description/proof of that for polynomials ring over integral domain, unit is nonzero constant
Topics
About: ring
The table of contents of this article
Starting Context
- The reader knows a definition of polynomials ring over commutative ring.
- The reader knows a definition of integral domain.
- The reader knows a definition of units of ring.
Target Context
- The reader will have a description and a proof of the proposition that for the polynomials ring over any integral domain, any unit is a nonzero constant.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(R\): \(\in \{\text{ the integral rings }\}\)
\(R [x]\): \(= \text{ the polynomials ring over } R\)
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Statements:
\(\{\text{ the units of } R [x]\} \subseteq \{\text{ the nonzero constants in } R [x]\}\)
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2: Note
As \(R\) is not necessarily a field, a constant may not be any unit: for a \(p (x) = p_0 \in R [x]\), \(p_0 \in R\) may not have any inverse, then, \(p (x) = p_0 \in R [x]\) will not have any inverse: compare with the proposition that for the polynomials ring over any field, the units are the nonzero constants.
3: Proof
Whole Strategy: Step 1: see that each unit is a nonzero constant.
Step 1:
\(0 \in R [x]\) is not any unit.
Let \(p (x) = p_n x^n + ... + p_0 \in R [x]\), where \(p_n \neq 0\), be any unit.
There is an inverse, \(p' (x) = p'_m x^n + ... + p'_0 \in R [x]\), where \(p'_m \neq 0\), such that \(p (x) p' (x) = p' (x) p (x) = 1\).
\(p (x) p' (x)\) has the \(p_n p'_m x^{n + m}\) term, but \(p_n p'_m \neq 0\), because \(R\) is an integral domain.
So, \(n + m = 0\), because \(p (x) p' (x) = 1\). That means that \(n = 0\) and \(m = 0\), which means that \(p (x)\) is a nonzero constant.
