description/proof of that for field, if field has non-1 prime-number-th root of 1, root is primitive root
Topics
About: field
The table of contents of this article
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.
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.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(p\): \(\in \{\text{ the prime numbers }\}\)
//
Statements:
\(\exists \alpha \in F \setminus \{1\} (\alpha^p = 1)\)
\(\implies\)
\(\forall j \in \{1, ..., p - 1\} (\alpha^j \neq 1)\)
//
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 \(\alpha = e^{2 \pi i / 3} \text{ or } e^{2 \pi i 2 / 3} \in F\).
\(p\) 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 \(\alpha \neq 1\) and \(\alpha^p = 1\), but \(\alpha^2 = 1\).
3: Proof
Whole Strategy: Step 1: take the smallest positive natural number, \(n\), such that \(\alpha^n = 1\); Step 2: suppose that \(n \lt p\), and find a contradiction.
Step 1:
Let us take the smallest positive natural number, \(n\), such that \(\alpha^n = 1\). That obviously exists as \(1 \lt n \le p\).
Step 2:
Let us suppose that \(n \lt p\).
\(\alpha^p = \alpha^n\).
\(\alpha \neq 0\), because if \(\alpha = 0\), \(\alpha^q = 0 \neq 1\): 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 \(n \neq p - n\), because otherwise, \(n = p - n\) would mean that \(p = 2 n\), a contradiction against \(p\)'s being a prime number, so, \(n \lt p - n\). But as \(\alpha^{p - n} = \alpha^n\), \(\alpha^{p - 2 n} = 1\), and \(n \lt p - 2 n\): \(n \neq p - 2 n\) lest \(p = 3 n\). Likewise, \(n \lt p - j n\) for any positive natural number, \(j\). That would be impossible, because \(p - j n\) would get smaller and smaller while \(n\) is fixed.
So, \(n = p\).
That means that \(\forall j \in \{1, ..., p - 1\} (\alpha^j \neq 1)\), which is what is meant by "\(\alpha\) is a primitive \(p\)-th root of 1".
