Consider a projective morphism of Noetherian schemes $p:X\to \mathrm{Spec}(A)$. Let $\mathcal{E},\mathcal{F}$ be coherent $\mathcal{O}\_X$-modules flat over $A$. For every (Noetherian) ring map $A\to B$, denote $X\_B:=X\times\_{\mathrm{Spec}(A)}\mathrm{Spec}(B)$, $\mathcal{E}\_B:=(X\_B\to X)^\*\mathcal{E}$, and $p\_B:X\_B\to\mathrm{Spec}(B)$.


Choose $i\in\mathbb{Z}$. We say $\mathrm{Ext}^i\_p(\mathcal{E},\mathcal{F})$ commutes with base change if for all $A\to B$, the base change map is an isomorphism
$$\tau^i:\mathrm{Ext}^i\_p(\mathcal{E},\mathcal{F})\otimes\_AB\xrightarrow{\cong} \mathrm{Ext}^i\_{p\_B}(\mathcal{E}\_B,\mathcal{F}\_B). $$


I have the following conjecture.



> 
> If $\mathrm{Ext}^i\_p(\mathcal{E},\mathcal{F})$ commutes with base change, then there exists a finitely generated $A$-module $Q$ such that $\mathrm{Ext}^{i+1}\_p(\mathcal{E}\_B,\mathcal{F}\_B)\cong \mathrm{Hom}\_A(Q,B)$ functorial for $A\to B$.
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> 
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Is this conjecture ture? If yes, then can we always get a module $Q$ and a natural transformation
$$\mathrm{Hom}\_A(\mathcal{Q},B)\to \mathrm{Ext}\_p^{i+1}(\mathcal{E}\_B,\mathcal{F}\_B)$$
without the base change assumption on $\mathrm{Ext}^i$.


Similar results hold true for sheaf cohomologies, derived functors of $\Gamma(X,-)$, which can be found in Section III.12 in *Algebraic Geometry* by Hartshorne. However, it is not clear to me whether the method can deform to one argument for $\mathrm{Ext}^i\_p$.



If $\mathrm{Ext}^i(E,F)$ commutes with base change, then is $\mathrm{Ext}^{i+1}(E,F)$ representable?
**Edit.** I started this answer earlier, but then I had to do something else.


**Positive answer.**
Without any further hypothesis, the answer is positive if $i$ equals $0$. This is one of the theorems the follows from the "Exchange Property" in Section 7.7 of EGA III\_2. If $i$ equals $1$, at least there is a well-defined functor: because $\mathcal{E}$ is $A$-flat, for every short exact sequence in the Yoneda-Ext group $\text{Ext}^1\_{\mathcal{O}\_{X}}(\mathcal{E},\mathcal{F})$, the base change to $B$ is still a short exact sequence.


When $X$ is $A$-flat, there are well-defined functors for every $i$. Moreover, there is a bounded below complex whose terms are finitely generated, locally free $A$-modules whose homology modules compute the Ext groups compatibly with arbitrary base change.
Since $p$ is projective, there exists a $p$-ample invertible sheaf $\mathcal{O}\_X(1)$. Thus, there exists a bounded above resolution of $\mathcal{E}$ by $\mathcal{O}\_X$-modules that are finite direct sums of twists $\mathcal{O}\_X(-d)$ with $d$ so positive that $\mathcal{F}(d)$ has vanishing higher direct image sheaves for $p$. Thus, we can compute $R\textit{Hom}\_{\mathcal{O}\_X}(\mathcal{E},\mathcal{F})$ by forming the usual sheaf Hom of this resolution into $\mathcal{F}$. By construction, this new bounded below complex has terms that are finite direct sums of coherent sheaves $\mathcal{F}(d)$ having vanishing higher direct image sheaves. Thus, $Rp\_\*$ is just $p\_\*$. Altogether, this gives a bounded below complex of finite free $A$-modules whose homologies compute the Ext groups, compatible with arbitrary base change (since everything, including the sheaves $\mathcal{O}\_X(-d)$, is $A$-flat).


**Negative answer.**
For $i\geq 2$, the conjecture is false without further hypotheses, e.g., flatness of $p$. If you assume that $p$ is flat, then I believe the conjecture is true (I will try to return to this soon).


Consider the case where $A$ equals $k[s]$, where $X$ is the effective Cartier divisor of $\mathbb{P}^3\_A = \text{Proj}\ A[T,U,V,W]$ with defining equation $s(T^3+U^3+V^3+W^3)$, where $\mathcal{E}$ equals the structure sheaf of the closed subscheme $C$ of $X$ with defining ideal $\langle T+U,V+W \rangle$, and where $\mathcal{F}$ equals the pushforward to $X$ of the dualizing sheaf of $C$. For $i=0$, the formation of $\text{Hom}\_{\mathcal{O}\_X}(\mathcal{E},\mathcal{F}) = H^0(C,\omega\_C)$ is compatible with arbitrary base change: it is always zero since $C$ is just $\mathbb{P}^1\_A$ and $\omega\_C$ is just $\mathcal{O}\_{\mathbb{P}^1}(-2)$.


