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

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Topics

About: ring

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

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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.

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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.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$R$: $\in \{\text{ the integral domains }\}$
$R[x]$: $=\text{ the polynomials ring over }R$
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Statements:
$\mathrm{\forall }p(x)\in \{\text{ the irreducibles on }R[x]\}$
(
$p(x)=p^{\prime }(x)p^{″}(x)$
$⟹$
$p^{\prime }(x)\in \{\text{ the constants on }R[x]\}\vee p^{″}(x)\in \{\text{ the constants on }R[x]\}$
)
//

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2: Note

If you are wondering how that Statements amounts to "can be factorized with at most 1 nonconstant", supposing $p(x)=p^{\prime }(x)p^{″}(x)p^{‴}(x)$, if there were 2 nonconstant factors, say $p^{\prime }(x)$ and $p^{‴}(x)$, $p(x)=(p^{\prime }(x)p^{″}(x))p^{‴}(x)$, and $p^{\prime }(x)p^{″}(x)$ 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 $p_0=1p_0$, 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.

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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 $p(x)=p^{\prime }(x)p^{″}(x)$ where both $p^{\prime }(x)$ 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 $p^{\prime }(x)$ and $p^{″}(x)$ were some non-units, a contradiction against $p(x)$'s being an irreducible.

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References

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