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