984: For Integers Modulo Prime Number Field, Prime-Number-th Power of Element Is Element

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description/proof of that for integers modulo prime number field, prime-number-th power of element is element

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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 integers modulo prime number field.

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Target Context

• The reader will have a description and a proof of the proposition that for the integers modulo prime number field for any prime number, the-prime-number-th power of each element is the element.

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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:
$\mathbb{Z}/p$: $=\text{ the integers modulo prime number field }$
$[z]$: $\in \mathbb{Z}/p$
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Statements:
$[z{]}^p=[z]$
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2: Proof

Whole Strategy: Step 1: see that for each $[z],[z^{\prime }]\in \mathbb{Z}/p$, $([z]+[z^{\prime }]{)}^p=[z{]}^p+[z^{\prime }{]}^p$; Step 2: prove it inductively.

Step 1:

Let us see that for each $[z],[z^{\prime }]\in \mathbb{Z}/p$, $([z]+[z^{\prime }]{)}^p=[z{]}^p+[z^{\prime }{]}^p$.

The binomial theorem holds for any commutative ring, because it is only about distributability and commutativity.

So, $([z]+[z^{\prime }]{)}^p=\sum _{j\in \{0,...,p\}}{}_p{C}_j[z{]}^{p-j}[z^{\prime }{]}^j$, where ${}_p{C}_j=p!/(j!(p-j)!)$. But for each $j\in \{1,...,p-1\}$, ${}_p{C}_j$ is a multiple of $p$, because the denominator does not contain any $p$ as a factor, which implies that ${}_p{C}_j[z{]}^{p-j}[z^{\prime }{]}^j=[0]$: for each $[z^{″}]\in \mathbb{Z}/p$, $p[z^{″}]=p[1][z^{″}]=[p][z^{″}]=[0][z^{″}]=[0]$ and $lp[z^{″}]=l(p[z^{″}])=l[0]=[0]$.

So, $([z]+[z^{\prime }]{)}^p={}_p{C}_0[z{]}^p[z^{\prime }{]}^0+{}_p{C}_p[z{]}^0[z^{\prime }{]}^p=[z{]}^p+[z^{\prime }{]}^p$.

Step 2:

Let us prove it inductively.

$[0{]}^p=[0]$.

Let us suppose that for each $0\le j\le k-1$, $[j{]}^p=[j]$.

$[k{]}^p=[k-1+1{]}^p=([k-1]+[1]{)}^p=[k-1{]}^p+[1{]}^p$, by Step 1, $=[k-1]+[1]$, by the induction hypothesis, $=[k-1+1]=[k]$.

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References

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