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.
If $F_{1}$ is a subfield of $F_{2}$, is $GL(N,F_{1})$ contained in $GL(N,F_{2})$?
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abstract-algebra
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4Absolutely. What's unclear about this? – 2010-12-02
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1I'm not sure how to answer your question. To me, this seems clear from the definitions. Perhaps you could edit your question to reflect a specific point of confusion. – 2010-12-02
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0Yes. This is pretty obvious. Brain fart – 2010-12-02
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1More generally, if $R \to S$ is a morphism of rings, then there is induced a morphism of sets $GL_n(R) \to GL_n(S)$. (Here $GL_n$ for a general ring means that the determinant is *invertible;* this is necessary for the functoriality to work out.) This describes a covariant functor from commutative rings to sets, and in fact it is representable by a scheme! (The scheme is just $\mathrm{Spec} \mathbb{Z}[X_{1,1} \dots X_{1,n}, \dots, X_{n,n}]_{\det M}$ where $M$ is the matrix of the $X_{i,j}$.) – 2010-12-02
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8To be fair to Myke, when you are first learning a subject, sometimes it can be very helpful to ask other people questions like "X is true, right?" even when you very much expect the answer to be "Right, obviously." – 2010-12-03
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1@Akhil: I admire your enthusiasm, but it seems unlikely to me that the OP will know what a representable functor is (or even need to learn about it in the near future). – 2010-12-03
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2@Pete: The possibility had occurred to me, but comments aren't just for the OP, no? – 2010-12-03
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3@Akhil: _commutative_ rings! – 2010-12-03
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0@Qiaochu: yes, I didn't make that clear; thanks. – 2010-12-04