Units of Ring of Polynomial Forms over Commutative Ring
This page has been identified as a candidate for refactoring of basic complexity. Until this has been finished, please leave {{Refactor}} in the code.
New contributors: Refactoring is a task which is expected to be undertaken by experienced editors only. Because of the underlying complexity of the work needed, it is recommended that you do not embark on a refactoring task until you have become familiar with the structural nature of pages of $\mathsf{Pr} \infty \mathsf{fWiki}$.To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Refactor}} from the code. |
This article needs to be linked to other articles. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{MissingLinks}} from the code. |
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 $\mathfrak p$.
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:
- $R / \mathfrak p \sqbrk X$ 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 $R$.
$\blacksquare$