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A field with elements is the splitting field of over its prime subfield
Statement
If is a field with elements, then every satisfies , and is the splitting field of over its prime subfield.
Facts & Assumptions
Given: A finite field of order .
The group is cyclic (The multiplicative group of a finite field is cyclic).
A splitting field is generated over the base by all roots of the polynomial (Polynomials that split and splitting fields of a polynomial or a family of polynomials).
A nonzero degree- polynomial over a domain has at most distinct roots (A nonzero polynomial of degree over an integral domain has at most distinct roots).
Proof
The cyclic group has order , so every nonzero satisfies and hence . The equality also holds for .
Thus all elements of are roots of . By [L4] there are no other distinct roots in any extension, so the polynomial splits into its linear factors over .
The set of roots is all of , so it generates over its prime subfield. By [L3], is the splitting field.
Depends on
Used by
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 42 results over 13 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.
Sources
- K. Conrad, Finite Fields, Section 2 (standard reference, not scraped)