# Confusion about a proof of equivalence in reducibility of polynomials over a UFD and over it's quotient space.

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.
• I see, thanks. So it's not true that $R[X]^*=R^*$. Is there a condition on $R$ which does make this statement true? Because I seem to remember a case in which it did hold. EDIT: Found it. Apparently this is true whenever $R$ is a field. Jul 5 '17 at 20:58
• @Tyron it is indeed true that $R[X]^* = R^*$, if $R$ is a reduced commutative ring, so in particular for domains. In this case $\mathbb Z[X]^*= \mathbb Z^* = \{\pm 1\}$. The problem is that if $R$ is not a field, then $R^*$ is smaller than $R \setminus \{0\}$ which gives rise to non-unit polynomials of degree $0$. Jul 5 '17 at 21:02
• @Tyron - You write "the units of $R[X]$ are precisely those of $R$". This is ok. You add "it follows that the factors must both be at least linear". This is not ok. Jul 5 '17 at 21:09
• @MatheiBoulomenos Ah! So if $R$ is not a field, then I can pick an $a\in ((R-\{0\})-R^*$), which is a polynomial of degree $0$ in $R[X]$, but not a unit in $R$, thus not a unit in $R[X]$. Jul 5 '17 at 21:15