Field theory and polynomials Galois theory Theorem
Finite subgroups of the multiplicative group of a field are cyclic
Builds on Group of units of a ringA polynomial over a field has at most as many roots as its degree
Statement
Let be a field and let be a finite subgroup of the multiplicative group . Then is cyclic.
Remarks
The proof: for each , the elements with are roots of . So there are at most of them, by the bound on the number of roots, and a finite group with this property is cyclic. For , the finite subgroups of are the groups of the -th roots of unity. The statement fails for division rings: the quaternion group is a finite subgroup of the nonzero quaternions that is not cyclic.
Reading path
To understand this page from the ground up, read these 20 pages in order. The foundations come first.
Show the path as a graph
- no evidence level
- Group concept
- Ring definition
- Commutative ring definition
- Polynomial ring definition
- Unit of a ring definition
- Zero divisor definition
- Degree of a polynomial definition
- Group of units of a ring definition
- Polynomial evaluation definition
- Integral domain definition
- Leading coefficient definition
- Root of a polynomial definition
- Monic polynomial definition
- Multiplicity of a root definition
- Degree of a product statement
- Division algorithm for polynomials statement
- Remainder theorem statement
- Factor theorem statement
- A polynomial over a field has at most as many roots as its degree statement
- Finite subgroups of the multiplicative group of a field are cyclic statement
Where this leads
3 pages build on this one.
Further reading
- Web https://en.wikipedia.org/wiki/Finite_field, Multiplicative structure
About this page
| Tag | 0CKGBEJX6B |
|---|---|
| URI | m2://0CKGBEJX6B |
| Kind | statement, theorem |
| Writing | W0 stub |
| MSC | 12E20 |
| Status | This copy of the site has no build data. Run m2 status 0CKGBEJX6B. |
| Uses | Group of units of a ring, A polynomial over a field has at most as many roots as its degree |
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