Ytukyg

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?

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.

It is not difficult to deduce a much stronger result, namely

Theorem $ $ Suppose $,D,$ is a domain, and $,0ne a,b in D,$ satisfy $,a,bmid d, Rightarrow, abmid d,$ 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,din D^*end{align}qquad $$

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 $R[X]$