannot be expressed as a product of irreducibles in $\frac{1}{2}x*2$ so that $\frac{1}{2}x*2$ is not a UFD. This surely contradicts 3.7.14 (every PID is a UFD)? [This example is due to P M Cohn]
I'm assuming that the error lies in the part $x=\frac{1}{2}x\cdot 2=\frac{1}{4}x\cdot 2^2=...$ But I'm having a hard time actually figuring out what it is, since to me neither $2^n$ nor $\frac{1}{2^n}x$ is a unit in $\mathbb{Q}_{\mathbb{Z}}[x]$. The book did not give a solution to this excercise and instead referred to the book Algebra 1:st Edition Vol 1, 1974 by P M Cohn, a book which I do not own.
Any help on what might be the error would be very useful.
(less important)
I also never managed to prove that $f_0,g_0$ have gcd 1 in $\mathbb{Q}[x]$, only that its degree is 0. So any help with that would also be great. Thanks
$\DeclareMathOperator{\QZ}{\mathbb{Q}_\mathbb{Z}} \DeclareMathOperator{\Q}{\mathbb{Q}} \DeclareMathOperator{\Z}{\mathbb{Z}}$ First, I'll show that if $d$ is a common divisor of greatest degree for $f, g\in \QZ[x]$, then $d$ is a greatest common divisor of $f$ and $g$ in $\Q[x]$. Such an element $d$ divides $f$ and $g$ in $\Q[x]$, so it divides their GCD. Further, if $d'$ is a greatest common divisor of $f$ and $g$ in $\Q[x]$, then $nd' \in \QZ[x]$ for some $n \in \Z$, so the degree of $d$ is at least the degree of $d'$.
The given proof shows by induction that any finitely generated ideal of $\QZ[x]$ is principal. However, the ring $\QZ[x]$ is not Noetherian. Consider the ascending chain of ideals $$ (x) \subseteq \left(\frac{1}{2}x\right) \subseteq \left( \frac{1}{2^2}x\right) \subseteq \cdots$$ Each inclusion is proper, as can be seen by taking the degree 1 part of each ideal. Hence, the union of these ideals is an ideal of $\QZ[x]$ which is not principal.