2025-01-19

962: For Polynomials Ring over Integral Domain, Unit Is Nonzero Constant

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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



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: { the integral rings }
R[x]: = the polynomials ring over R
//

Statements:
{ the units of R[x]}{ the nonzero constants in R[x]}
//


2: Note


As R is not necessarily a field, a constant may not be any unit: for a p(x)=p0R[x], p0R may not have any inverse, then, p(x)=p0R[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:

0R[x] is not any unit.

Let p(x)=pnxn+...+p0R[x], where pn0, be any unit.

There is an inverse, p(x)=pmxn+...+p0R[x], where pm0, such that p(x)p(x)=p(x)p(x)=1.

p(x)p(x) has the pnpmxn+m term, but pnpm0, 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.


References


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