description/proof of that for field, positive natural number, and nonzero element of field, if field has primitive natural-number-th root of 1 and natural-number-th root of element, roots of element are products of root and roots of 1
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, if the field has a primitive positive-natural-number-th root of 1, the 1 to the-natural-number powers of the primitive root are the the-natural-number-th roots of 1.
- 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 over any field, any n-degree polynomial has at most n roots.
Target Context
- The reader will have a description and a proof of the proposition that for any field, any positive natural number, and any nonzero element of the field, if the field has a primitive the-number-th root of 1 and a the-number-th root of the element, the the-number-th roots of the element are the products of the root and the the-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:
\(F\): \(\in \{\text{ the fields }\}\)
\(n\): \(\in \mathbb{N} \setminus \{0\}\)
\(r\): \(\in F \setminus \{0\}\)
\(R_{r, n}\): \(= \{\alpha \in F \vert \alpha^n = r\}\)
//
Statements:
(
\(\exists \omega \in F \setminus \{1\} (\omega^n = 1 \land \forall j \in \{1, ..., n - 1\} (\alpha^j \neq 1))\)
\(\land\)
\(\exists \alpha \in F (\alpha^n = r)\)
)
\(\implies\)
\(R_{r, n} = \{\alpha \omega, ..., \alpha \omega^n\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(\{\omega, ..., \omega^n\}\) are the \(n\)-th roots of \(1\); Step 2: see that \(\{\alpha \omega, ..., \alpha \omega^n\}\) is distinct; Step 3: see that each element of \(\{\alpha \omega, ..., \alpha \omega^n\}\) is in \(R_{r, n}\); Step 4: see that there is no other element in \(R_{r, n}\).
Step 1:
\(\{\omega, ..., \omega^n\}\) are the \(n\)-th roots of \(1\), by the proposition that for any field, if the field has a primitive positive-natural-number-th root of 1, the 1 to the-natural-number powers of the primitive root are the the-natural-number-th roots of 1.
Step 2:
Let us see that \(\{\alpha \omega, ..., \alpha \omega^n\}\) is distinct.
Let us suppose that \(\alpha \omega^j = \alpha \omega^k\) where \(1 \le j, k \le n\) with \(j \le k\) without loss of generality.
\(\alpha \neq 0\), because if \(\alpha = 0\), \(\alpha^n = 0 \neq r\): the proposition that for any field, any positive-natural-number-th root of 0 is 0, a contradiction.
So, \(\alpha\) has a inverse, \(\alpha^{-1}\).
\(\omega^j = \alpha^{-1} \alpha \omega^j = \alpha^{-1} \alpha \omega^k = \omega^k\), which means that \(j = k\), because we already know that \(\{\omega, ..., \omega^n\}\) is distinct.
Step 3:
Let us see that each element of \(\{\alpha \omega, ..., \alpha \omega^n\}\) is in \(R_{r, n}\).
\((\alpha \omega^j)^n = \alpha^n (\omega^j)^n = r \omega^{j n} = r (\omega^n)^j = r 1^j = r 1 = r\).
Step 4:
\(\{\alpha \omega, ..., \alpha \omega^n\}\) has \(n\) elements, and \(R_{r, n}\) cannot have any other element, by the proposition that over any field, any n-degree polynomial has at most n roots.
