Let $R$ be an unique factorization domain (UFD) and $K$ its quotient field, i.e. $Q(R)=K$. Further, let $P\in R[X]$ be a primitive polynomial, that is if $$P(X)=\sum_{i=0}^{n}a_iX^i,$$ then $\text{gcd}(a_1,...,a_n)=1$.
Under these assumptions, the following statement holds:
$P(X)$ is irreducible in $K[X]\Longleftrightarrow P(X)$ is irreducible in $R[X]$
My lecture notes prove this statement by proving that the polynomial is reducible in $K[X]$ iff it is reducible in $R[X]$.
I understand the proof of one direction of the implication, but the other one confuses me. I will quote that part of the proof:
Suppose $P(X)$ is reducible in $R[X]$. Since it is primitive it factors into two factors of lower degree than $P(X)$. Hence $P(X)$ is reducible in $K[X]$.
First of all, I don't understand why it is stated like this at all. Isn't this direction of the biconditional trivially true? Also, I don't understand why the author explictly states that $P$ is primitive. The way I see it, if $P$ is reducible, then by definition it can be factored into two non-units. Since the units of $R[X]$ are precisely those of $R$ (is this actually true?), it follows that the factors must both be at least linear, therefore $P(X)=S(X)T(X)$ for some $S,T\in R[X]$ with $\text{deg}(S),\text{deg}(T)\ge1$. Further, in general we have $\text{deg}(ST)=\text{deg}(S)+\text{deg}(T)$ for any polynomials over any domain, so naturally both $S$ and $T$ have lower degree than $P$. As you can see, I haven't used the notion of primitiveness anywhere, so am I missing something here or does the author simply uses some shortcut (which I'm clearly missing then) to aviod putting up with a similar proof?
We don't have that any non-unit in $R[X]$ has degree $\geq 1$, for example, in $\mathbb Z[X]$ $2$ is not a unit, thus $2x+2$ is reducible in $\mathbb Z [X]$, but not in $\mathbb Q[X]$. Of course this kind of behaviour cannot happen for primitive polynomials, that's we need this assumption.