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