TITLE: Can the homological dimension of a coherent sheaf explode along a formal deformation? (is the resolution property hereditary for formal deformations?)
QUESTION [6 upvotes]: Let $X_0$ be a locally noetherian scheme and $\mathcal{F}_0$ a coherent $\mathcal{O}_{X_0}$-module. Let $C$ be an artin ring with residue field $k$ and let $X \to Spec C$ be a (flat) deformation of $X_0$ over $C$ (meaning the closed fiber isomorphic to $X_0$). 

Definition 1: A deformation of $\mathcal{F}_0$ over $X$ consists of the following data:

A coherent $\mathcal{O}_X$-module $\mathcal{F}$ flat over $C$
An epimorphism $q:\mathcal{F} \to \mathcal{F}_0$ inducing isomorphism $\mathcal{F}\otimes_{\mathcal{O}_X
} \mathcal{O}_{X_0} \cong \mathcal{F}_0$

Definition 2: The homological dimension of a coherent sheaf $\mathcal{F}$ is the minimal length of coherent locally free resolutions of $\mathcal{F}$. If $\mathcal{F}$ doesn't have a coherent locally free resolution or all it's resolutions are infinite then define the $hd(\mathcal{F})=\infty$.

The basic question is:

Question: Suppose in the above situation $q:\mathcal{F} \to \mathcal{F}_0$ is a deformation of $\mathcal{F}_0$ which is of finite homological dimension $n < \infty$. Could $hd(\mathcal{F}) > hd(\mathcal{F}_0)$ ? In other words can homological dimension jump in a formal deformation?

In the affine case this is not possible:

Proposition: If $X_0$ is affine homological dimension can't jump.

Proof: Let $\mathcal{F} \to \mathcal{F}_0$ be a deformation of $\mathcal{F}_0$ over $X$. The kernel of $\mathcal{O}_X \to \mathcal{O}_{X_0}$ is nilpotent. By factoring $Speck \to SpecC$ into small extensions we may assume that the kernel $J$ of $\mathcal{O}_X \to \mathcal{O}_{X_0}$ is square zero where $X_0 \to SpecC_0$ is over an artin ring now and s.t. $J$ is annihilated by the maximal ideal of $C_0$. Then by the local criteria for flatness over noetherian rings we conclude that $\mathcal{F}$ sits in an exact sequence:
$$ 0 \to \mathcal{F}_0 \otimes_k J \to \mathcal{F} \to \mathcal{F}_0 \to 0$$
By the long exact sequence for the functor $Ext^{j}(-,Q)$ (with $Q$ arbitrary) we know that the $pd(\mathcal{F})$ is at most $pd(\mathcal{F}_0)$ and in fact they are equal since a resolution of $\mathcal{F}$ will give a resolution of $\mathcal{F}_0$. 
I will now use a somewhat strengthened version of the characterization of projective dimension which is a special case of what's proved here:

$(*)$ Over a noetherian ring projective dimension (defined by the vanishing of Ext groups) equals the
  minimal possible length of a resolution by finitely generated proejctives. In other words $hd=pd$.

We have therefore $hd(\mathcal{F})=hd(\mathcal{F}_0)$. Q.E.D.

In the non-affine case one can use the above proof to show that all deformations $\mathcal{F}$ that admit coherent locally free resolutions have the same homological dimension as $\mathcal{F}_0$. 
So in fact if everything I said until now is correct then either homological dimension is constant along a formal deformation or it explodes. Therefore the main problem that arises in the non-affine case is the following:

Not enough vector bundles: There might not exist any resolution of $\mathcal{F}$ by finitely generated locally free sheaves (even an infinite one). 

If the scheme $X$ has the resolution property (meaning every coherent sheaf is a quotient of a coherent locally free sheaf) then this problem disappears and we are left with the following questions:


Is the resolution property inherited by formal deformations? Suppose $X_0$ has the resolution property and $X$ is a deformation of $X_0$ over an artin ring $C$, does $X$ have the resolution property?


