963: For Polynomials Ring over Field, Units Are Nonzero Constants

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description/proof of that for polynomials ring over field, units are nonzero constants

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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 field.
• The reader knows a definition of units of 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 field, the units are the nonzero constants.

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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:
$F$: $\in \{\text{ the fields }\}$
$F[x]$: $=\text{ the polynomials ring over }F$
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Statements:
$\{\text{ the units of }F[x]\}=\{\text{ the nonzero constants in }F[x]\}$
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2: Note

When $F$ is just an integral domain, a nonzero constant may not be any unit: for a $p(x)=p_0\in F[x]$, $p_0\in F$ may not have any inverse, then, $p(x)=p_0\in F[x]$ will not have any inverse: compare with the proposition that for the polynomials ring over any integral domain, any unit is a nonzero constant.

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3: Proof

Whole Strategy: Step 1: see that each unit is a nonzero constant; Step 2: see that each nonzero constant is a unit.

Step 1:

As any field is an integral domain, each unit is a nonzero constant, by the proposition that for the polynomials ring over any integral domain, any unit is a nonzero constant.

Step 2:

Let $p(x)=p_0{x}^0\in F[x]$ be any nonzero constant: as we use notations like "$p_0{x}^0$", it is usually written as $p_0$: we use such notations in order to distinguish $p_0\in F$ and $p_0\in F[x]$.

As $p_0\in F$ and $F$ is a field, there is the inverse, $p_0^{-1}\in F$. $p_0^{-1}{x}^0\in F[x]$ and $p_0{x}^0{p}_0^{-1}{x}^0=p_0{p}_0^{-1}{x}^0=1x^0$ and $p_0^{-1}{x}^0{p}_0{x}^0=p_0^{-1}{p}_0{x}^0=1x^0$.

As $1x^0$ is the identity element of $F[x]$, $p_0^{-1}{x}^0$ is an inverse of $p_0{x}^0$.

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References

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