# Can the homological dimension of a coherent sheaf explode along a formal deformation? (is the resolution property hereditary for formal deformations?)

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:

1. 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:

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

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.

• What a great answer! Thank you! Regarding the resolution property, do you happen to know if it's inherited by deformations? – Saal Hardali May 15 '17 at 8:21
• I suspect that the resolution property is not inherited by deformations, but I believe that this is an open problem already for proper morphisms. In Hartshorne's "Algebraic Geometry", Exercise III.5.9 suggests that already nonreduced double structures can destroy the resolution property. That exercise proves that there do not exist resolutions by direct sums of invertible sheaves. However, higher rank locally free sheaves are more difficult to understand. – Jason Starr May 15 '17 at 9:18