If $X$ doesn't have the resolution property everything seems to be stuck. Therefore I must ask the following imprecise question:


Is there a deformation theoretic way to detect the resolution property?

REPLY [8 votes]: You ask many questions.  I will answer the question that you labelled "question".  The way that you phrase the question is ambiguous.  When you write "finite homological dimension $n<\infty$", you do not specify whether you are assuming that $\mathcal{F}$ has finite homological dimension, or whether you are assuming that $\mathcal{F}_0$ has finite homological dimension.  I will answer both formulations.
 If $hd(\mathcal{F})$ is finite, then it equals $hd(\mathcal{F}_0)$. As you explain, if $\mathcal{F}$ has finite homological dimension, then the homological dimension of $\mathcal{F}$ equals the finite homological dimension of $\mathcal{F}_0$.  As you point out, the homological dimension of $\mathcal{F}$ is at least as large as the homological dimension of $\mathcal{F}_0$.  If the homological dimension of $\mathcal{F}_0$ equals $n$, then for every exact complex of locally free $\mathcal{O}_X$-modules, $$ \mathcal{E}_{n-1}\xrightarrow{d_{n-1}} \mathcal{E}_{n-2} \xrightarrow{d_{n-2}} \dots \xrightarrow{d_2} \mathcal{E}_1\xrightarrow{d_1} \mathcal{E}_0\xrightarrow{\eta} \mathcal{F}\to 0,$$ the kernel of $d_{n-1}$ is locally free.  This can be checked locally on open affines.  Then you can use the argument that you outline.  Thus, if the homological dimension of $\mathcal{F}$ is finite, then the homological dimension of $\mathcal{F}$ equals the homological dimension of $\mathcal{F}_0$.
On a nonseparated scheme, even if $hd(\mathcal{F}_0)$ is finite, $hd(\mathcal{F})$ may be infinite. There do exist examples where $\mathcal{F}_0$ has finite homological dimension, yet $\mathcal{F}$ has infinite homological dimension.  Let $C$ be $k[\epsilon]/\langle \epsilon^2 \rangle$.  Consider polynomial rings $A=C[a,b]$ and $R=C[r,s]$.  Denote by $\phi$ the isomorphism of $C$-algebras, $$ C[a,b]\leftrightarrow C[r,s], \ \ a\leftrightarrow r, \ \ b\leftrightarrow s.$$  This gives rise to a $C$-isomorphism $f$ between $\text{Spec}(A)$ and $\text{Spec}(R)$ that identifies the closed points $\langle \epsilon, a,b\rangle$ and $\langle \epsilon, r,s\rangle$.  Thus, $f$ identifies the open complements of these closed points, say $U=\text{Spec}(A)\setminus \langle \epsilon,a,b\rangle$ and $V=\text{Spec}(R)\setminus\langle \epsilon,r,s\rangle$.  Denote by $X$ the locally Noetherian (non-separated) flat $C$-scheme containing $\text{Spec}(A)$ and $\text{Spec}(R)$ as open subschemes obtained by glueing $U$ and $V$ via $f$.  Denote this common open by $O$.
