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The set-up

  1. $k$ is a field
  2. $T_i = \operatorname{Spec} A_i$ is an inverse system of affine $k$-schemes, where $i<j$ if $\operatorname{Spec} A_j \subset \operatorname{Spec} A_i$ (inclusion).
  3. $X$ is a $k$-variety (hence locally of finite presentation).

Then, by this porposition, we have

$$ \operatorname{Mor}(\varprojlim_i T_i, X) = \varinjlim_i \operatorname{Mor}(T_i, X)$$

The confusion

In the proof, we use the direct limit $A = \varinjlim_i A_i$ and then use it to create the inverse system $T_i$ using the fact that $\varphi : A \to B$ induces $\operatorname{Spec} \phi: \operatorname{Spec} B \to \operatorname{Spec} A$ via contraction. This should work sice the functor between rings and schemes is contravariant. However, the converse relation isn't true: spec of inverse limit of rings is not same as direct limit of spec of rings.

The question

How can we get the direct system $(A_i, \varphi_{ij})$ from the above inverse system $(T_i, f_{ij})$? In particular, what is the meaning of this statement

$$\varprojlim_i \operatorname{Spec} A_i = \operatorname{Spec} \varinjlim_i A_i$$

Edit

I have realized that this was a dumb question. The scheme morphism $\operatorname{Spec} A \to \operatorname{Spec} B$ by definition gives the ring homomorphism $B \to A$. Moreover, as commented below by Martin Brandenburg and Zhen Lin on the linked question, spectrum commutes with filtered colimits (direct limit).

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  • $\begingroup$ It's really hard to tell what you do and don't understand here and what you are asking for an explanation of. How you get the direct system is completely trivial: a morphism $f_{ij}:\operatorname{Spec}A_j\to\operatorname{Spec} A_i$ of schemes gives you a morphism $\varphi_{ij}:A_i\to A_j$ of rings. $\endgroup$ Jul 16, 2021 at 1:13
  • $\begingroup$ It's also not clear why you think math.stackexchange.com/a/241473 is relevant at all. $\endgroup$ Jul 16, 2021 at 1:18
  • $\begingroup$ @EricWofsey Given the inclusion morphism $f_{ij}: \operatorname{Spec} A_j \to \operatorname{Spec} A_i$ what is the defintion of the map $\varphi_{ij}: A_i \to A_j$? $\endgroup$ Jul 16, 2021 at 1:55
  • $\begingroup$ @EricWofsey I linked to the other answer since I think it says that $\varinjlim_i \operatorname{Spec} A_i \neq \operatorname{Spec} \varprojlim_i A_i$, something I would have expected to hold due to contravarience. $\endgroup$ Jul 16, 2021 at 1:58
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    $\begingroup$ The spectrum commutes not with all colimits, yes, but: we have a filtered colimit, and there it works. $\endgroup$ Jul 16, 2021 at 7:31

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