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A representation of a group $G$ on a vector space $V$ over a field $K$ is a group homomorphism from $G$ to $GL(V)$, the general linear group on $V$. That is, a representation is a map $ \rho \colon G \to GL(V) \,\! $

such that $\rho(g_1 g_2) = \rho(g_1) \rho(g_2) , \qquad \text{for all }g_1,g_2 \in G $

Here $V$ is called the representation space and the dimension of $V$ is called the dimension of the representation.

In the case where V is of finite dimension $n$ it is common to choose a basis for V and identify $GL(V)$ with $GL (n, K)$ the group of $n$-by-$n$ invertible matrices on the field $K$.

An integral representation of $G$ is a map $\rho : G \to GL_n(\mathbb Z)$, what is the vector space $V$ (the representation space) in this case of integral representation? and what is the field $F$ that makes $V$ an $F$-vector space?

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I think it is not a vector space. Rather, it is a $\mathbb{Z}$-module. – user17090 Nov 23 '11 at 19:55
This is a representation (using the term in a more general sense) of $G$ on $\mathbb{Z}^n$ which @AliBleybel pointed out is a $\mathbb{Z}$-module. No field/vector space here. Something a little more general. – Bill Cook Nov 23 '11 at 19:58
Any integral representation extends to, but carries more information than, a representation over $\mathbb{Q}$. – Qiaochu Yuan Nov 23 '11 at 20:19

As noted in the comments, there is no vector space here, but the natural generalisation, a $\mathbb{Z}$-module. In the case of a map $\rho:G\to \operatorname{GL}_n(\mathbb{Z})$ this is the free $\mathbb{Z}$-module $\mathbb{Z}^n$.

You can make $\mathbb{Z}^n$ into a vector space over a field by tensoring with this field. In general an integral representation carries more information than that.

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