MHB If R is a domain , then units in R[x] are non-zero constants

[Math Amateur]

I am reading Joseph Rotman's book Advanced Modern Algebra.

I need help with Problem 2.21 Part (i) on page 94.

Problem 2.21 Part (i) reads as follows:

--------------------------------------------------------------------

Let $$R$$ be a domain. Prove that if a polynomial in $$R[x]$$ is a unit, then it is a nonzero constant.

--------------------------------------------------------------------

Now presumably the proof goes something like the following:

$$p(x)$$ is a unit $$ \Longrightarrow \exists \ q(x) \text{ such that } p(x)q(x) =1 $$

$$ \Longrightarrow p(x) \text{ possesses an inverse } [p(x)]^{-1}$$

$$ \Longrightarrow p(x) $$ is a nonzero constant

... ... BUT ...? ... what is a rigorous way to show that $$ p(x) $$ possesses an inverse $$ \Longrightarrow p(x) $$ is a nonzero constant

I would appreciate help in this matter.

Peter

 

[Deveno]

Suppose that $p(x)q(x) = 1$.

Write: $p(x) = a_0 + a_1x + a_2x^2 + \cdots + a_mx^m$ and:

$q(x) = b_0 + b_1x + b_2x^2 + \cdots + b_nx^n$

and furthermore stipulate that we have "no extra terms" so that $a_m,b_n \neq 0$

Then $p(x)q(x) = a_0b_0 + (a_0b_1 + a_1b_0)x + (a_0b_2 + a_1b_1 + a_2b_0)x^2 + \cdots + a_mb_nx^{m+n}$.

Thus:

$a_0b_0 = 1$
$a_0b_1 + a_1b_0 = 0$
$a_0b_2 + a_1b_1 + a_2b_0 = 0$
...
$a_mb_n = 0$

The last equation is impossible in an integral domain, by our restriction on $a_m,b_n$...UNLESS...$m = n = 0$.

But this means we have:

$p(x) = a_0$
$q(x) = b_0$

and since $a_0b_0 = 1$, these are both units in $R$, and the units in $R$ are of necessity non-zero (they cannot even be zero divisors).