If $f(x)$ has positive degree and is irreducible then it's primitive

Question

The author of the book I'm reading defines reducibility as follows: $g(x)$ is reducible if there are $g_1$ and $g_2$ of positive degree such that $g=g_1g_2$.

It is claimed that:

If $D$ is a UFD and $f(x) \in D[x]$ has positive degree and is irreducible, then it is primitive.

I am having trouble seeing why this is true. If we allow zero-degree factors the statement is trivial, but I am 100% sure that what I mentioned above is the definition the author is using, as it is explicitly stated in the book.

This claim is used in a lemma towards proving $D[x]$ a UFD if $D$ is a UFD. The lemma is that such $f(x)$ is irreducible in $F[x]$, where $F$ is the field of fractions.

Answer 1

In Jacobson's book, Theorem 2.16, this notion of irreducibility is given for $F[x]$, where $F$ is specifically a field. The more general definition for irreducibility in $R[x]$, where $R$ is an arbitrary ring, is that $g(x)$ is irreducible if whenever we have $$ g(x)=h_1(x)h_2(x), $$ then either $h_1(x)$ or $h_2(x)$ is a unit. This returns the definition of irreducibility over a field, since any constant in a field is invertible. Thus the polynomial $2x+2$ is irreducible over $\mathbb{Q}[x]$, but is reducible over $\mathbb{Z}[x]$, as $$ 2x+2=2(x+1) $$ and $2$ is not invertible.

Answer 2

Let $ d $ be the gcd of the coefficients of $ f$. Then we can write $ f=df_1$ where $ f_1$ has the same degree of $ f $. Since $ f $ is irreducible it follows that $ d $ must be necessarily an unit. Hence $ f $ is primitive.