# Direct limit, inverse limit and Spec

The set-up

1. $$k$$ is a field
2. $$T_i = \operatorname{Spec} A_i$$ is an inverse system of affine $$k$$-schemes, where $$i if $$\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).

• 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. Jul 16, 2021 at 1:13
• It's also not clear why you think math.stackexchange.com/a/241473 is relevant at all. Jul 16, 2021 at 1:18
• @EricWofsey Given the inclusion morphism $f_{ij}: \operatorname{Spec} A_j \to \operatorname{Spec} A_i$ what is the defintion of the map $\varphi_{ij}: A_i \to A_j$? Jul 16, 2021 at 1:55
• @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. Jul 16, 2021 at 1:58
• The spectrum commutes not with all colimits, yes, but: we have a filtered colimit, and there it works. Jul 16, 2021 at 7:31