description/proof of that for polynomials ring over integral domain, irreducible can be factorized with at most 1 nonconstant
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 irreducible element of commutative ring.
- The reader admits the proposition that for the polynomials ring over any integral domain, any unit is a nonzero constant.
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:
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Statements:
(
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2: Note
If you are wondering how that Statements amounts to "can be factorized with at most 1 nonconstant", supposing
There may be no nonconstant factor: as
The reverse of this proposition does not necessarily hold: any unit (which is inevitably a constant) on
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
Step 1:
Let
Let us suppose that
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