I am working through Gauss Lemma and various corollaries of it. In the book Algebra of Michael Artin, I have a question to the proof of the following Theorem:
Theorem.
(a) Let $f,g$ be polynomials in $\mathbb Q[x]$, and let $f_0, g_0$ be the associated primitive polynomials in $\mathbb Z[x]$. If $f$ divides $g$ in $\mathbb Q[x]$, then $f_0$ divides $g_0$ in $\mathbb Z[x]$.
(b) Let $f$ be a primitive polynomial in $\mathbb Z[x]$, and let $g$ be any polynomial with integer coefficients. Suppose that $f$ divides $g$ in $\mathbb Q[x]$, say $g = fq$, with $q \in \mathbb Q[x]$. Then $q \in \mathbb Z[x]$, and hence $f$ divides $g$ in $\mathbb Z[x]$.
(c) Let $f, g$ be polynomials in $\mathbb Z[x]$. If they have a common nonconstant factor in $\mathbb Q[x]$, then they have a common nonconstant factor in $\mathbb Z[x]$ too.
Proof: To prove (a), we may clear denominators so that $f$ and $g$ become primitive. Then (a) is a consequence of (b). To prove (b), we apply (3.1) in order to write the quotient in the form $q = cq_0$, where $q_0$ is primitive and $c \in \mathbb Q$. By Gauss's Lemma, $fq_0$ is primitive, and the equation $g = cfq_0$ shows that it is the primitive polynomial $g_0$ associated to $g$. Therefore $g = c g_0$ is the expression for $g$ referred to in Lemma (3.1), and $c$ is the content of $g$. Since $g \in \mathbb Z[x]$, it follows that $c \in \mathbb Z$, hence that $q \in \mathbb Z[x]$. Finally, to prove (c), suppose that $f,g$ have a common factor $h$ in $\mathbb Q[x]$. We may assume that $h$ is primitive, and then by (b) $h$ divides both $f$ and $g$ in $\mathbb Z[x]$.
The following theorem are proofed in the chapter before.
(3.1) Lemma. Every nonzero polynomial $f(x) \in \mathbb Q[x]$ can be written as a product
$$
f(x) = cf_0(x),
$$
where $c$ is a rational number and $f_0(x)$ is a primitive polynomial in $\mathbb Z[x]$.
(3.3) Theorem. Gauss's Lemma: A product of primitive polynomials in $\mathbb Z[x]$ is primitive.
I have a question to the proof of part (c) of the Theorem. Why we can assume that the common factor $h$ is primitive? The notion of primitive polynomial is just definied for polynomials in $\mathbb Z[x]$, and $h$ is in $\mathbb Q[x]$? So to me this steps makes no sense…
Best Answer
If we multiply $h \in \mathbb{Q}[x]$ by an integer constant with just enough prime factors to clear denominators, then $h$ becomes a primitive polynomial in $\mathbb{Z}[x]$. This new $h$ is still a common factor of $f$ and $g$ in $\mathbb{Q}[x]$ (because integer constants are units in $\mathbb{Q}[x]$). So we can assume we started with such a primitive $h$ as a common factor of $f$ and $g$.