Notation and Setting: let $\operatorname{Aff}$ denote the category of affine schemes whose objects are covariant representable functors $\operatorname{X}:\operatorname{Ring}\rightarrow\operatorname{Set}$ and $\operatorname{Spec}:\operatorname{Ring^{op}}\rightarrow\operatorname{Func(Ring, Set)} $ be the contravariant Yoneda embedding of $\operatorname{Ring^{op}}$ in its category of presheaves so that $\operatorname{Aff}\simeq\operatorname{Ring^{op}}$.

In addition, let $\mathcal{O}:\operatorname{Func(Ring, Set)}\rightarrow\operatorname{Ring^{op}}$ be the functor that sends a functor $\operatorname{X}:\operatorname{Ring}\rightarrow\operatorname{Set}$ to the ring of maps $\operatorname{X}\rightarrow \mathbb{A}^1$ (where $\mathbb{A}^1$ is the forgetful functor) so that $\operatorname{Spec}$ and $\mathcal{O}$ are inverse of one another.

Let $\widehat{\operatorname{Aff}}$ be the indization of $\operatorname{Aff}$, i.e. the category whose objects are functors $\operatorname{X}:\operatorname{Ring}\rightarrow\operatorname{Set}$ that are small filtered colimits of affine schemes.

My question: I am looking for (simple) examples of functors which are objects of $\widehat{\operatorname{Aff}}$ but are not affine schemes.

I am particularly interested in examples of the following form: Let $\operatorname{X}$ be an affine scheme, $I\subseteq\mathcal{O}_{X}$ an ideal and consider the following diagram in $\operatorname{Func(Ring, Set)}$ over $\mathbb{Z_{\geq0}}$


Since $\operatorname{Func(Ring, Set)}$ admits small colimits, $\mathrm{colimit}_{n\in\mathbb{Z}_{\geq0}}\operatorname{Spec(\mathcal{O}_{X}/I^{n})}$ exists. Thus I am looking for examples of affine schemes $\operatorname{X}$ and ideals $I\subseteq\mathcal{O}_{X}$ for which $(\mathrm{colimit}_{n\in\mathbb{Z}_{\geq0}}\operatorname{Spec(\mathcal{O}_{X}/I^{n})})\neq \mathrm{colimit}_{n\in\mathbb{Z}_{\geq0}}(\operatorname{Spec(\mathcal{O}_{X}/I^{n})})$

The only example that I could find so far was that of $\operatorname{Spec(\mathbb{Z}[x])}$ and the ideal $(x)$ which give the functor $\operatorname{Nil}\simeq\mathrm{colimit}_{n\in\mathbb{Z}_{\geq0}}\operatorname{Spec(\mathbb{Z}[x]/(x)^{n})}\neq(\mathrm{colimit}_{n\in\mathbb{Z}_{\geq0}}\operatorname{Spec(\mathbb{Z}[x]/(x)^{n})})\simeq\operatorname{Spec}(\mathbb{Z}[\![ x ]\!])$.

To see that you can either show that $\operatorname{Nil}$ is not representable1 or check, for example, that $\operatorname{Spec}(\mathbb{Z}[\![ x ]\!])(\mathbb{Z})\neq\operatorname{Nil}(\mathbb{Z})$.

  • 2
    $\begingroup$ It looks like you are asking about formal schemes. Your example Nil is the formal spectrum $\operatorname{Spf}\mathbb{Z}[[x]]$. They are discussed at the end of EGA1. $\endgroup$
    – S. Carnahan
    Commented Jan 24, 2020 at 12:59
  • 2
    $\begingroup$ I don't think that the question is really about formal schemes (since here no "global space" is given). A better keyword is "ind-schemes". $\endgroup$ Commented Jan 24, 2020 at 20:33

1 Answer 1


A basic standard example is the colimit of $\mathbb{A}^0 \to \mathbb{A}^1 \to \mathbb{A}^2 \to \cdots$ with transition maps $x \mapsto (x,0)$. The $R$-valued points are finite sequences in $R$. This functor is not representable.

More generally, let $A$ be a commutative ring with a sequence of ideals $I_0 \supseteq I_1 \supseteq I_2 \supseteq \cdots$. (In the mentioned example, $A = \mathbf{Z}[x_0,x_1,\dotsc]$ and $I_n = \langle x_n,x_{n+1},\dotsc\rangle$.) Then the colimit $X$ of $\mathrm{Spec}(A/I_0) \to \mathrm{Spec}(A/I_1) \to \cdots$ is the subfunctor of $\mathrm{Spec}(A)$ whose $R$-valued points are those $A \to R$ whose kernel contains some $I_n$. We have $\mathcal{O}(X) = \lim_n A/I_n =: \widehat{A}$.

Then $X$ is representable aka affine iff the canonical morphism $X \to \mathrm{Spec}(\widehat{A})$ is an isomorphism. It is injective anyway, and it is surjective on $R$-valued points iff every homomorphism $\widehat{A} \to R$ factors through some projection $\widehat{A} \to A/I_n$. So $X$ is representable iff the identity $\widehat{A} \to \widehat{A}$ factors through some projection $\widehat{A} \to A/I_n$. But this clearly implies $I_n = I_{n+1} = \dotsc$ and the sequence is stationary.

Conversely this means that for every non-stationary sequence $I_0 \supseteq I_1 \supseteq I_2 \supseteq \cdots$ the functor $X$ is not representable.


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