Let $T$ be an algebraic torus defined over $\mathbb Q$, $T_\infty$ be its real points and $\pi_0(T_\infty)$ be the group of connected components of $T_\infty$.

Why is the homomorphism $T(\mathbb Q)\to \pi_0(T_\infty)$ surjective?

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Let $T$ be an algebraic torus defined over $\mathbb Q$, $T_\infty$ be its real points and $\pi_0(T_\infty)$ be the group of connected components of $T_\infty$. Why is the homomorphism $T(\mathbb Q)\to \pi_0(T_\infty)$ surjective? Thanks |
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cyclicdecomposition groups in $\Pi$, the group $T(k)$ is dense in $\prod_{v\in S}T(k_v)$. (This result is also due to Serre). Now take $k=\mathbf{Q}$, $S=\{\infty\}$, then a decomposition group of $\infty$ is of order 2, hence cyclic, whence we obtain that $T(\mathbf{Q})$ is dense in $T(\mathbf{R})$. – Mikhail Borovoi Mar 7 '12 at 19:42