# Algebraic Group with two components over algebraically closed field

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.

• 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. – Jyrki Lahtonen Jul 1 at 20:19
• @jgon Sorry, I meant 2$\mathbb{Z}$ - will correct it. – nilradical1 Jul 1 at 20:38
• @JyrkiLahtonen Thanks, this works! – nilradical1 Jul 2 at 12:38
• You could take the constant group $\underline{\mathbb{Z}/2\mathbb{Z}}$. – Alex Youcis Jul 24 at 13:53