Linear representation is realizable over principal ideal domain iff it is realizable over field of fractions

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Statement

Suppose is a Principal ideal domain (?) and is its field of fractions. Suppose is a linear representation of a finite group . Then, we can choose a basis for , such that, in this new basis, all the entries of the matrices are from .

Related facts

Applications

In particular, this result applies to the case , and shows that for any rational representation group, we can find a representation where all the matrix entries of all the representing matrices are from .

Facts used

  1. Structure theorem for finitely generated modules over principal ideal domains

Proof

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

Given: A linear representation of a finite group over the field of fractions of a principal ideal domain .

To prove: There is a choice of basis of in which all the matrices for have entries from .

Proof: We let be the vector space acted upon.

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 There exists a finite spanning set for (as a -vector space) such that is -invariant is finite-dimensional, is finite [SHOW MORE]
2 Let be the -submodule generated by . Then, is a -invariant -module is contained in . Step (1) [SHOW MORE]
3 Every element of has a nonzero -multiple in is the field of fractions of (implicitly, is an integral domain) Step (1) [SHOW MORE]
4 is a finitely generated free -module Fact (1) is a principal ideal domain Step (2) [SHOW MORE]
5 Let be a freely generating set for as a -module. Then, is a basis for is the field of fractions of (implicitly, is an integral domain) Steps (3), (4) [SHOW MORE]
6 For any , the matrix for the action of in the basis has all its entries in Steps (2), (5) [SHOW MORE]