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Let $K$ be a field of characteristic 0, $n$ a positive integer, and $G$ the group $GL_n(K)$ of invertible linear transformations on $K^n$. Show that the following subsets of $G$ are $\emptyset$-definable: (a) the set of all scalar matrices; (b) the set of all matrices which are similar to a diagonal matrix with distinct scalars down the main diagonal; (c) the set of diagonalisable matrices.

My question is this. Are we viewing $G$ a model of the theory of groups and being asked to define subsets of $G$ in the language of groups, or are we viewing $K$ as a definable subset of $K^{n^2}$ and being asked to define subsets of $G$ in the language of fields? My first thought was the former, but I'm beginning to think the latter is intended.

Adam Gutter
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    My impression is that the structure is the field $K$ of characteristic $0$ and $G$ is considered as a definable subset of $K^{n^2}$. – Levon Haykazyan Jul 05 '13 at 23:18
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    Since this is a starred exercise, and is easy under the second interpretation, I'm inclined to think the first is correct. – Chris Eagle Jul 06 '13 at 18:56
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    Idea for (b): try to define the relation that holds between $kI$ and $A$ iff $k$ is an eigenvalue of $A$. – Chris Eagle Jul 06 '13 at 18:59

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I'm pretty sure it's the latter. This exercise is about linear algebraic groups, which are specific kind of algebraic groups, which in turn are definable groups in (algebraically closed) fields. If there are definable groups anywhere near this exercise, there's little room left for doubt.

I don't think the exercise is even doable in the other interpretation. It's easy to define scalar matrices (just the center), but I doubt you can define diagonal matrices using only group language in $GL_n$.

tomasz
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    You certainly can't define diagonal matrices (not closed under group automorphisms). I'm not sure if you can define diagonalisable ones. – Chris Eagle Jul 06 '13 at 18:48