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

Statement

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

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Applications

In particular, this result applies to the case $\text{[math]}$ , 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 $\text{[math]}$ .

Facts used

Structure theorem for finitely generated modules over principal ideal domains

Proof

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Given: A linear representation $\text{[math]}$ of a finite group $\text{[math]}$ over the field of fractions $\text{[math]}$ of a principal ideal domain $\text{[math]}$ .

To prove: There is a choice of basis of $\text{[math]}$ in which all the matrices for $\text{[math]}$ have entries from $\text{[math]}$ .

Proof: We let $\text{[math]}$ 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 $\text{[math]}$ for $\text{[math]}$ (as a $\text{[math]}$ -vector space) such that $\text{[math]}$ is $\text{[math]}$ -invariant		$\text{[math]}$ is finite-dimensional, $\text{[math]}$ is finite		[SHOW MORE] Start with any finite spanning set $\text{[math]}$ of $\text{[math]}$ . Let $\text{[math]}$ . Then, $\text{[math]}$ is finite, because it is a finite union of finite subsets. It is a spanning set because it contains the spanning set $\text{[math]}$ of $\text{[math]}$ . and it is invariant under the $\text{[math]}$ -action because for any $\text{[math]}$ and any element $\text{[math]}$ , the element $\text{[math]}$ is also in $\text{[math]}$ .
2	Let $\text{[math]}$ be the $\text{[math]}$ -submodule generated by $\text{[math]}$ . Then, $\text{[math]}$ is a $\text{[math]}$ -invariant $\text{[math]}$ -module		$\text{[math]}$ is contained in $\text{[math]}$ .	Step (1)	[SHOW MORE] This directly follows from step (1), specifically, the $\text{[math]}$ -invariance follows from the $\text{[math]}$ -invariance of its generating set, and the fact that the $\text{[math]}$ -action, being $\text{[math]}$ -linear, is also $\text{[math]}$ -linear because $\text{[math]}$ is a subset of $\text{[math]}$ .
3	Every element of $\text{[math]}$ has a nonzero $\text{[math]}$ -multiple in $\text{[math]}$		$\text{[math]}$ is the field of fractions of $\text{[math]}$ (implicitly, $\text{[math]}$ is an integral domain)	Step (1)	[SHOW MORE] Since $\text{[math]}$ spans $\text{[math]}$ , any element $\text{[math]}$ of $\text{[math]}$ can be written as $\text{[math]}$ where $\text{[math]}$ and $\text{[math]}$ . By the definition of field of fractions, each $\text{[math]}$ is expressible as $\text{[math]}$ where $\text{[math]}$ and $\text{[math]}$ . Let $\text{[math]}$ be the product. Then, the element $\text{[math]}$ is a linear combination of the $\text{[math]}$ s with coefficients all coming from $\text{[math]}$ (because we have cleared the denominators).
4	$\text{[math]}$ is a finitely generated free $\text{[math]}$ -module	Fact (1)	$\text{[math]}$ is a principal ideal domain	Step (2)	[SHOW MORE] Since $\text{[math]}$ is a submodule of a vector space over the field of fractions, it is torsion-free. Also, by construction, it is finitely generated. Hence, $\text{[math]}$ is finitely generated and torsion-free, so by fact (1), it is finitely generated and free.
5	Let $\text{[math]}$ be a freely generating set for $\text{[math]}$ as a $\text{[math]}$ -module. Then, $\text{[math]}$ is a basis for $\text{[math]}$		$\text{[math]}$ is the field of fractions of $\text{[math]}$ (implicitly, $\text{[math]}$ is an integral domain)	Steps (3), (4)	[SHOW MORE] The existence of $\text{[math]}$ is guaranteed by step (4). To see that $\text{[math]}$ generates $\text{[math]}$ over $\text{[math]}$ , note that $\text{[math]}$ generates $\text{[math]}$ , and by Step (3), every element of $\text{[math]}$ is some element of $\text{[math]}$ times the reciprocal of an element of $\text{[math]}$ , so $\text{[math]}$ generates all of $\text{[math]}$ over $\text{[math]}$ . Finally, to see that $\text{[math]}$ is a basis: suppose there is a linear relation between the elements of $\text{[math]}$ over $\text{[math]}$ , of the form $\text{[math]}$ . Suppose $\text{[math]}$ , $\text{[math]}$ , and $\text{[math]}$ . Multiply both sides by the element $\text{[math]}$ . We now get a relation between the $\text{[math]}$ s with coefficients in $\text{[math]}$ , contradicting the assumption that they freely generate $\text{[math]}$ .
6	For any $\text{[math]}$ , the matrix for the action of $\text{[math]}$ in the basis $\text{[math]}$ has all its entries in $\text{[math]}$			Steps (2), (5)	[SHOW MORE] By step (2), $\text{[math]}$ is $\text{[math]}$ -invariant. In particular, since $\text{[math]}$ by Step (5), the image of any element of $\text{[math]}$ is in $\text{[math]}$ . Hence, it is a $\text{[math]}$ -linear combination of elements of $\text{[math]}$ . This forces all the matrix entries to be from $\text{[math]}$ .

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

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