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I can not think of any example of a definable group in algebraically closed fields?

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When you say definable, can you specify the language and theory? I assume that you mean language of rings and the theory is ACF, but the more you write the better answers you might get. – Asaf Karagila Mar 21 '12 at 9:12

Algebraic groups are the most relevant examples that are definable over alegebraically closed fields. So you can take $\mathrm{GL}(\mathfrak{n},\mathbb{C})$ can serve as an example.

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yes, the language is the ring language and the theory is ACF. But what is the formula defining GL(n,C)? – sagara Mar 21 '12 at 9:17
Its nothing but the General Linear Group of set of invertible matrices over $\mathbb{C}$. So if you are looking for some more examples see this where the author gives numerous examples of such groups in $8.1$ @sagara – Iyengar Mar 21 '12 at 9:24
thank you very much – sagara Mar 21 '12 at 9:28
Iyengar, definable does not mean that you can wildly define. It means that there is a formula in the language of rings that the set it defines forms a group. I don't know whether your answer is correct, but your reasoning is not, just being "the set of invertible matrices" is not necessarily being definable. If that was the case, why won't the positive integers be definable as well? Those are just "the integers". You should perhaps read some short introductory text on model theory. – Asaf Karagila Mar 21 '12 at 9:30
Ummm, Iyengar is right. "The set of invertible matrices" easily unpacks to the first order definition $\{ (x_{11}, \ldots, x_{nn}) : \exists (y_{11}, \ldots, y_{nn}) \mbox{ such that } (x_{ij}) (y_{ij}) = \mathrm{Id}_n \}$. Here the matrix identity must be written out as $n^2$ explicit equations of the form $\sum_k x_{ik} y_{kj} = \delta_{ij}$. (There are more efficient ways to define $GL_n$, using determinants, but this is the most straightforward interpretation of what Iyengar wrote.) – David Speyer Mar 21 '12 at 14:07

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