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.