969: For Field, if Field Has Primitive Positive-Natural-Number-th Root of 1, 1 to Natural-Number Powers of Primitive Root Are Roots

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description/proof of that for field, if field has primitive positive-natural-number-th root of 1, 1 to natural-number powers of primitive root are roots

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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: Note
• 3: Proof

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Starting Context

• The reader knows a definition of field.
• 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 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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Target Context

• The reader will have a description and a proof of 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.

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

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2: Note

There may not be such any $\omega$. For example, when $F=\mathbb{R}$ and $n=3$, $x^3=1$ has only 1 root, $1$, which does not satisfy the condition: $x^1=1$. Of course, if we take $F=\mathbb{C}$, there is $\omega =e^{2\pi i/3}\in F$.

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3: Proof

Whole Strategy: Step 1: see that $\{\omega ,...,{\omega }^n=1\}$ is distinct; Step 2: see that each ${\omega }^j\in R_{1,n}$; Step 3: see that there is no other element in $R_{1,n}$.

Step 1:

Let us see that $\{\omega ,...,{\omega }^n=1\}$ is distinct.

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

$\omega \ne 0$, because if $\omega =0$, ${\omega }^n=0$, by the proposition that for any field, any positive-natural-number-th root of 0 is 0, but $0\ne 1$ and ${\omega }^n=0\ne 1$, a contradiction.

So, there is an inverse, ${\omega }^{-1}$. $1={\omega }^j{\omega }^{-j}={\omega }^k{\omega }^{-j}={\omega }^{k-j}$. As $0\le k-j<n$, $k-j=0$, which means that $k=j$.

Step 2:

Let us see that each ${\omega }^j$ where $1\le j\le n$ is in $R_{1,n}$.

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

Step 3:

Let us see that there is no other element in $R_{1,n}$.

Let us think of the polynomials ring over $F$, $F[x]$.

Take $p(x)=x^n-1\in F[x]$.

$p(\omega )={\omega }^n-1=0$, so, $p(x)=(x-\omega )q_1(x)$ for a $q_1(x)\in F[x]$, an ($n-1$)-degree polynomial, 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.

$0=({\omega }^2{)}^n-1=p({\omega }^2)=({\omega }^2-\omega )q_1({\omega }^2)$, but ${\omega }^2-\omega \ne 0$, and so, $q_1({\omega }^2)=0$. So, $q_1(x)=(x-{\omega }^2)q_2(x)$ for a $q_2(x)\in F[x]$, an ($n-2$)-degree polynomial, 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.

And so on, after all, $p(x)=(x-\omega )...(x-{\omega }^n)q_n(x)$ for a $q_n(x)\in F[x]$, an ($n-n=0$)-degree polynomial, in fact, a nonzero constant, $q_n(x)=c$.

Then, for any $a\in F\setminus \{\omega ,...,{\omega }^n=1\}$, $p(a)\ne 0$, because each $a-{\omega }^j\ne 0$: if $(a-\omega )...(a-{\omega }^n)c=0$, $a-\omega =(a-\omega )...(a-{\omega }^n)c{c}^{-1}(a-{\omega }^n{)}^{-1}...(a-{\omega }^2{)}^{-1}=0c^{-1}(a-{\omega }^n{)}^{-1}...(a-{\omega }^n{)}^{-1}=0$, a contradiction.

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References

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