108.69 A counter example to Grothendieck's existence theorem
Let $k$ be a field and let $A = k[[t]]$. Let $X$ be the glueing of $U = \mathop{\mathrm{Spec}}(A[x])$ and $V = \mathop{\mathrm{Spec}}(A[y])$ by the identification
\[ U \setminus \{ 0_ U\} \longrightarrow V \setminus \{ 0_ V\} \]
sending $x$ to $y$ where $0_ U \in U$ and $O_ V \in V$ are the points corresponding to the maximal ideals $(x, t)$ and $(y, t)$. Set $A_ n = A/(t^ n)$ and set $X_ n = X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A_ n)$. Let $\mathcal{F}_ n$ be the coherent sheaf on $X_ n$ corresponding to the $A_ n[x]$-module $A_ n[x]/(x) \cong A_ n$ and the $A_ n[y]$ module $0$ with obvious glueing. Let $\mathcal{I} \subset \mathcal{O}_ X$ be the sheaf of ideals generate by $t$. Then $(\mathcal{F}_ n)$ is an object of the category $\textit{Coh}_{\text{support proper over } A}(X, \mathcal{I})$ defined in Cohomology of Schemes, Section 30.27. On the other hand, this object is not in the image of the functor Cohomology of Schemes, Equation (30.27.0.1). Namely, if it where there would be a finite $A[x]$-module $M$, a finite $A[y]$-module $N$ and an isomorphism $M[1/t] \cong N[1/t]$ such that $M/t^ nM \cong A_ n[x]/(x)$ and $N/t^ nN = 0$ for all $n$. It is easy to see that this is impossible.
Lemma 108.69.1. Counter examples to algebraization of coherent sheaves.
Grothendieck's existence theorem as stated in Cohomology of Schemes, Theorem 30.27.1 is false if we drop the assumption that $X \to \mathop{\mathrm{Spec}}(A)$ is separated.
The stack of coherent sheaves $\mathcal{C}\! \mathit{oh}_{X/B}$ of Quot, Theorems 97.6.1 and 97.5.12 is in general not algebraic if we drop the assumption that $X \to S$ is separated
The functor $\mathrm{Quot}_{\mathcal{F}/X/B}$ of Quot, Proposition 97.8.4 is not an algebraic space in general if we drop the assumption that $X \to B$ is separated.
Proof.
Part (1) we saw above. This shows that $\textit{Coh}_{X/A}$ fails axiom [4] of Artin's Axioms, Section 96.14. Hence it cannot be an algebraic stack by Artin's Axioms, Lemma 96.9.5. In this way we see that (2) is true. To see (3), note that there are compatible surjections $\mathcal{O}_{X_ n} \to \mathcal{F}_ n$ for all $n$. Thus we see that $\mathrm{Quot}_{\mathcal{O}_ X/X/A}$ fails axiom [4] and we see that (3) is true as before.
$\square$
Comments (0)