Algebraic Torus is a group scheme

Question

I am taking a course on toric varieties this semester, and I am a little confused by how the algebraic torus is a group scheme, as we didn't really define what a group scheme is. I was given the following definition of the algebraic torus: $$T^n=\operatorname{Spec}(\mathbb C[x_1,\dots, x_n,x^{-1}_1,\dots, x^{-1}-n]\cong (\mathbb (A^1\smallsetminus\{0\})^n$$ Now, a group scheme is defined after fixing a base scheme, and since my instructor stated that everything we are doing is over $\mathbb C$ I assume that this base scheme is $\operatorname{Spec}\mathbb C$. A group scheme over the base scheme $\operatorname{Spec}\mathbb C$ is given by scheme morphisms: \begin{align} m:&T^n\times_{\operatorname{Spec} \mathbb C }T^n\longrightarrow T^n\\ e:&\operatorname{Spec}\mathbb C\longrightarrow T^n\\ i:&T^n\longrightarrow T^n \end{align} which satisfy the axioms of a group object in the category of schemes over $\operatorname{Spec}\mathbb C$.

Now my professor said that these maps were given by identifying $T^n$ with $(\mathbb C^*)^n$, and then multiplication and inversion are the usual maps, and $e:\operatorname{Spec}\mathbb C\longrightarrow T^n$ is given by $\cdot\mapsto 1$. However, what's confusing me is that all these schemes have the generic point $\eta$ corresponding to the zero ideal, so $\cdot \mapsto 1$ really means that $\eta\mapsto 1$ as $\eta$ is the only point in $\operatorname{Spec}\mathbb C$. But what ring morphism induces this scheme morphism? Moreover, how do we multiply an element in $(\mathbb C^*)^n$ with the generic point? How do we invert the generic point? It would seem reasonable if the generic point were the identity, but that certainly has to be the prime ideal $\langle x_1-1,\dots, x_n-1\rangle$ right?

Any help would be greatly appreciated.

Answer 1

The morphism $e$ is induced by the morphism of rings which sends all $x_i$ to $1$, which is exactly what you expected it to be, the quotient by the ideal generated by all the $x_i-1$.

You are wondering how the generic point of $T^n$ acts by multiplication, but there is yet another surprise: there are points of $T^n\times_{\operatorname{Spec}\mathbb{C}}T^n$ which are not just tuples of points of the factors. For example, there are generic points of various curves on this variety. And evaluating this morphism at these additional points of the product doesn't look anything like acting by an element of a group on a different element of the same group.

The good news is, we are not expecting the underlying set of a group scheme to be a group, so this is not really a problem! The forgetful functor from the category of $\mathbb{C}$-schemes to sets does not preserve group objects. What is true, however, is that over an algebraically closed field the closed points of a variety which is a group object as a scheme will inherit the structure of a group. So what your professor said or meant is that we are identifying the closed points of $T^n$ with $(\mathbb C^*)^n$, and then the multiplication and inversion are really the obvious ones.