description/proof of that for field and prime number, if characteristic of field is 0 or larger than prime number, field can be extended to have primitive prime-number-th root of 1
Topics
About: field
The table of contents of this article
Starting Context
- The reader knows a definition of field.
- The reader knows a definition of primitive n-th root of 1 on field.
- The reader knows a definition of characteristic of ring.
- The reader admits the proposition that for any field, the polynomials ring over the field, and any larger-than-1-degree irreducible polynomial, the field can be extended to have a root of the polynomial in the polynomials ring over the extended field, which (the extended field) is 'fields - homomorphisms' isomorphic to any smallest such.
Target Context
- The reader will have a description and a proof of the proposition that for any field and any prime number, if the characteristic of the field is 0 or larger than the prime number, the field can be extended to have a primitive prime-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:
\(F\): \(\in \{\text{ the fields }\}\)
\(Ch (F)\): \(= \text{ the characteristic of } F\)
\(p\): \(\in \{\text{ the prime numbers }\}\)
//
Statements:
\(Ch (F) = 0 \lor p \lt Ch (F)\)
\(\implies\)
\(\exists F' \in \{\text{ the extended fields of } F\} (\exists \omega_p \in F' (p \text{ is the smallest } j \in \mathbb{N} \setminus \{0\} \text{ such that } \omega_p^j = 1))\)
//
2: Proof
Whole Strategy: Step 1: take \(p (x) = x^p - 1 \in F [x]\); Step 2: take an extended field of \(F\), \(F'\), such that for the extended polynomial of \(p (x)\), \(\overline{p (x)} (x) \in F' [x]\), \(\overline{p (x)} (x) = (x - \alpha_1) ... (x - \alpha_p)\); Step 3: let the smallest positive natural number power to which \(\alpha_j\) is \(1\) be \(k_j\), and see that \(k_j\) is a divisor of \(p\); Step 4: suppose that all the \(k_j\) s were \(1\), and find a contradiction; Step 5: conclude the proposition.
Step 1:
Let us take \(p (x) = x^p - 1 \in F [x]\).
Step 2:
Let us take an extended field of \(F\), \(F'\), such that for the extended polynomial of \(p (x)\), \(\overline{p (x)} (x) \in F' [x]\), \(\overline{p (x)} (x) = (x - \alpha_1) ... (x - \alpha_p)\), which is possible by the proposition that for any field, the polynomials ring over the field, and any larger-than-1-degree irreducible polynomial, the field can be extended to have a root of the polynomial in the polynomials ring over the extended field, which (the extended field) is 'fields - homomorphisms' isomorphic to any smallest such: refer to its Note 2.
Step 3:
Each \(\alpha_j\) satisfies \(\alpha_j^p = 1\), because \(\overline{p (x)} (\alpha_j) = \alpha_j^p - 1 = 0\).
Let the smallest \(l \in \mathbb{N} \setminus \{0\}\) such that \(\alpha_j^l = 1\) be \(k_j\). \(1 \le k_j \le p\).
\(k_j\) is a divisor of \(p\), because taking \(p = k_j q + r\) where \(0 \le r \lt k_j\), \(\alpha_j^p = \alpha_j^{k_j q + r} = \alpha_j^{k_j q} \alpha_j^r = (\alpha_j^{k_j})^q \alpha_j^r = 1^q \alpha_j^r = 1 \alpha_j^r = \alpha_j^r = 1\), which implies that \(r = 0\): otherwise, \(r\) would be a smaller than the smallest \(k_j\).
So, as \(p\) is a prime number, \(k_j = 1 \text{ or } p\).
Step 4:
Let us suppose that all the \(k_j\) s were \(1\).
That would mean that \(\alpha_j^1 = \alpha_j = 1\).
\(\overline{p (x)} (x) = x^p - 1 = (x - \alpha_1) ... (x - \alpha_p) = (x - 1) ... (x - 1) = x^p - (1 + ... + 1) x^{p - 1} + ...\), where "\(1 + ... + 1\)" is the addition of \(p\) \(1\) s. But as \(Ch (F) = 0\) or \(p \lt Ch (F)\), \(1 + ... + 1 \neq 0\), a contradiction against its being \(x^p - 1\).
So, there is at least 1 \(k_j = p\).
Step 5:
Then, \(\alpha_j\) is a primitive root of \(1\) on \(F'\).
