1084: For Field and Prime Number, if Characteristic of Field Is 0 or Larger Than Prime Number, Field Can Be Extended to Have Primitive Prime-Number-th Root of 1

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description/proof of that for field and prime number, if characteristic of field is 0 or larger than prime number, field can be extended to have primitive prime-number-th root 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 knows a definition of primitive n-th root of 1 on field.
• The reader knows a definition of characteristic of ring.
• The reader admits the proposition that for any field, the polynomials ring over the field, and any larger-than-1-degree irreducible polynomial, the field can be extended to have a root of the polynomial in the polynomials ring over the extended field, which (the extended field) is 'fields - homomorphisms' isomorphic to any smallest such.

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Target Context

• The reader will have a description and a proof of the proposition that for any field and any prime number, if the characteristic of the field is 0 or larger than the prime number, the field can be extended to have a primitive prime-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:
$F$: $\in \{\text{ the fields }\}$
$Ch(F)$: $=\text{ the characteristic of }F$
$p$: $\in \{\text{ the prime numbers }\}$
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Statements:
$Ch(F)=0\vee p<Ch(F)$
$⟹$
$\mathrm{\exists }{F}^{\prime }\in \{\text{ the extended fields of }F\}(\mathrm{\exists }{\omega }_p\in F^{\prime }(p\text{ is the smallest }j\in \mathbb{N}\setminus \{0\}\text{ such that }{\omega }_p^j=1))$
//

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2: Proof

Whole Strategy: Step 1: take $p(x)=x^p-1\in F[x]$; Step 2: take an extended field of $F$, $F^{\prime }$, such that for the extended polynomial of $p(x)$, $\overline{p(x)}(x)\in F^{\prime }[x]$, $\overline{p(x)}(x)=(x-{\alpha }_1)...(x-{\alpha }_p)$; Step 3: let the smallest positive natural number power to which ${\alpha }_j$ is $1$ be $k_j$, and see that $k_j$ is a divisor of $p$; Step 4: suppose that all the $k_j$ s were $1$, and find a contradiction; Step 5: conclude the proposition.

Step 1:

Let us take $p(x)=x^p-1\in F[x]$.

Step 2:

Let us take an extended field of $F$, $F^{\prime }$, such that for the extended polynomial of $p(x)$, $\overline{p(x)}(x)\in F^{\prime }[x]$, $\overline{p(x)}(x)=(x-{\alpha }_1)...(x-{\alpha }_p)$, which is possible by the proposition that for any field, the polynomials ring over the field, and any larger-than-1-degree irreducible polynomial, the field can be extended to have a root of the polynomial in the polynomials ring over the extended field, which (the extended field) is 'fields - homomorphisms' isomorphic to any smallest such: refer to its Note 2.

Step 3:

Each ${\alpha }_j$ satisfies ${\alpha }_j^p=1$, because $\overline{p(x)}({\alpha }_j)={\alpha }_j^p-1=0$.

Let the smallest $l\in \mathbb{N}\setminus \{0\}$ such that ${\alpha }_j^l=1$ be $k_j$. $1\le k_j\le p$.

$k_j$ is a divisor of $p$, because taking $p=k_{j}q+r$ where $0\le r<k_j$, ${\alpha }_j^p={\alpha }_j^{k_{j}q+r}={\alpha }_j^{k_{j}q}{\alpha }_j^r=({\alpha }_j^{k_j}{)}^q{\alpha }_j^r=1^q{\alpha }_j^r=1{\alpha }_j^r={\alpha }_j^r=1$, which implies that $r=0$: otherwise, $r$ would be a smaller than the smallest $k_j$.

So, as $p$ is a prime number, $k_j=1\text{ or }p$.

Step 4:

Let us suppose that all the $k_j$ s were $1$.

That would mean that ${\alpha }_j^1={\alpha }_j=1$.

$\overline{p(x)}(x)=x^p-1=(x-{\alpha }_1)...(x-{\alpha }_p)=(x-1)...(x-1)=x^p-(1+...+1)x^{p-1}+...$, where "$1+...+1$" is the addition of $p$ $1$ s. But as $Ch(F)=0$ or $p<Ch(F)$, $1+...+1\ne 0$, a contradiction against its being $x^p-1$.

So, there is at least 1 $k_j=p$.

Step 5:

Then, ${\alpha }_j$ is a primitive root of $1$ on $F^{\prime }$.

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References

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