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I want to find an algebraic group with two components over, say, $\mathbb{C}$ for simplicity. This more or less means that we want an index 2 subgroup. Wikipedia says that the character group $X^*(\mathbb{G}_m) \approx \mathbb{Z}$, so this should work since we have the subgroup $2\mathbb{Z}$. However, I do not understand why this is correct (I think the isomorphism is true since all of the maps are power maps of some degree) - in particular, a variety should be the zero locus of a finite set of (finite degree) polynomials, so I do not see how $\mathbb{Z}$, or even $\text{Spec}\mathbb{Z}$, could satisfy this.

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    $\begingroup$ Would the group of $2\times2$-matrices $X$ such that $\det(X)^2=1$ fit the bill? Adjust size of the matrices to your liking. $\endgroup$ – Jyrki Lahtonen Jul 1 at 20:19
  • $\begingroup$ @jgon Sorry, I meant 2$\mathbb{Z}$ - will correct it. $\endgroup$ – nilradical1 Jul 1 at 20:38
  • $\begingroup$ @JyrkiLahtonen Thanks, this works! $\endgroup$ – nilradical1 Jul 2 at 12:38
  • $\begingroup$ You could take the constant group $\underline{\mathbb{Z}/2\mathbb{Z}}$. $\endgroup$ – Alex Youcis Jul 24 at 13:53

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