Repository navigation
Improve IrreducibleModules for permutation groups - #6832
Draft
ThomasBreuer wants to merge 2 commits into
Draft
ThomasBreuer wants to merge 2 commits into
ThomasBreuer wants to merge 2 commits into
Conversation
As far as I understand, the number of absolutely irreducible constituents of an irreducible module over a finite field is determined by the degree of the minimal field extension that is needed to split the module into absolutely irreducible constituents. Thus one need not construct and decompose the module over the larger field.
ThomasBreuer
added
kind: enhancement
ThomasBreuer
marked this pull request as draft
November 11, 2026 09:57
Contributor
Author
|
(why is this a draft: another improvement should be added ...) |
This branch has not been deployed
This file contains hidden or bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
Sign up for free
to join this conversation on GitHub.
Already have an account?
Sign in to comment
Add this suggestion to a batch that can be applied as a single commit.This suggestion is invalid because no changes were made to the code.Suggestions cannot be applied while the pull request is closed.Suggestions cannot be applied while viewing a subset of changes.Only one suggestion per line can be applied in a batch.Add this suggestion to a batch that can be applied as a single commit.Applying suggestions on deleted lines is not supported.You must change the existing code in this line in order to create a valid suggestion.Outdated suggestions cannot be applied.This suggestion has been applied or marked resolved.Suggestions cannot be applied from pending reviews.Suggestions cannot be applied on multi-line comments.Suggestions cannot be applied while the pull request is queued to merge.Suggestion cannot be applied right now. Please check back later.
As far as I understand, the number of absolutely irreducible constituents of an irreducible module over a finite field is determined by the degree of the minimal field extension that is needed to split the module into absolutely irreducible constituents, see the comment in the code.
Thus one need not construct and decompose the module over the larger field.
(The
MTX.IsAbsolutelyIrreduciblecall happens now insideMTX.DegreeFieldExt. Note that the old code did in fact construct the abs. irred. module over an even bigger field than is needed, sinceMTX.DegreeSplittingFieldmeans the degree over the prime field.)