description/proof of that for integers modulo prime number field, prime-number-th power of element is element
Topics
About: field
The table of contents of this article
Starting Context
- The reader knows a definition of integers modulo prime number field.
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.
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.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(\mathbb{Z} / p\): \(= \text{ the integers modulo prime number field }\)
\([z]\): \(\in \mathbb{Z} / p\)
//
Statements:
\([z]^p = [z]\)
//
2: Proof
Whole Strategy: Step 1: see that for each \([z], [z'] \in \mathbb{Z} / p\), \(([z] + [z'])^p = [z]^p + [z']^p\); Step 2: prove it inductively.
Step 1:
Let us see that for each \([z], [z'] \in \mathbb{Z} / p\), \(([z] + [z'])^p = [z]^p + [z']^p\).
The binomial theorem holds for any commutative ring, because it is only about distributability and commutativity.
So, \(([z] + [z'])^p = \sum_{j \in \{0, ..., p\}} {}_pC_j [z]^{p - j} [z']^j\), where \({}_pC_j = p! / (j! (p - j)!)\). But for each \(j \in \{1, ..., p - 1\}\), \({}_pC_j\) is a multiple of \(p\), because the denominator does not contain any \(p\) as a factor, which implies that \({}_pC_j [z]^{p - j} [z']^j = [0]\): for each \([z''] \in \mathbb{Z} / p\), \(p [z''] = p [1] [z''] = [p] [z''] = [0] [z''] = [0]\) and \(l p [z''] = l (p [z'']) = l [0] = [0]\).
So, \(([z] + [z'])^p = {}_pC_0 [z]^p [z']^0 + {}_pC_p [z]^0 [z']^p = [z]^p + [z']^p\).
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, \(= [k - 1 + 1] = [k]\).
