972: Root of Polynomial in Polynomials Ring over Commutative Ring

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definition of root of polynomial in polynomials ring over commutative ring

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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

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Starting Context

• The reader knows a definition of polynomials ring over commutative ring.

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Target Context

• The reader will have a definition of root of polynomial in polynomials ring over commutative ring.

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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:
$R$: $\in \{\text{ the commutative rings }\}$
$R[x]$: $=\text{ the polynomials ring over }R$
$p(x)$: $\in R[x]$
$\ast r$: $\in R$
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Conditions:
$\mathrm{\exists }q(x)\in R[x](p(x)=(x-r)q(x))$
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2: Note

We have not defined 'root' as $p(r)=0$.

We need to distinguish what are implied when $R$ is a field and when $R$ is not necessarily a field.

When $R$ is a field, this definition equals $p(r)=0$: if $p(x)=(x-r)q(x)$, $p(r)=(r-r)q(r)=0q(r)=0$; if $p(r)=0$, $p(x)=(p-r)q(x)$, by 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: when $p(x)$ is any constant, $p(x)=0$ and $p(x)=(x-r)0$ anyway.

But when $R$ is not necessarily a field, if $p(x)=(x-r)q(x)$, $p(r)=(r-r)q(r)=0q(r)=0$; but 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 cannot be applied because it uses the fact that $R[x]$ is a Euclidean domain, which (the fact) has been proved based on the requirement that $R$ is a field.

So, in general, this definition guarantees that $p(r)=0$, but we have not proved that $p(r)=0$ guarantees this definition.

For a $p(x)$, there may be no root and there may be some multiple roots.

It is crucial to be aware that what $R[x]$ we are thinking in: $p(x)=x^2-2$ has no root with $p(x)$ regarded in $\mathbb{Q}[x]$, but it has the roots, $\sqrt{2},-\sqrt{2}$, with $p(x)$ regarded in $\mathbb{R}[x]$.

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References

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