Algebraic torus and generic points

Question

Many definitions of the multiplicative group scheme $G_m$ define it as $\operatorname{Spec}\mathbb{C}[x, x^{-1}]$. But the Spec of that ring is actually $\mathbb{C}^*\cup\{\eta\}$, with an extra generic point. So when you want to define multiplication, etc., it seems obvious what to do on the rest, but how do you deal with the generic point?

If you define your multiplication map to be the one induced by $x\mapsto x\otimes x$ on the level of rings, it seems like if you "multiply" the generic point of $G_m$ with any other element, your output will be the generic point again.

Furthermore, many basic examples in GIT have the group scheme $G_m$ acting on some space, but people seem to largely ignore the fact that this extra generic point $\eta$ is there, and just define how the elements of $\mathbb{C}^*$ act. What allows us to do this?

Answer 1

$\newcommand{\Sch}{\operatorname{Sch}}\newcommand{\Spec}{\operatorname{Spec}}$This is confusing in general because products of schemes are weird. Let's fix a base field to work over, $k$, which you are more than fine to think of as $\mathbb C$. The category $\Sch/k$ has objects which are schemes $X$ equipped with morphism to $\Spec k$, i.e. the objects are 'arrows' $f_X:X\rightarrow \Spec k$. The morphisms are scheme morphisms $g:X\rightarrow Y$ such that $f_Y\circ g=f_X$ (draw the commutative diagram out to make sense of this).

Now, products in the category $\Sch/k$ are as I am sure you are aware not the product $X\times Y$, but instead the fibre product $X\times_kY$. The fibre product in this case is set theoretically different from the cartesian product, and the topology is also not the product topology. In particular, given an affine open covering of $\Spec A_i\subset X$ and affine open covering of $\Spec B_j\subset Y$, we have that $X\times_kY$ is covered by affine opens of the form $\Spec A_i\otimes_k B_j$. Now in nice enough situations, i.e. where $k=\bar k$, and the $A_i$ and $B_j$ are finitely generated $k$ algebras, the $k$-points of the fibre product are in natural bijection with the cartesian product of the $k$-points of $X$ and the $k$-points of $Y$, but this is not the case in full generality, and obviously does not hold non closed points.

Now, a group scheme over $k$ is a group object in the category $\Sch/k$, meaning it is a scheme $G$, with a morphisms $e:\Spec k\rightarrow G$, $m:G\times_kG\rightarrow G$, and $i:G\rightarrow G$ which make the relevant diagrams commute (i.e. the group axiom diagrams). This is actually how you define group objects in any category with finite products, and terminal objects. For any such category $\mathcal C$, we obtain a new category $\operatorname{Grp}(\mathcal C)$ consisting of objects which are group objects of $\mathcal C$, and morphisms which are morphisms in $\mathcal C$ which respect $m$. If the objects of $\mathcal C$ are sets with extra structure, it is natural to ask the following question: when does the forgetful functor $\mathcal C\rightarrow \operatorname{Set}$ induce a forgetful functor $\operatorname{Grp}(\mathcal C)\rightarrow \operatorname{Grp}$, and the answer is only when the forgetful functor sends your terminal object to $\{\star\}$, and when it preserves products. As we remarked this not the situation in the category of schemes over $k$, hence the underlying sets of group schemes are not literally groups.

There is, however, some groupiness going on here as you remark. After all, if we forget about the generic point, i.e. only consider the closed points of $\Spec \mathbb C[x,x^{-1}]$, then we do obtain the group $\mathbb C^*$. Let $G$ be a group object in $\Sch/k$ as before, and define $h_G$ to be the contravariant functor: \begin{align*} h_G:\Sch/k&\longrightarrow \operatorname{Set}\\ X&\longmapsto \operatorname{Hom}_{\Sch/k}(X,G) \end{align*} The maps $m$, $i$, and $e$ give each $\operatorname{Hom}_{\Sch/k}(X,G)$ a group structure, so each group scheme can be thought of as a representable functors $\Sch/k\rightarrow \operatorname{Grp}$. Moreover, if a functor $\Sch/k\rightarrow \operatorname{Grp}$ is representable, then there exists moprhisms $m$, $i$, and $e$ which make the scheme a group object. In particular the $k$-points, or when $k=\bar k$ the closed points, are in natural bijection with $\operatorname{Hom}_{\Sch/k}(\Spec k,G)$, hence form a group.

Now when we work over a base field $k$, since everything is locally determined by a $k$-algebra, and all morphisms come group gluing morphisms induced by $k$-algebra morphisms, it actually suffices to define a contravariant functor: $$\operatorname{Aff}\Sch/k\longrightarrow \operatorname{Grp}$$ And here the group scheme $k^*$ in $\Sch/k$ is the one representing the functor that takes $\Spec A$ to $A^*$, the invertible elements in $A$. Similarly, the group scheme $GL_n(k)$ represents the functor $A\mapsto GL_n(A)$.

To describe a group scheme action $G\times_kX\rightarrow X$, it suffices to describe group actions: $$\operatorname{Hom}_{Sch/k}(\Spec A,G)\times \operatorname{Hom}_{\Sch/k}(\Spec A,X)\longrightarrow \operatorname{Hom}_{\Sch/k}(\Spec A,X)$$ which satisfy some compatibility condition so that you actually get a natural transformation of functors from $\operatorname{Aff}\Sch/k\rightarrow \operatorname{Set}$. So, in general you cannot just ignore non closed points, and define a morphism only on $k$-valued points.

However, when $k=\bar k$, and your group objects, and your schemes in $\Sch/k$ are reduced, finite type, and separated (i.e. varieties) it actually does suffices to only define morphism on $k$ points, so long as it is `locally given by polynomials'. This is most likely what the authors of your GIT text are implicitly assuming (and I agree that they should make this type of thing explicit).

Answer 2

Q: "So when you want to define multiplication, etc., it seems obvious what to do on the rest, but how do you deal with the generic point?"

A: Let $A:=k[x,1/x]$ with $k$ the complex number field and let $\mathfrak{m_i}:=(x-a_i)$ with $a_i \in k^*$. Let $G:=Spec(A)$ and let $\mathfrak{p}:=(0)$ be the generic point. Given two points $a_1,a_2 \in k^*$ you add them as follows: The comultiplication $\Delta$ gives the sequence

$$ A \rightarrow A\otimes_k A \rightarrow^{f_1 \otimes f_2} k \otimes_k k \rightarrow^m k$$

where $f_i:A \rightarrow A/\mathfrak{m}_i$ is the canonical map and $m$ is the multiplication. You get elements $f_1,f_2 \in Hom_{k-alg}(A, k):=G(k)$. The multiplication of these two points is by definition the map

$$ f_1*f_2:= m \circ (f_1 \otimes f_2) \circ \Delta \in G(k).$$

(You may check that this turns the set $G(k)$ into an abelian group isomorphic to the multiplicative group $k^*$.)

By definition when you have an affine group scheme $G$ over a field $k$ with coordinate ring $k[G]$, you may only multiply points with the same residue field. You cannot multiply the generic point with a point on the form $\mathfrak{m}_i$ since the residue fields differ. For any field extension $k \subseteq K$ it follows the set $G(K):=Hom_{k-alg}(k[G], K)$ is a group with group multiplication as defined above.