I am trying to understand the proof of this corollary to Gauss's lemma found in Abstract Algebra (Theory and Applications) by Thomas W. Judson.
Here is the link to a screenshot of the proof
What Gauss's Lemma says is that if one has a monic polynomial $p(x) \in \mathbb{Z}[x]$ that factors into two polynomials $\alpha(x)$ and $\beta(x)$ in $\mathbb{Q}[x]$, then there exist monic polynomials $a(x)$ and $b(x)$ in $\mathbb{Z}[x]$ with $\deg \alpha(x) = \deg a(x)$ and $\deg \beta(x) = \deg b(x)$ such that $p(x) = a(x)b(x)$.
I understand that if $a \in \mathbb{Q}$ is a zero of $p(x)$, then $x-a$ is a factor of $p(x)$. The problem I have with understanding this proof is that the lemma doesn't say that the linear factor in $\mathbb{Z}[x]$ must have the form $x-\alpha$. The general form for linear factors is $a'x+b$ for $a', b \in \mathbb{Z}$. Now, if I assume this, I can write $$p(x) = (a'x+b)q(x),$$ where $\deg q(x) = n-1$. Since $\mathbb{Z}$ is not a field, I'm not assured that there exists $(a')^{-1}$ such that $a'(a')^{-1} = (a')^{-1}a = 1$, otherwise I would be able to write $p(x)= (x-\alpha )q'(x)$ where $\alpha = -b(a')^{-1}$ and $q'(x) = (a')^{-1}q(x)$. I'm aware of Monic polynomial reducible over rationals (Gauss's Lemma) but this doesn't say exactly why the linear factor has the form $x-\alpha$ instead of the usual $a'x+b$ for linear factors. Is the Polynomial Factor Theorem true over all commutative rings? discuss Factor Theorem's validity in arbitrary commutative rings, which I think is broader than want I'm asking. I'd appreciate any explanations.
