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

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Statement

Suppose R is a Principal ideal domain (?) and K is its field of fractions. Suppose \varphi:G \to GL(n,K) is a linear representation of a finite group G. Then, we can choose a basis for K^n, such that, in this new basis, all the entries of the matrices \varphi(g), g \in G are from R.

Related facts

Applications

In particular, this result applies to the case R = \mathbb{Z}, 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 \mathbb{Z}.

Facts used

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

Proof

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Given: A linear representation \varphi:G \to GL(n,K) of a finite group G over the field of fractions K of a principal ideal domain R.

To prove: There is a choice of basis of K^n in which all the matrices for \varphi(g) have entries from R.

Proof: We let V = K^n 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 S for V (as a K-vector space) such that S is G-invariant V is finite-dimensional, G is finite [SHOW MORE]
2 Let M be the R-submodule generated by S. Then, M is a G-invariant R-module R is contained in K. Step (1) [SHOW MORE]
3 Every element of V has a nonzero R-multiple in M K is the field of fractions of R (implicitly, R is an integral domain) Step (1) [SHOW MORE]
4 M is a finitely generated free R-module Fact (1) R is a principal ideal domain Step (2) [SHOW MORE]
5 Let T be a freely generating set for M as a R-module. Then, T is a basis for V K is the field of fractions of R (implicitly, R is an integral domain) Steps (3), (4) [SHOW MORE]
6 For any g \in G, the matrix for the action of g in the basis T has all its entries in R Steps (2), (5) [SHOW MORE]