The flattening stratification of $X$ is the locally closed partition of $\text{Spec}(A)$ into two subsets: the basic open subset $D(s) = \text{Spec}(A[1/s])$ and the closed point $\text{Zero}(s) = \text{Spec}(A/sA)$. Over each of these locally closed subschemes, the restriction of $X$ is smooth: either a smooth cubic surface over $D(s)$ or all of $\mathbb{P}^2$ over $\text{Zero}(s)$. Moreover, the curve $C$ is smooth over $\text{Spec}(A)$; just a line. Thus, over each of these locally closed subschemes, the curve $C$ is a "regular embedding" in $X$ with locally free normal sheaf $N\_{C/X}$ (the sheaf Hom into $\mathcal{O}\_C$ of the ideal sheaf of $C$ in $X$). Over $D(s)$, the locally free normal sheaf $N\_{C/X}$ is just $\mathcal{O}\_{\mathbb{P}^1}(-1)$. Over $\text{Zero}(s)$, the locally free normal sheaf $N\_{C/X}$ is just $\mathcal{O}\_{\mathbb{P}^1}(+1)^{\oplus 2}$.


In fact, in this example $C$ is a transverse intersection of divisors: it is a divisor in $X$ over $D(s)$, and it is an intersection of two hyperplanes in $X$ over $\text{Zero}(s)$. This means that the object $R\textit{Hom}\_{\mathcal{O}\_X}(\mathcal{O}\_C,\mathcal{F})$ is particularly simple when restricted over $D(s)$ and over $\text{Zero}(s)$: it is quasi-isomorphic to a complex that is simply the direct sum of its homology sheaves (i.e., the differentials are all zero), and these homology sheaves are just $\mathcal{H}^q = \bigwedge^q\_{\mathcal{O}\_C}(N\_{C/X})\otimes\_{\mathcal{O}\_C} \mathcal{F}$. Thus, $Rp\_\*$ (over each stratum) is also just the direct sum of the complexes $Rp\_\*(\bigwedge^q\_{\mathcal{O}\_C}(N\_{C/X})\otimes\_{\mathcal{O}\_C} \mathcal{F})[-q]$. That makes it particularly easy to compute the ranks of the Ext groups over each stratum.


More precisely, over $D(s)$, the complex $R\textit{Hom}\_{\mathcal{O}\_X}(\mathcal{E},\mathcal{F})$ is just $\mathcal{O}\_{\mathbb{P}^1}(-2)[0]\oplus \mathcal{O}\_{\mathbb{P}^1}(-3)[-1]$. Thus $Rp\_\*R\textit{Hom}\_{\mathcal{O}\_X}(\mathcal{E},\mathcal{F})$ is just $H^1(\mathbb{P}^1,\mathcal{O}\_{\mathbb{P}^1}(-2))[-1]\oplus H^1(\mathbb{P}^1,\mathcal{O}\_{\mathbb{P}^1}(-3))[-2]$.


Similarly, over $\text{Zero}(s)$, the complex $R\textit{Hom}\_{\mathcal{O}\_X}(\mathcal{E},\mathcal{F})$ is just $\mathcal{O}\_{\mathbb{P}^1}(-2)[0]\oplus \mathcal{O}\_{\mathbb{P}^1}(-1)^{\oplus 2}[-1]\oplus \mathcal{O}\_{\mathbb{P}^1}(0)[-2]$. Thus $Rp\_\*R\textit{Hom}\_{\mathcal{O}\_X}(\mathcal{E},\mathcal{F})$ is just $H^1(\mathbb{P}^1,\mathcal{O}\_{\mathbb{P}^1}(-2))[-1]\oplus H^0(\mathbb{P}^1,\mathcal{O}\_{\mathbb{P}^1})[-2]$.


Now consider the functor $\text{Ext}^1\_{p\_B}(\mathcal{E},\mathcal{F})$. For the base change to $A[1/s]$, it is representable by a locally free sheaf $Q\_{1/s}$ of rank $1$. Also, for the base change to $A/sA$, it is representable by a locally free sheaf of rank $1$. So if your conjecture is correct, then the functor is representable by a locally free sheaf $Q$ of rank $1$.


But then the same hypotheses apply for the functor $\text{Ext}^2\_{p\_B}(\mathcal{E},\mathcal{F})$. However, now the base change to $A[1/s]$ is representable by a locally free sheaf of rank $2$, whereas the base change to $A/sA$ is representable by a locally free sheaf of rank $1$. There is no finitely generated $A$-module $R$ such that $R\otimes\_A A[1/s]$ is free of rank $2$ yet $R/sR$ is free of rank $1$.


