# Why is $\mathbb{G}_m$ is a representable functor?

What does it mean that multiplication $$\mathbb{G}_m$$ is a representable functor, with $$\mathbb{G}_m = \text{Spec}(\mathbb{Z}[x,x^{-1}])$$ ?

When I looked at the stacks project page on $$\mathbb{G}_m$$ they have for any scheme [1]

$$\mathbb{G}_m : T \mapsto \Gamma (T, \mathcal{O}^*_T)$$

So I tested this definition on an scheme, $$\mathbb{A}_\mathbb{Q} = \text{Spec}( \mathbb{Q}[x])$$ (the affine line can be defined for any ring). Then

$$\mathbb{G}_m : \mathbb{Q} \mapsto \Gamma(\mathbb{Q},\mathcal{O}_\mathbb{Q}^*) \simeq \mathbb{Q}^\times$$

The units of $$\mathbb{Q}$$ is everything but zero, $$\mathbb{Q}^\times = \mathbb{Q} \backslash \{ 0 \}$$. This could work for now.

Next I looked up the definition of "representable functor". We'd have that:

$$\begin{eqnarray*} \mathcal{F} &\simeq& \text{Hom}_{\mathcal{A}} (A, -) \\ \\ \mathbb{G}_m &\simeq& \text{Hom}_{k-\text{Alg}}(\text{Spec}[\mathbb{Z}[x, x^{-1}], -) \end{eqnarray*}$$

I don't know a lot about affine group schemes, but I know I should definitely expect that $$\mathbb{G}_m(\mathbb{Q}) = \mathbb{Q}^\times$$ and $$\mathbb{G}_m(\mathbb{Q}(i)) = \mathbb{Q}(i)^\times$$. I just wanted to see any clarification of how the category theory fits into here.

Example Here is a morphism $$\phi:\mathbb{Q}(i)\to \mathbb{Q}$$ with $$\phi:a + bi \mapsto a$$. What is $$\mathbb{G}_m(\phi)$$? Certainly it is a morphism of groups.

• Is $\phi$ even a morphism in this category? – cactus314 Dec 18 '18 at 19:40

$$\mathbb{G}_m$$ is a functor from the opposite of the category of schemes to the category of groups: $$\mathbb{G}_m : \mathsf{Sch}^{op}\to\mathsf{Grp}$$ (or even to $$\mathsf{Ab},$$ as schemes are built from commutative rings). You can also define $$\mathbb{G}_m$$ as a functor on the category of commutative rings first via \begin{align*} \mathbb{G}_m : \mathsf{CRing}&\to\mathsf{Ab}\\ R&\mapsto R^\times. \end{align*} This extends to the original definition of $$\mathbb{G}_m$$ because affine schemes form a basis of the topology for schemes (and $$\mathsf{Aff}\simeq\mathsf{CRing}^{op}$$).

Now, $$\mathbb{G}_m$$ is not the functor $$\operatorname{Hom}_{k\textrm{-}\mathsf{Alg}}(\operatorname{Spec}(\Bbb Z[x,x^{-1}]),-);$$ indeed, this does not even make sense! $$\operatorname{Spec}(\Bbb Z[x,x^{-1}])$$ is not an object in $$k\textrm{-}\mathsf{Alg}$$ - it's a scheme! The correct statement would be either $$\mathbb{G}_m\cong\operatorname{Hom}_{\mathsf{Sch}}(-,\operatorname{Spec}(\Bbb Z[x,x^{-1}]))$$ as functors $$\mathsf{Sch}^{op}\to\mathsf{Grp},$$ or $$\mathbb{G}_m\cong\operatorname{Hom}_{\mathsf{CRing}}(\Bbb Z[x,x^{-1}],-)\cong\operatorname{Hom}_{\mathsf{Aff}}(\operatorname{Spec}(-),\operatorname{Spec}(\Bbb Z[x,x^{-1}]))$$ as functors $$\mathsf{CRing}\to\mathsf{Grp}.$$ (So depending on which perspective you take, it is either representable or corepresentable. You might hear both abusively referred to as representable. Exercise: a representable functor is equivalent to the data of a universal element - work out what that universal element is in this case!)