Let $\mathcal{F}_0$ be the ideal sheaf whose restriction to $\text{Spec}(A/\epsilon A)$ equals $\widetilde{\langle a,b\rangle}$ and whose restriction to $\text{Spec}(R/\epsilon R)$ equals $\widetilde{\langle r,s\rangle}.$  This $\mathcal{O}_{X_0}$-module has homological dimension $1$.  Indeed, let $\eta:\mathcal{O}_{X_0}^{\oplus 2}\to \mathcal{F}_0$ be the homomorphism of coherent sheaves that on $\text{Spec}(A)$ equals $$\eta_A:(A/\epsilon A)^{\oplus 2} \to \langle a,b\rangle, \ \ (g,h) \mapsto g \cdot a + h \cdot b$$ and that on $\text{Spec}(R)$ equals $$\eta_R:(R/\epsilon R)^{\oplus 2} \to \langle r,s\rangle, \ \ (g,h) \mapsto g \cdot r + h \cdot s.$$  Then $\eta$ is surjective, and the kernel is a locally free sheaf of rank $1$.  Via the usual Koszul complex, this kernel is actually isomorphic to $\bigwedge^2_{\mathcal{O}_{X_0}}(\mathcal{O}_{X_0}^{\oplus 2}) \cong \mathcal{O}_{X_0}$.  At any rate, $\mathcal{F}_0$ has homological dimension $1$.
Now let $\mathcal{F}$ be the ideal sheaf in $\mathcal{O}_X$ whose restriction to $\text{Spec}(A)$ equals $\widetilde{\langle a,b\rangle}$ and whose restriction to $\text{Spec}(R)$ equals $\widetilde{\langle r,s-\epsilon \rangle}$.  Observe that the restriction of each of these ideal sheaves to $U$, resp. to $V$, is the full structure sheaf.  Thus, there is a glueing of these sheaves compatible with $f$.  So $\mathcal{F}$ is an $\mathcal{O}_X$-module.  Checking locally, $\mathcal{F}$ is a coherent $\mathcal{O}_X$-module that is $C$-flat.
I claim that there is no surjection from a locally free sheaf to $\mathcal{F}$.  First, there is an observation about trivializations of locally free sheaves on small opens that contain both $\langle \epsilon,a,b\rangle$ and $\langle \epsilon,r,s\rangle$.   For every open affine neighborhood $\text{Spec}(A[1/\alpha])$ of $\langle \epsilon,a,b\rangle$, then $\text{Spec}(R[1/\phi(\alpha)])$ is an open affine neighborhood of $\langle \epsilon,r,s\rangle$.  Thus, for every locally free $\mathcal{O}_X$-module, there exists $\alpha \in A\setminus \langle \epsilon,a,b\rangle$ such that the locally free sheaf is trivialized on $\text{Spec}(A[1/\alpha])$ and on $\text{Spec}(R[1/\phi(\alpha)])$.  Denote by $W$ the union of these two open affines.  By the $S2$ property, every section of $\mathcal{O}_X$ on $U\cap D(\alpha)$ extends to a section on $\text{Spec}(A[1/\alpha])$, and similarly for sections on $V\cap D(\phi(\alpha))$.  It follows that every automorphism of $\mathcal{O}_X^{\oplus m}$ on $U\cap D(\alpha)$ extends to an automorphism on $\text{Spec}(A[1/\alpha])$, and similarly for $V\cap D(\phi(\alpha))$.  Thus, the locally free sheaf on $W$ is isomorphic to $\mathcal{O}_W^{\oplus m}$ for some integer $m$.
Finally, let $\theta:\mathcal{O}_W^{\oplus m}\to \mathcal{F}|_W$ be a homomorphism of coherent sheaves.  The composition with the inclusion $\mathcal{F}_W\subset \mathcal{O}_W$ defines a homomorphism $\theta':\mathcal{O}_W^{\oplus m} \to \mathcal{O}_W$.  Again using the $S2$ property, it follows that $\theta'$ is uniquely determined by the restriction of $\theta'$ to the open $W\cap O$.  Thus, on the open $\text{Spec}(A)$, it follows that $\theta'_A:A^{\oplus m}\to A$ factors through both $\langle a,b\rangle$ and through $\langle \phi^{-1}(r),\phi^{-1}(s+\epsilon)\rangle = \langle a,b+\epsilon \rangle$.  Thus, $\theta'$ factors through $$\langle a,b\rangle \cap \langle a,b+\epsilon \rangle = \langle a, b^2,\epsilon b\rangle.$$  In particular, $\theta'$ is not surjective to $\langle a,b\rangle$.  This contradiction proves that $\mathcal{F}$ is not a quotient of a locally free $\mathcal{O}_X$-module.