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

## 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

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 $\varphi:G \to GL(n,K)$ 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]