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

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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] Start with any finite spanning set $U$ of $V$. Let $S = \bigcup_{g \in G} \varphi(g)U$. Then, $S$ is finite, because it is a finite union of finite subsets. It is a spanning set because it contains the spanning set $U$ of $V$. and it is invariant under the $G$-action because for any $x \in G$ and any element $\varphi(g)u \in S$, the element $\varphi(x)(\varphi(g)u) = \varphi(xg)u$ is also in $S$.
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] This directly follows from step (1), specifically, the $G$-invariance follows from the $G$-invariance of its generating set, and the fact that the $G$-action, being $K$-linear, is also $R$-linear because $R$ is a subset of $K$.
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] Since $S$ spans $V$, any element $v$ of $V$ can be written as $v = \sum a_is_i$ where $a_i \in K$ and $s_i \in S$. By the definition of field of fractions, each $a_i$ is expressible as $b_i/c_i$ where $b_i, c_i \in R$ and $c_i \ne 0$. Let $c = \prod_i c_i$ be the product. Then, the element $cv$ is a linear combination of the $s_i$s with coefficients all coming from $R$ (because we have cleared the denominators).
4	$M$ is a finitely generated free $R$-module	Fact (1)	$R$ is a principal ideal domain	Step (2)	[SHOW MORE] Since $M$ is a submodule of a vector space over the field of fractions, it is torsion-free. Also, by construction, it is finitely generated. Hence, $M$ is finitely generated and torsion-free, so by fact (1), it is finitely generated and free.
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] The existence of $T$ is guaranteed by step (4). To see that $T$ generates $V$ over $K$, note that $T$ generates $M$, and by Step (3), every element of $V$ is some element of $M$ times the reciprocal of an element of $R$, so $T$ generates all of $V$ over $K$. Finally, to see that $T$ is a basis: suppose there is a linear relation between the elements of $T$ over $K$, of the form $\sum a_i t_i = 0$. Suppose $a_i = b_i/c_i$, $c_i \ne 0$, and $b_i,c_i \in R$. Multiply both sides by the element $c = \prod_i c_i$. We now get a relation between the $t_i$s with coefficients in $R$, contradicting the assumption that they freely generate $M$.
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] By step (2), $M$ is $G$-invariant. In particular, since $T \subseteq M$ by Step (5), the image of any element of $T$ is in $M$. Hence, it is a $R$-linear combination of elements of $T$. This forces all the matrix entries to be from $R$.