Units of Ring of Polynomial Forms over Commutative Ring

Theorem

Let $\struct {R, +, \circ}$ be a non-null commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.

Let $R \sqbrk X$ be the ring of polynomial forms in an indeterminate $X$ over $R$.

Let $\map P X = a_0 + a_1 X + \cdots + a_n X^n \in R \sqbrk X$.

Then:

$\map P X$ is a unit of $R \sqbrk X$
$a_0$ is a unit of $R$

Also, for $i = 1, \ldots, n$, $a_i$ is nilpotent in $R$.

Corollary

Let $R$ be a reduced ring.

Let $R \sqbrk X$ be the ring of polynomial forms in an indeterminate $X$ over $R$.

The group of units of $R \sqbrk X$ is precisely the group of elements of $R \sqbrk X$ of degree zero that are units of $R$.

Proof

Necessary condition

Let $a_0$ be a unit of $R$.

For $i = 1, \ldots, n$, let $a_i$ be nilpotent in $R$.

Because the nilradical is an ideal of $R$, it follows that:

$Q = -a_1 X + \dotsb + -a_n X^n$

is nilpotent.

Moreover, multiplying through by $a_0^{-1}$ we may as well assume that $a_0 = 1_R$.

Then:

$P = 1_R - Q$

and from Unity plus Negative of Nilpotent Ring Element is Unit $P$ is a unit of $R \sqbrk X$.

$\Box$

Sufficient condition

Let $\map P X$ be a unit of $R \sqbrk X$.

That is, there exists:

$Q = b_0 + b_1 X + \dotsb + b_m X^m \in R \sqbrk X$

such that $P Q = 1$.

By the definition of polynomial multiplication the degree zero term of $P Q$ is $a_0 b_0$.

Therefore $a_0 b_0 = 1_R$.

So $a_0$ is a unit of $R$.

$\Box$

Next we show that $a_1, \dotsc, a_n$ are nilpotent.

By Spectrum of Ring is Nonempty, $R$ has at least one prime ideal $\mathfrak p$.

$R / \mathfrak p$ is an integral domain.
$R / \mathfrak p \sqbrk X$ is also an integral domain.

For any polynomial $T \in R \sqbrk X$ let $\overline T$ denote the image of $T$ under the Induced Homomorphism of Polynomial Forms defined by the quotient mapping $R \to R / \mathfrak p$.

Now we have:

$\overline P \cdot \overline Q = 1_{R / \mathfrak p}$

By Units of Ring of Polynomial Forms over Integral Domain this implies that $\overline P$ has degree zero.

In particular for $i = 1, \dotsc, n$, the image of $a_i$ in $R / \mathfrak p$ is $0_{R / \mathfrak p}$.

By definition, this means that $a_i \in \mathfrak p$.

But this is true for every prime ideal $\mathfrak p$.

Thus by definition:

$a_i \in \Nil R$

where $\Nil R$ denotes the nilradical of $R$.

$\blacksquare$