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)


1 Answer 1


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 $\mathrm{Spec} R\to \mathrm{Spec }\mathbb{C}$.]

  • $\begingroup$ Just to fully understand: in any event, $\mathbf{A}^1_R$ carries an action of $\mathbf{C}\times\mathbf{C}$. Am I right? $\endgroup$
    – user128448
    Sep 3, 2018 at 21:20
  • 1
    $\begingroup$ I would say no, but some data missing in your question. To properly discuss such a question, you need to view $(\mathrm{Spec}?)\,\mathbb{C}\times\mathbb{C}$ as a ring object in some category which also contains $\mathbb{A}^1_R$. Then you can ask about the existence of a meaningful action (in the categorical sense). What is true, though, is that $\mathbb{A}^1_R\to \mathrm{Spec} R$ (which is $\mathbb{A}^1_{\mathbb{C}\times \mathbb{C}}\to \mathrm{Spec} R$) acts on itself in the category of $R$-schemes. $\endgroup$ Sep 4, 2018 at 6:04
  • 1
    $\begingroup$ I would also add that the fact that $R$ is a commutative $\mathbb{C}$-algebra, does not mean that $\mathrm{Spec} R\to \mathrm{Spec}\mathbb{C}$ is a ring object in the category of $\mathbb{C}$-schemes. $\endgroup$ Sep 4, 2018 at 6:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy