971: For Field, Positive Natural Number, and Nonzero Element of Field, if Field Has Primitive Natural-Number-th Root of 1 and Natural-Number-th Root of Element, Roots of Element Are Products of Root and Roots of 1

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description/proof of that for field, positive natural number, and nonzero element of field, if field has primitive natural-number-th root of 1 and natural-number-th root of element, roots of element are products of root and roots of 1

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Topics

About: field

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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 field.
• The reader admits the proposition that for any field, if the field has a primitive positive-natural-number-th root of 1, the 1 to the-natural-number powers of the primitive root are the the-natural-number-th roots of 1.
• The reader admits the proposition that for any field, any positive-natural-number-th root of 0 is 0.
• The reader admits the proposition that over any field, any n-degree polynomial has at most n roots.

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Target Context

• The reader will have a description and a proof of the proposition that for any field, any positive natural number, and any nonzero element of the field, if the field has a primitive the-number-th root of 1 and a the-number-th root of the element, the the-number-th roots of the element are the products of the root and the the-number-th roots of 1.

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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:
$F$: $\in \{\text{ the fields }\}$
$n$: $\in \mathbb{N}\setminus \{0\}$
$r$: $\in F\setminus \{0\}$
$R_{r,n}$: $=\{\alpha \in F|{\alpha }^n=r\}$
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Statements:
(
$\mathrm{\exists }\omega \in F\setminus \{1\}({\omega }^n=1\wedge \mathrm{\forall }j\in \{1,...,n-1\}({\alpha }^j\ne 1))$
$\wedge$
$\mathrm{\exists }\alpha \in F({\alpha }^n=r)$
)
$⟹$
$R_{r,n}=\{\alpha \omega ,...,\alpha {\omega }^n\}$
//

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2: Proof

Whole Strategy: Step 1: see that $\{\omega ,...,{\omega }^n\}$ are the $n$-th roots of $1$; Step 2: see that $\{\alpha \omega ,...,\alpha {\omega }^n\}$ is distinct; Step 3: see that each element of $\{\alpha \omega ,...,\alpha {\omega }^n\}$ is in $R_{r,n}$; Step 4: see that there is no other element in $R_{r,n}$.

Step 1:

$\{\omega ,...,{\omega }^n\}$ are the $n$-th roots of $1$, by the proposition that for any field, if the field has a primitive positive-natural-number-th root of 1, the 1 to the-natural-number powers of the primitive root are the the-natural-number-th roots of 1.

Step 2:

Let us see that $\{\alpha \omega ,...,\alpha {\omega }^n\}$ is distinct.

Let us suppose that $\alpha {\omega }^j=\alpha {\omega }^k$ where $1\le j,k\le n$ with $j\le k$ without loss of generality.

$\alpha \ne 0$, because if $\alpha =0$, ${\alpha }^n=0\ne r$: the proposition that for any field, any positive-natural-number-th root of 0 is 0, a contradiction.

So, $\alpha$ has a inverse, ${\alpha }^{-1}$.

${\omega }^j={\alpha }^{-1}\alpha {\omega }^j={\alpha }^{-1}\alpha {\omega }^k={\omega }^k$, which means that $j=k$, because we already know that $\{\omega ,...,{\omega }^n\}$ is distinct.

Step 3:

Let us see that each element of $\{\alpha \omega ,...,\alpha {\omega }^n\}$ is in $R_{r,n}$.

$(\alpha {\omega }^j{)}^n={\alpha }^n({\omega }^j{)}^n=r{\omega }^{jn}=r({\omega }^n{)}^j=r{1}^j=r1=r$.

Step 4:

$\{\alpha \omega ,...,\alpha {\omega }^n\}$ has $n$ elements, and $R_{r,n}$ cannot have any other element, by the proposition that over any field, any n-degree polynomial has at most n roots.

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References

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