9
$\begingroup$

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 $\mathbb{Z_{\geq0}}$

$0=\operatorname{Spec(\mathcal{O}_{X}/I^{0})}\hookrightarrow\ldots\hookrightarrow\operatorname{Spec(\mathcal{O}_{X}/I^{n-1})}\hookrightarrow\operatorname{Spec(\mathcal{O}_{X}/I^{n})}\hookrightarrow\ldots$

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 $\operatorname{Spec}(\mathbb{Z}[\![ x ]\!])(\mathbb{Z})\neq\operatorname{Nil}(\mathbb{Z})$.

$\endgroup$
2
  • 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
    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". $\endgroup$ Jan 24, 2020 at 20:33

1 Answer 1

10
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.