I am a little bit confused about two lemmas regarding Grothendieck's algebraization. Assume all algebras are defined over some field. Here is the short version of my question: Does Tag 09ZT ("Lemma 1") plus Tag 0A42 ("Lemma 2") imply that colimit of $Y_i$'s in Lemma 1 exists in the category of schemes under mild assumptions? (this would mean the construction of $Y$ in the proof of Lemma 1 does not depend on $X$ and is the colimit) If so what is the minimal requirement for that to happen.

The longer version:

Let $A$ be complete with respect to $I$. Let $S=\text{Spec}(A)$ and $S_n=\text{Spec}(A/I^n)$ Let $Y_1\rightarrow Y_2 \rightarrow Y_3 \rightarrow \ldots$ be an infinite sequence of morphism of schemes. Assume $Y_i$ has a structure of a scheme over $S_i$. Let $X$ be a scheme which receives morphisms from $Y_i$'s in a compatible manner. Because of the condition on $Y_i$'s we can deduce that $X\times S$ also receives same type of compatible morphisms from $Y_i$'s. This specially induces a diagram of commutative morphisms in the following form:

enter image description here

Assuming $X_i:=X\times S_i$ let's denote the maps from $Y_i$ to $X_i$ by $f_i$. Furthermore all $f_i$ are finite morphism and $Y_1\rightarrow S_1$ is proper then by Tag 09ZT, there is some scheme $Y$ such all the maps from $Y_i$ to $X\times S$ factors through $Y$ and $Y_i=Y\times S_i$.

Now I want to deduce that the scheme $Y$ does not depend on $X$ and is the colimit of the diagram of $Y_i$'s. Now assume the same situation, we are going to replace $X$ with $X'$ with exactly the same role. Assume $X'$ is receiving compatible morphisms from $Y_i$'s. Then we can form a compatible commutative diagram between $Y_i$ and $X'_i:=X'\times S_i$.

enter image description here

Applying Tag 0A42 (for some reason $X$ and $Y$ is switched in this lemma), implies that the map from $Y_i$'s to $X'\times S$ factors uniquely through $Y$. Then we can project $X'\times S$ on the first factor to get the same conclusion for $X'$. This means $Y$ is the colimit of the diagram of $Y_i$'s.

  • 1
    $\begingroup$ Yes, $Y$ is the colimit of the $Y_i$ in the category of separated, finite type $S$-schemes. However, in general you expect $Y$ only to exist as a formal scheme; the statement of the Grothendieck existence theorem is that a proper formal scheme with a finite morphism to a scheme is actually representable by a scheme. $\endgroup$ – R. van Dobben de Bruyn Mar 4 at 3:35
  • $\begingroup$ Amazing! can anything be said about the colimit in the category of $k$-schemes? ($k$ is the base field). I was thinking the same colimit should coincide with the colimit in the category of $k$-schemes. Since we can turn any $k$-scheme into a $S$ by simple tensoring and the later we can project onto the original factor. $\endgroup$ – user127776 Mar 4 at 4:05

Here's one way to see what's going on. I will use the Tannakian duality theorem of Hall and Rydh (see Theorem 1.1 here). It is stated for algebraic stacks, but if you replace the word "algebraic stack" with "scheme'' and forget about stabilizers, you're golden.

We will use your notation above. Let $T$ be an $S$-scheme (we will worry about finiteness conditions on $T$ later). Then $$\begin{eqnarray*}\text{Hom}(Y,T) &=& \text{Hom}(\text{Coh}(T),\text{Coh}(Y)) \\ &=& \text{Hom}(\text{Coh}(T),\varprojlim\text{Coh}(Y_n))\\ &=&\varprojlim \text{Hom}(\text{Coh}(T), \text{Coh}(Y_n)) \\ &=&\varprojlim \text{Hom}(Y_n, T).\end{eqnarray*}$$ The first line is Hall-Rydh. Since $Y$ is proper over $S$ (the spectrum of a complete Noetherian local ring), the second line is Grothendieck's existence theorem. The third line is from the definition of the inverse limit, and the fourth from Hall-Rydh again. This shows that we may define the direct limit $\varinjlim Y_n$ (in whatever category for the moment) to be $Y$.

Let us now worry about the relevant category of schemes/finiteness assumptions on $T$ that we must impose. First we check that the assumptions of Hall-Rydh are satisfied. The theorem assumes:

  1. $Y$ is locally excellent.
  2. $T$ is Noetherian.

Assumption (1) is automatically satisfied since $Y$ is finite type over a complete Noetherian local ring and thus is excellent. (2) is satisfied if we work in the category of finite type $S$-schemes (the category of Noetherian $S$-schemes is bad since it does not have fiber products).

Conclusion: The direct limit of the $Y_n$'s in the category of finite type $S$-schemes "is" $Y$.

P.S. The reader may be confused and think: "Wait a minute, the Hom scheme is only representable when the target scheme is separated. Doesn't this mean that the direct limit only exists in the category of finite type separated $S$-schemes?" Wrong! It turns out that the Hom scheme is representable even when the target scheme is not separated. This is Theorem 1.2 of the the same Hall-Rydh paper.

The usual representability theorems for Hom schemes assume separatedness of the target scheme for the reason that it has simply been tradition to check that formal deformations are algebraizable by a "graph trick." More precisely, we may realize a morphism of formal $\widehat{S}$-schemes $\widehat{T} \to \widehat{Y}$ as a closed formal subscheme of $\widehat{Y \times_S T}$ (crucially using separatedness!). Algebraizing the coherent ideal cutting out this closed formal subscheme gives the required morphism of $S$-schemes $Y \to T$.

My argument above shows that we can avoid all this ballyhoo with separatedness and instead invoke the deus ex machina of Tannakian duality. For more on why we may want to avoid separatedness, the reader may refer to the introduction here (specifically Sections 1.4 and 1.6).

Short answer: Stacks with infinite stabilizers are almost never separated.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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