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The Galois correspondence exchanges composita with subgroup intersections and field intersections with generated subgroups
Statement
Let be finite Galois, and let correspond to subgroups for . Then
and
Facts & Assumptions
Given: The compositum , the generated subgroup of The subgroup generated by a subset, the cyclic subgroup , and cyclic groups, and the subgroup intersection of The intersection of a nonempty family of subgroups of is a subgroup of .
The assignments and are mutually inverse inclusion-reversing bijections (The fundamental theorem of finite Galois theory).
Proof
An automorphism of fixes exactly when it fixes every element of both and , exactly when it belongs to both and . Therefore .
An element of is fixed by exactly when it is fixed by every element of both generating subgroups, so . Applying [L1] gives the second formula. These membership equivalences also cover equal fields, the base and top fields, and trivial or full subgroups.
Depends on
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
12 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- K. Conrad, The Galois Correspondence, Theorem 5.13 (standard reference, not scraped)
- J. S. Milne, Fields and Galois Theory, v5.10, Chapter 3 (standard reference, not scraped)