description/proof of that for field, if field has primitive positive-natural-number-th root of 1, 1 to natural-number powers of primitive root are roots
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.
- The reader admits the proposition that for the polynomials ring over any field and any nonconstant polynomial, if and only if the evaluation of the polynomial at a field element is 0, the polynomial can be factorized with x - the element.
Target Context
- The reader will have a description and a proof of the proposition that for any field, if the field has a primitive positive-natural-number-th root of 1, 1 to the natural-number powers of the primitive root are the natural-number-th roots 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:
\(F\): \(\in \{\text{ the fields }\}\)
\(n\): \(\in \mathbb{N} \setminus \{0\}\)
\(R_{1, n}\): \(= \{\alpha \in F \vert \alpha^n = 1\}\)
//
Statements:
\(\exists \omega \in F (\omega^n = 1 \land \forall j \in \{1, ..., n - 1\} (\omega^j \neq 1))\)
\(\implies\)
\(R_{1, n} = \{\omega, ..., \omega^n = 1\}\)
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When \(n = 1\), \(\forall j \in \{1, ..., n - 1\} (\omega^j \neq 1)\) is vacuously true.
2: Note
There may not be such any \(\omega\). For example, when \(F = \mathbb{R}\) and \(n = 3\), \(x^3 = 1\) has only 1 root, \(1\), which does not satisfy the condition: \(x^1 = 1\). Of course, if we take \(F = \mathbb{C}\), there is \(\omega = e^{2 \pi i / 3} \in F\).
3: Proof
Whole Strategy: Step 1: see that \(\{\omega, ..., \omega^n = 1\}\) is distinct; Step 2: see that each \(\omega^j \in R_{1, n}\); Step 3: see that there is no other element in \(R_{1, n}\).
Step 1:
Let us see that \(\{\omega, ..., \omega^n = 1\}\) is distinct.
Let us suppose that \(\omega^j = \omega^k\) where \(1 \le j, k \le n\) with \(j \le k\) without loss of generality.
\(\omega \neq 0\), because if \(\omega = 0\), \(\omega^n = 0\), by the proposition that for any field, any positive-natural-number-th root of 0 is 0, but \(0 \neq 1\) and \(\omega^n = 0 \neq 1\), a contradiction.
So, there is an inverse, \(\omega^{-1}\). \(1 = \omega^j \omega^{- j}= \omega^k \omega^{- j} = \omega^{k - j}\). As \(0 \le k - j \lt n\), \(k - j = 0\), which means that \(k = j\).
Step 2:
Let us see that each \(\omega^j\) where \(1 \le j \le n\) is in \(R_{1, n}\).
\((\omega^j)^n = \omega^{j n} = (\omega^n)^j = 1^j = 1\).
Step 3:
Let us see that there is no other element in \(R_{1, n}\).
Let us think of the polynomials ring over \(F\), \(F [x]\).
Take \(p (x) = x^n - 1 \in F [x]\).
\(p (\omega) = \omega^n - 1 = 0\), so, \(p (x) = (x - \omega) q_1 (x)\) for a \(q_1 (x) \in F [x]\), an (\(n - 1\))-degree polynomial, by the proposition that for the polynomials ring over any field and any nonconstant polynomial, if and only if the evaluation of the polynomial at a field element is 0, the polynomial can be factorized with x - the element.
\(0 = (\omega^2)^n - 1 = p (\omega^2) = (\omega^2 - \omega) q_1 (\omega^2)\), but \(\omega^2 - \omega \neq 0\), and so, \(q_1 (\omega^2) = 0\). So, \(q_1 (x) = (x - \omega^2) q_2 (x)\) for a \(q_2 (x) \in F [x]\), an (\(n - 2\))-degree polynomial, by the proposition that for the polynomials ring over any field and any nonconstant polynomial, if and only if the evaluation of the polynomial at a field element is 0, the polynomial can be factorized with x - the element.
And so on, after all, \(p (x) = (x - \omega) ... (x - \omega^n) q_n (x)\) for a \(q_n (x) \in F [x]\), an (\(n - n = 0\))-degree polynomial, in fact, a nonzero constant, \(q_n (x) = c\).
Then, for any \(a \in F \setminus \{\omega, ..., \omega^n = 1\}\), \(p (a) \neq 0\), because each \(a - \omega^j \neq 0\): if \((a - \omega) ... (a - \omega^n) c = 0\), \(a - \omega = (a - \omega) ... (a - \omega^n) c c^{-1} (a - \omega^n)^{-1} ... (a - \omega^2)^{-1} = 0 c^{-1} (a - \omega^n)^{-1} ... (a - \omega^n)^{-1} = 0\), a contradiction.
