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Let $G$ be a linear algebraic group, embedded as a Zariski-closed subgroup of $GL(n,\mathbb{C})$ with respect to the identification of $GL(n,\mathbb{C})$ as a subset of $\mathbb{C}^{n^2+1}$ via $M \mapsto (M,\det(M)^{-1})$.

Then $G$ is said to be defined over $\mathbb{Q}$ if it is the vanishing set of polynomials $f_1,\ldots,f_m \in \mathbb{Q}[X_{11},\ldots,X_{nn},T]$. All linear algebraic groups I know of are defined over $\mathbb{Q}$, for example $GL(n), SL(n), SO(n)$, invertible diagonal matrices $T_n$, $\mathbb{C}^\times, (\mathbb{C}^n,+)$ etc.

What is an instructive example of a group not defined over $\mathbb{Q}$?

abenthy
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    I'm not sure about affine algebraic groups but elliptic curves with defining equation $y^2=x(x-1)(x-\lambda)$, $\lambda \notin \mathbb{R}$ are examples of abelian varieties without $\mathbb{Q}$-structures – Exit path Jul 06 '17 at 16:18

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How about the group of matrices of the form $\begin{pmatrix} 1 & x & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & \pi x \\ 0 & 0 & 0 & 1 \end{pmatrix}$ for some $x \in \mathbf{C}$? It's obviously isomorphic to a group defined over $\mathbf{Q}$ -- the additive group -- but it's embedded into $\mathrm{GL}(4)$ in a weird, non-rational way.

David Loeffler
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  • Do you have an example where the group homomorphism sending it to the group over $\mathbf{Q}$ can't be holomorphic ? (as for $\mathbf{C}/(\mathbf{Z}+\pi \mathbf{Z})$) – reuns Jul 06 '17 at 18:13
  • I don't understand your question: the quotient $C / (Z + \pi Z)$ isn't an algebraic group. – David Loeffler Jul 07 '17 at 09:17