TAOCP 4.6.1 Exercise 7

Section 4.6.1: Division of Polynomials

Exercise 7. [M10] What are the units in the domain of all polynomials over a unique factorization domain?

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Let $S$ be a unique factorization domain, and let $f(x)\in S[x]$ be a unit. Then there exists $g(x)\in S[x]$ such that

$$ f(x)g(x)=1. $$

Taking degrees gives

$$ \deg(f)+\deg(g)=\deg(1)=0, $$

since property (i) of a unique factorization domain implies that nonzero polynomials have additive degrees under multiplication. Hence

$$ \deg(f)=\deg(g)=0. $$

Therefore every unit of $S[x]$ is a constant polynomial. If $f(x)=a$ is constant and $ag(x)=1$, then $a$ is a unit of $S$.

Conversely, if $a$ is a unit of $S$, there exists $b\in S$ with $ab=1$; viewed as constant polynomials, $a$ and $b$ are reciprocals in $S[x]$. Hence every unit of $S$ is a unit of $S[x]$.

Therefore the units in the domain of all polynomials over a unique factorization domain are precisely the constant polynomials whose coefficients are units of the coefficient domain. Thus the set of units of $S[x]$ is exactly the set of units of $S$, regarded as constant polynomials.

$$ \boxed{\text{The units of }S[x]\text{ are precisely the constant unit elements of }S.} $$