970: For Field, if Field Has Non-1 Prime-Number-th Root of 1, Root Is Primitive Root

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description/proof of that for field, if field has non-1 prime-number-th root of 1, root is primitive root

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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.

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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 non-1 prime-number-th root of 1, the root is a primitive the-number-th root 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:
$F$: $\in \{\text{ the fields }\}$
$p$: $\in \{\text{ the prime numbers }\}$
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Statements:
$\mathrm{\exists }\alpha \in F\setminus \{1\}({\alpha }^p=1)$
$⟹$
$\mathrm{\forall }j\in \{1,...,p-1\}({\alpha }^j\ne 1)$
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2: Note

There may not be such any $\alpha$. For example, when $F=\mathbb{R}$ and $p=3$, $x^3=1$ has only 1 root, $1$, which is not non-1. Of course, if we take $F=\mathbb{C}$, there is $\alpha =e^{2\pi i/3}\text{ or }{e}^{2\pi i2/3}\in F$.

$p$ is a prime number in $\mathbb{N}$, not any prime in $F$: taking an an-element-of-$F$-th power of $\alpha$ does not make sense; the power is about taking the a-natural-number-times self-product of $\alpha$, so, the power needs to be a natural number.

$p$ needs to be a prime number for this proposition: if $F=\mathbb{Q}$ and $p=4$, there is $\alpha =-1$, which satisfies $\alpha \ne 1$ and ${\alpha }^p=1$, but ${\alpha }^2=1$.

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3: Proof

Whole Strategy: Step 1: take the smallest positive natural number, $n$, such that ${\alpha }^n=1$; Step 2: suppose that $n<p$, and find a contradiction.

Step 1:

Let us take the smallest positive natural number, $n$, such that ${\alpha }^n=1$. That obviously exists as $1<n\le p$.

Step 2:

Let us suppose that $n<p$.

${\alpha }^p={\alpha }^n$.

$\alpha \ne 0$, because if $\alpha =0$, ${\alpha }^q=0\ne 1$: the proposition that for any field, any positive-natural-number-th root of 0 is 0, a contradiction.

So, $\alpha$ would have an inverse, ${\alpha }^{-1}$, and ${\alpha }^{p-n}={\alpha }^p{\alpha }^{-n}={\alpha }^n{\alpha }^{-n}=1$. But $n$ was supposed to be the smallest such, and $n\le p-n$, but $n\ne p-n$, because otherwise, $n=p-n$ would mean that $p=2n$, a contradiction against $p$'s being a prime number, so, $n<p-n$. But as ${\alpha }^{p-n}={\alpha }^n$, ${\alpha }^{p-2n}=1$, and $n<p-2n$: $n\ne p-2n$ lest $p=3n$. Likewise, $n<p-jn$ for any positive natural number, $j$. That would be impossible, because $p-jn$ would get smaller and smaller while $n$ is fixed.

So, $n=p$.

That means that $\mathrm{\forall }j\in \{1,...,p-1\}({\alpha }^j\ne 1)$, which is what is meant by "$\alpha$ is a primitive $p$-th root of 1".

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References

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