I am starting to study algebraic groups and I came across the following statement: let $S=\mathrm{Res}_{\mathbb{C}/\mathbb{R}}(\mathbb{G}_{m,\mathbb{C}})$ be the restriction of scalars of the multiplicative group over $\mathbb{C}$. Then, there is a canonical map $w:\mathbb{G}_{m,\mathbb{R}}\to S$ which on real points is defined as the inclusion of $\mathbb{R}^\ast$ in $\mathbb{C}^\ast$.
I don’t understand how can we obtain a map of schemes/varieties by just defining it over the set of real points. Moreover, since $\mathbb{R}$ is not algebraically closed, the set of real points doesn’t necessarily equals the whole set of points of the underlying topological space and I can’t seem to understand how an extension to the whole space can be obtained.
I am not really experienced about these topics, so if you could provide some sort of reference to learn about it (and to get more comfortable working with these kind of statements), that would be very much appreciated.
Thanks in advance
Just to get this off the unanswered list.
Yes, it's not true that the functor from linear algebraic groups (or even reductive groups) over $\mathbb{R}$ to real Lie groups is full. For example, let $U(1)$ denote the algebraic group given by
$$U(1)=\mathrm{Spec}(\mathbb{R}[x,y]/(x^2+y^2+1))$$
with the obvious multiplication (one can think of this as the norm $1$ elements of $\mathsf{Res}_{\mathbb{C}/\mathbb{R}}\mathbb{G}_{m,\mathbb{C}}$ if you'd like). Then, $U(1)(\mathbb{R})\cong S^1$ as real Lie groups. Note then that the map
$$\mathrm{Hom}(\mathbb{G}_{m,\mathbb{R}},U(1))\to \mathrm{Hom}(\mathbb{G}_{m,\mathbb{R}}(\mathbb{R}),U(1)(\mathbb{R}))=\mathrm{Hom}(\mathbb{R}^\times,S^1)$$
is not surjective. Indeed, every map $\mathbb{G}_{m,\mathbb{R}}\to U(1)$ is constant since the former is a split torus and the latter is an $\mathbb{R}$-anisotropic torus. That said, there is certainly a map of Lie groups
$$\mathbb{R}^\times\cong \mathbb{R}^{>0}\times \mathbb{Z}/2\mathbb{Z}\to \mathbb{R}^{>0}\cong \mathbb{R}\twoheadrightarrow \mathbb{R}/\mathbb{Z}\cong S^1$$
or more explicitly
$$x\mapsto \exp(2\pi i\log(|x|))$$
NB: This fully-faithfulness, and in fact equivalence, for $\mathbb{R}$-anisotropic groups to compact Lie groups is true. For example, see [Con, Theorem D.2.4].
That said, the map
$$\mathbb{G}_{m,\mathbb{R}}=\mathbb{R}^\times\hookrightarrow \mathbb{C}^\times=(\mathsf{Res}_{\mathbb{C}/\mathbb{R}}\mathbb{G}_{m,\mathbb{C}})(\mathbb{R})$$
can be made functorial, and thus algebraic. Namely, for every $\mathbb{R}$-algebra $R$ one has an inclusion of $R$-algebras
$$R\hookrightarrow R\otimes_\mathbb{R}\mathbb{C}$$
and thus an inclusion of groups
$$\mathbb{G}_{m,\mathbb{R}}(R)=R^\times\hookrightarrow (R\otimes_\mathbb{R}\mathbb{C})^\times=(\mathsf{Res}_{\mathbb{C}/\mathbb{R}}\mathbb{G}_{m,\mathbb{C}})(R)$$
Explicitly, one can model
$$\mathsf{Res}_{\mathbb{C}/\mathbb{R}}\mathbb{G}_{m,\mathbb{C}}=\left\{\begin{pmatrix}a & -b\\b & a\end{pmatrix}\right\}\subseteq \text{GL}_{2,\mathbb{R}}$$
and the map $\mathbb{G}_{m,\mathbb{R}}\to\text{GL}_{2,\mathbb{R}}$ is just the inclusion of the center of $\text{GL}_{2,\mathbb{R}}$ (the diagonal matrices).
[Con] Conrad, B., 2014. Reductive group schemes. Autour des schémas en groupes, 1, pp.93-444.