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If $F_{1}$ is a subfield of $F_{2}$, is $GL(N,F_{1})$ contained in $GL(N,F_{2})$? I am particular interested in the case of when $F_{1} =\mathbb{Q}(\omega)$ and $F_{2} = \mathbb{C}$. Here $\omega$ is a complex root of unity.

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    @Qiaochu: yes, I didn't make that clear; thanks.2010-12-04

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For a field $\mathbb{F}$, $M(n, \mathbb{F})$ is the set of $n\times n$ matrices with entries in $\mathbb{F}$, and $GL(n, \mathbb{F}) \subset M(n, \mathbb{F})$ is the set of $n\times n$ invertible matrices with entries in $\mathbb{F}$.

If $\mathbb{F}_1, \mathbb{F}_2$ are fields with $\mathbb{F}_1 \subseteq \mathbb{F}_2$, then $M(n, \mathbb{F}_1) \subseteq M(n, \mathbb{F}_2)$ because every $n\times n$ matrix with entries in $\mathbb{F}_1$ is an $n\times n$ matrix with entries in $\mathbb{F}_2$ (because $\mathbb{F}_1 \subseteq \mathbb{F}_2$).

Now, if $A \in GL(n, \mathbb{F}_1)$, there is $A^{-1} \in GL(n, \mathbb{F}_1)$ with $AA^{-1} = A^{-1}A = I$. As $GL(n, \mathbb{F}_1) \subset M(n, \mathbb{F}_1) \subseteq M(n, \mathbb{F}_2)$, we have $A, A^{-1} \in M(n, \mathbb{F}_2)$ and $AA^{-1} = A^{-1}A = I$, so $A \in GL(n, \mathbb{F}_2)$. Therefore, $GL(n, \mathbb{F}_1) \subseteq GL(n, \mathbb{F}_2)$.