Units of Ring of Polynomial Forms over Commutative Ring

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

Let $R \left[{X}\right]$ be the ring of polynomial forms in an indeterminate $X$ over $R$.

Let $P \left({X}\right) = a_0 + a_1 X + \cdots + a_n X^n \in R \left[{X}\right]$.

Then:
 * $P \left({X}\right)$ is a unit of $R \left[{X}\right]$


 * $a_0$ is a unit of $R$
 * $a_0$ 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 $R \left[{X}\right]$ be the ring of polynomial forms in an indeterminate $X$ over $R$.

The group of units of $R \left[{X}\right]$ is precisely the group of elements of $R \left[{X}\right]$ of degree zero that are units of $R$.

Necessary condition
Let $a_0$ be a unit of $R$

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

Since 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 we have for each $k \ge 1$:
 * $P \cdot \left({1_R + Q + \dotsb + Q^{k-1} }\right) = 1_R - Q^k$

Choosing $k$ sufficiently large that $Q^k = 0$, we have:
 * $P \cdot \left({1 + Q + \dotsb + Q^{k-1} }\right) = 1_R$

and therefore $P$ is a unit of $R \left[{X}\right]$.

Sufficient condition
Suppose that $P$ is a unit of $R \left[{X}\right]$.

That is, there exists:
 * $Q = b_0 + b_1 X + \dotsb + b_m X^m \in R \left[{X}\right]$

such that $P Q = 1$.

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

Therefore $a_0 b_0 = 1$.

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

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

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

By Prime Ideal iff Quotient Ring is Integral Domain:
 * $R / \mathfrak p$ is an integral domain.

By Ring of Polynomial Forms over Integral Domain is Integral Domain:
 * $R / \mathfrak p \left[{ X }\right]$ is also an integral domain.

For any polynomial $T \in R \left[{X}\right]$ let $\overline T$ denote the image of $T$ under the Induced Homomorphism of Polynomial Forms defined by the Canonical Surjection $R \to R / \mathfrak p$.

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, $a_i \in \operatorname {Nil} \left({R}\right)$.