Let $R$ be the ring $\mathbf{C}\times\mathbf{C}$, and consider the affine line $\mathbf{A}^1\_R$.


$\mathbf{A}^1\_R$ can be given the structure of additive group scheme over $R$, denoted $(\mathbf{G}\_a)\_R$. 


$(\mathbf{G}\_a)\_R$ carries a functorial multiplicative action of $R$ making it into an $R$-module object in schemes: the free $R$-module scheme of rank $1$.


On the other hand, $R$ is a $\mathbf{C}$-vector space of dimension $2$.



> 
> Is $(\mathbf{G}\_a)\_R$ a $\mathbf{C}$-vector space object in schemes, of dimension $2$?
> 
> 
> 


I am having difficulties to gain intuition about this, because $(\mathbf{G}\_a)\_R$ is, as a scheme, a disjoint union of two copies of $\mathbf{A}^1\_{\mathbf{C}}$. How does one endow it with a $\mathbf{C}$-vector space object structure?


(likely a very simple question. I’m just a little too confused about something trivial)



Vector space objects in schemes - confusion
Briefly, $\mathbb{A}^1\_R$ is not a vector space over $\mathbb{A}^1\_\mathbb{C}$ in a natural way. 


Strictly speaking, saying that $\mathbb{A}^1\_R$ is a ring object in schemes in not precise. What is correct is that the morhpism $\mathbb{A}^1\_R\to \mathrm{Spec} R$ (adjoint to $R\to R[T]$) is a ring object in the category of schemes over $R$ (i.e. schemes equipped with a morphism into $\mathrm{Spec} R$). [This means that there are addition and product morphisms $+,\cdot:\mathbb{A}^1\_R\times\_R\mathbb{A}^1\_R\to \mathbb{A}^1\_R$ and $R$-sections $0,1:\mathrm{Spec} R\to \mathbb{A}^1\_R$ such that the ring axioms (phrased as diagrams) hold.] 


In order to speak about $\mathbb{A}^1\_R$ as a "module over $\mathbb{A}^1\_{\mathbb{C}}$" in some sense, you need to work in the category of schemes over $\mathbb{C}$, where $\mathbb{A}^1\_{\mathbb{C}}\to\mathrm{Spec } \mathbb{C}$ is a ring object.
With this notation, the $2$-dimensional ($\mathbb{A}^1\_{\mathbb{C}}\to \mathrm{Spec}\mathbb{C}$)-vector space object in schemes over $\mathbb{C}$ would be 
$$
\mathbb{A}^1\_{\mathbb{C}}\times\_{\mathrm{Spec}\mathbb{C}}\mathbb{A}^1\_{\mathbb{C}}=\mathbb{A}^2\_{\mathbb{C}}
$$
equipped with the evident morphism into $\mathrm{Spec}\mathbb{C}$.
In particular, it is not isomorphic to $\mathbb{A}^1\_R\to \mathrm{Spec}\mathbb{C}$ (adjoint to $x\mapsto (x,x):\mathbb{C}\to R[T]$) in the category of schemes over $\mathbb{C}$. Therefore, $\mathbb{A}^1\_R\to \mathrm{Spec} \mathbb{C}$ is not a 2-dimensional vector space over $\mathbb{A}^1\_{\mathbb{C}}\to \mathrm{Spec}\mathbb{C}$
(or a vector space of any dimension in the said category).


**A more advanced way** to see what is happening is by looking on:


* the sheaf on (the large Zariski site of) $\mathrm{Spec} R$ represented by $\mathbb{A}^1\_R\to \mathrm{Spec} R$,


compared to:


* the sheaf on $\mathrm{Spec} \mathbb{C}$ represented by $\mathbb{A}^1\_R\to \mathrm{Spec} \mathbb{C}$.


The first is easily seen to be naturally isomorphic to the sheaf of rings $\mathcal{O}\_{\mathrm{Spec} R}$ (it assigns $Y\to \mathrm{Spec R}$ to $\Gamma(Y,\mathcal{O}\_Y)$), which is essentially
the same as saying that $\mathbb{A}^1\_R\to \mathrm{Spec} R$ is a ring object in the category of $R$-schemes. 


The second one is just a sheaf of sets and does not posses any natural structure of an $\mathcal{O}\_{\mathbb{A}^1\_{\mathbb{C}}}$-module (which amounts to an ($\mathbb{A}^1\_{\mathbb{C}}\to \mathrm{Spec}\mathbb{C}$)-vector space structure on $\mathbb{A}^1\_R\to \mathrm{Spec}\mathbb{C}$). Rather, it is the set sheaf $\mathcal{O}\_{\mathbb{A}^1\_{\mathbb{C}}}\coprod \mathcal{O}\_{\mathbb{A}^1\_{\mathbb{C}}}$.


[**Edit:** You can obtain from $\mathbb{A}^1\_R$ a two-dimensional
"$\mathrm{A}^1\_\mathbb{C}$-vector space in the category of $\mathbb{C}$-schemes", i.e. $\mathbb{A}^2\_{\mathbb{C}}$, by applying *Weil restriction* relative
to $\mathrm{Spec} R\to \mathrm{Spec }\mathbb{C}$.]


