966: For Polynomials Ring over Field and Nonconstant Polynomial, iff Evaluation of Polynomial at Field Element Is 0, Polynomial Can Be Factorized with x - Element

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description/proof of that for polynomials ring over field and nonconstant polynomial, iff evaluation of polynomial at field element is 0, polynomial can be factorized with x - element

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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: Proof

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Starting Context

• The reader knows a definition of polynomials ring over commutative ring.
• The reader knows a definition of field.
• The reader admits the proposition that the polynomials ring over any field is a Euclidean domain.

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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 field and any nonconstant polynomial, if and only if the evaluation of the polynomial at a field element is 0, the polynomial can be factorized with x - the element.

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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:
$F$: $\in \{\text{ the fields }\}$
$F[x]$: $=\text{ the polynomials ring over }F$
$p(x)$: $\in F[x]$
$\alpha$: $\in F$
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Statements:
$p(\alpha )=0$
$⟺$
$\mathrm{\exists }q(x)\in F[x](p(x)=(x-\alpha )q(x))$
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2: Proof

Whole Strategy: use the fact that $F[x]$ is a Euclidean domain; Step 1: suppose that $p(\alpha )=0$, express $p(x)$ as $=(x-\alpha )q(x)+r(x)$, and see that $r(x)=0$; Step 2: suppose that $p(x)=(x-\alpha )q(x)$, and see that $p(\alpha )=0$.

Step 1:

Let us suppose that $p(\alpha )=0$.

$F[x]$ is a Euclidean domain with the size function as taking the degree of the polynomial, by the proposition that the polynomials ring over any field is a Euclidean domain.

So, $p(x)=(x-\alpha )q(x)+r(x)$ where $r(x)$ is with degree smaller than 1, so, 0, which means that $r(x)=r$, a constant.

$0=p(\alpha )=(\alpha -\alpha )q(\alpha )+r=0q(\alpha )+r=r$.

So, $p(x)=(x-\alpha )q(x)$.

Step 2:

Let us suppose that $p(x)=(x-\alpha )q(x)$.

$p(\alpha )=(\alpha -\alpha )q(\alpha )=0q(\alpha )=0$.

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References

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