Take $X = \mathbb{A}^1$ and $Y = \{0\}$. I want to take the formal group scheme at $Y \subset X$. This is a locally ringed space, $(Y, \mathcal{O}_{ \hat{X}})$ where $\mathcal{O}_{\hat{X}}$ is the $(x)$-adic completion of $k[x]$, i.e. $k[[x]]$.

This might be vague, but why formal schemes, i.e. what is the difference between this formal scheme and $Spec (k[[x]])$? For example, is the category of coherent sheaves (defined on any ringed space) equivalent for the two? I suppose the underlying topological space is different, but why would you prefer one over the other?

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    $\begingroup$ It's something like the difference between a module for $k [[x]]$ as an ordinary ring and a module for $k [[x]]$ as a topological ring. $\endgroup$
    – Zhen Lin
    Sep 17, 2014 at 10:25

1 Answer 1


Here's an answer that's somewhat what I'm looking for. First, it seems that they have equivalent categories of coherent sheaves, given by Lemma 2.2 in the following reference.

However there is a difference in the functor of points, namely, considering $\hat{\mathbb{G}}_a(R)$ as a direct limit, one wants $$\lim Hom(k[t]/t^i, R)$$ which is exactly the set of nilpotent elements of $R$. In particular, $k[[t]]$ has no nilpotent elements, but $$Hom(k[[t]], k[[t]])$$ is nonempty.

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    $\begingroup$ More generally, the category of affine formal schemes is anti-equivalent to the category of topological rings (with some condition, e.g. admissible) with morphisms being continuous homomorphisms. For example, any continuous hom $k[[t]] \to k[[x]]$ is determined by where it sends $t$, and for this map to be continuous we only need the image of $t$ to be topologically nilpotent. $\endgroup$
    – ggg
    Sep 6, 2019 at 15:00

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