Unit in an Integral Domain $R$ implying a polynomial is irreducible in $R[X]$ I saw this theorem in another post with the comment that it is "easy to prove", and yet I'm struggling to see how it's simple.
Theorem: In an Integral Domain R[x] 
If $a \in U(R) \Rightarrow ax+b$ is Irreducible in $R[x]$.
My thought would be to use the idea that a unit is reducible, but I also don't see how that helps us with the polynomial $ax+b$. Does anyone have any tips or hints to help me figure out this proof?
 A: An irreducible element element in a ring is an element that cannot be written as a product of two non-unit elements from the ring.
In your case the ring is $R[x]$ and if your poly was the product of two other polys then one of them would have to be a constant and the other poly would have to be linear(simply by degree considerations). But then by equating coefficients you discover that the constant would have to divide the "$a$" in your $aX+b$ so the constant would actually have to be a unit.
A: It is not difficult to deduce a much stronger result, namely
Theorem $\ $ Suppose  $\,D\,$ is a domain, and $\,0\ne a,b \in D\,$ satisfy  $\,a,b\mid d\, \Rightarrow\, ab\mid d\,$ for all $\,d \in D.\,$  Then $\,  f = ax+b\,$  is prime (so irreducible) in $\,D[x].$
This generalizes to the following, where $K = $ fraction field of $D$.
$$\begin{align}& f\,\ {\rm is\ prime\ in}\ D[x]\iff  f\,\ {\rm is\ prime (= irreducible)\ in}\ K[x]\ {\rm and}\,\ f\,\ {\rm is\ superprimitive}\\[.3em]
&{\rm where}\,\ f\,{\rm\ is\ {\bf superprimitive}\ in}\ D[x]\,\ :=\,\  d\,|\,cf\, \Rightarrow\, d\,|\,c\,\ \ {\rm for\ all}\,\ c,d\in D^*\end{align}\qquad $$
A: We first note that if $R[x]$ is an integral domain, so also is $R$, since
$0 \ne a, b \in R, \; ab = 0 \Longrightarrow 0 \ne ax, ab \in R[x], \; (ax)(bx) = (ab)x^2 = 0, \tag 1$
so if $R[x] \ni abx^2 \ne 0$, then $R \ni ab \ne 0$.  Now since $R$ is an integral domain, the product of the leading terms of any two polynomials $p(x), q(x) \in R[x]$ cannot vanish, for if
$p(x) = \displaystyle \sum_0^{\deg p} p_i x^i, \; q(x) = \sum_0^{\deg q} q_j x^j, \tag 2$
then
$p(x)q(x) = \displaystyle \sum_{i + j = \deg p + \deg q} p_i q_j x^{i + j}, \tag 3$
the leading term of which is $p_{\deg p}q_{\deg q} x^{\deg p + \deg q} \ne 0$ provided $p_{\deg p}, q_{\deg q} \ne 0$.  
If $ax + b \in R[x]$ were reducible, then by definition  we would have
$ax + b = p(x)q(x), \tag 4$
where
$p(x), q(x) \in R[x], \; \deg p, \deg q \ge 1; \tag 5$
but by what we have seen above the degree of $p(x)q(x)$ must then be at least $2$; therefore we conclude that $ax + b \in R[x]$ is irreducible, as are all polynomials of degree one in $R[x]$.
We observe that the irreducibility of $ax + b$ is in fact independent of whether $a \in U(R)$ or not; of course the implication
$a \in U(R) \Longrightarrow ax + b \; \text{is irreducible in} \; R[x] \tag 6$
is in fact true, since $ax + b$ is always irreducible when $R[x]$ is an integral domain, as has been shown; but (6) is really a consequence of the logic of propositions, viz. $\theta \longrightarrow \phi$ is true whenever $\phi$ is; we haven't really used any algebraic facts about $U(R)$ in establisihing the validity of (6).
Unit in an Integral Domain $R$ implying a polynomial is irreducible in $R[X]$
