Vector space objects in schemes - confusion 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)
 A: 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 $\mathrm{Spec} R\to \mathrm{Spec }\mathbb{C}$.]