Once you know the contravariant correspondence between rings and affine schemes is given by $$\operatorname{Spec},$$ you can see why $$\mathbb{G}_m(S) := \Gamma(S,\mathcal{O}_S)^\times$$ is "the same" as $$\operatorname{Hom}_{\mathsf{CRing}}(\Bbb Z[x,x^{-1}],-).$$ Indeed, it suffices to look at the situation when $$S = \operatorname{Spec}(R)$$ is affine. In this case, $$\Gamma(R,\mathcal{O}_{\operatorname{Spec}(R)})^\times\cong R^\times.$$ On the other hand, a morphism of rings $$\Bbb Z[x,x^{-1}]\to R$$ is given by choosing an invertible element of $$R$$ which $$x$$ maps to - so $$\operatorname{Hom}_{\mathsf{CRing}}(\Bbb Z[x,x^{-1}],R)$$ is naturally in bijection with $$R^\times.$$

You got the right answer for $$\mathbb{G}_m(\mathbb{A}^1_{\Bbb{Q}}),$$ but let me make a few comments (some things you've written appear off). Since $$\mathbb{A}^1_{\Bbb{Q}} = \operatorname{Spec}(\Bbb Q[t])$$ is affine, compute $$\mathbb{G}_m(\mathbb{A}^1_{\Bbb{Q}})$$ as the units in $$\Bbb Q[t]$$ (or equivalently, units in $$\Gamma(\Bbb A^1_{\Bbb Q},\mathcal{O}_{\Bbb A^1_{\Bbb Q}})$$). This gives $$\mathbb{G}_m(\mathbb{A}^1_{\Bbb{Q}}) = \mathbb{G}_m(\Bbb Q[t]) = \left(\Bbb Q[t]\right)^\times = \Bbb Q^\times,$$ as the units of the polynomial ring over a field are simply the units of the field itself. What you've written appears to be the computation of $$\Bbb G_m(\operatorname{Spec}\Bbb Q) = \Bbb G_m(\Bbb Q) = \Bbb Q^\times.$$ Perhaps you're thinking of $$\Bbb A^1_{\Bbb Q}$$ as being identified with $$\Bbb Q,$$ but this only accounts for some of the points in the scheme $$\Bbb A^1_{\Bbb Q}.$$ There are also points corresponding to [conjugacy classes of] Galois extensions of $$\Bbb Q,$$ and a generic point. When working over a non-algebraically closed field $$k,$$ one needs to be careful about identifying schemes $$X/k$$ with their $$k$$-rational points $$X(k).$$

To answer your last question, the map $$\phi : \Bbb Q(i)\to\Bbb Q$$ you've defined is not a morphism of rings - so it does not make sense to apply $$\mathbb{G}_m$$ to it. On the other hand, the inclusion $$\varphi : \Bbb Q\to\Bbb Q(i)$$ is a morphism of rings. Applying $$\mathbb{G}_m$$ gives the natural map $$\Bbb Q^\times\to\Bbb Q(i)^\times.$$

More generally, for any map of rings $$R\to S,$$ there is an induced group homomorphism on the groups of units $$R^\times\to S^\times.$$ This induced morphism is precisely the one you get when you apply $$\mathbb{G}_m$$ to the map $$R\to S.$$

A final remark: any scheme $$X$$ naturally defines a representable functor in the same way. That is, the scheme $$X$$ may be thought of as a functor \begin{align*} X : \mathsf{Sch}^{op}&\to\mathsf{Set}\\ S&\mapsto X(S) := \operatorname{Hom}_{\mathsf{Sch}}(S,X). \end{align*} This is the so-called "functor of points" perspective of scheme theory. An "$$S$$-valued point of $$X$$" is a morphism $$S\to X.$$ One can recover the set-theoretic points of the scheme $$X$$ by looking at morphisms from the spectra of various fields (which as topological spaces are simply points) to the scheme. The reason that $$\mathbb{G}_m$$ is a functor to the category of groups rather than sets is that it is naturally a group object in $$\mathsf{Sch},$$ which gives the hom-sets $$\operatorname{Hom}(-,\mathbb{G}_m)$$ the structure of a group.

This is not unique to scheme theory - this may be done in any locally small category by the Yoneda embedding - but it is a particularly fruitful perspective here. This allows us to think of schemes as presheaves on the category $$\mathsf{Sch},$$ and in many of the typical topologies one considers on this category (Zariski, etale, fpqc, fppf, see also here, here, and here), the functors defined by schemes are in fact sheaves. Thinking about sheaves on the category of schemes (with respect to some topology) leads to the theory of algebraic spaces, stacks, and other exotic objects that arise in modern algebraic geometry.