2025-01-19

965: For Polynomials Ring over Integral Domain, Irreducible Can Be Factorized with at Most 1 Nonconstant

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description/proof of that for polynomials ring over integral domain, irreducible can be factorized with at most 1 nonconstant

Topics


About: ring

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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 irreducible can be factorized with at most 1 nonconstant.

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 domains }
R[x]: = the polynomials ring over R
//

Statements:
p(x){ the irreducibles on R[x]}
(
p(x)=p(x)p(x)

p(x){ the constants on R[x]}p(x){ the constants on R[x]}
)
//


2: Note


If you are wondering how that Statements amounts to "can be factorized with at most 1 nonconstant", supposing p(x)=p(x)p(x)p(x), if there were 2 nonconstant factors, say p(x) and p(x), p(x)=(p(x)p(x))p(x), and p(x)p(x) or p(x) would not be any constant, a contradiction against the Statements.

There may be no nonconstant factor: as R is not necessarily a field, a constant, p(x)=p0R[x], may not be any unit, and so, p0 may be an irreducible, and p0=1p0, which is a factorization with no nonconstant factor.

The reverse of this proposition does not necessarily hold: any unit (which is inevitably a constant) on R[x] can be factorized with at most 1 (in fact, 0) nonconstant, but it is not any irreducible, because being irreducible requires being non-unit.

Compare with the proposition that for the polynomials ring over any field, the irreducibles are the nonconstants that can be factorized only with only 1 nonconstant.


3: Proof


Whole Strategy: Step 1: let p(x)R[x] be any irreducible, suppose that p(x) is factorized with 2 nonconstants, and find a contradiction.

Step 1:

Let p(x)R[x] be any irreducible.

Let us suppose that p(x)=p(x)p(x) where both p(x) and p(x) were nonconstants.

By the proposition that for the polynomials ring over any integral domain, any unit is a nonzero constant, any unit is a nonzero constant, and by its contraposition, any nonconstant is never any unit.

So, both p(x) and p(x) were some non-units, a contradiction against p(x)'s being an irreducible.


References


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