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:
\(R\): \(\in \{\text{ the integral domains }\}\)
\(R [x]\): \(= \text{ the polynomials ring over } R\)
//
Statements:
\(\forall p (x) \in \{\text{ the irreducibles on } R [x]\}\)
(
\(p (x) = p' (x) p'' (x)\)
\(\implies\)
\(p' (x) \in \{\text{ the constants on } R [x]\} \lor p'' (x) \in \{\text{ the constants on } R [x]\}\)
)
//
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, \(p (x) = p_0 \in R [x]\), may not be any unit, and so, \(p_0\) may be an irreducible, and \(p_0 = 1 p_0\), 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 \(p (x) \in R [x]\) be any irreducible, suppose that \(p (x)\) is factorized with 2 nonconstants, and find a contradiction.
Step 1:
Let \(p (x) \in R [x]\) 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.
