# T^i functors are isomorphic for analytically isomorphic isolated singular points

I've been having trouble proving the following: Let $B$ and $B'$ be local rings, essentially of finite type over $k$, having isolated singularities at the closed points. Suppose that they are analytically isomorphic, i.e. have isomorphic completions, then the deformation theory modules $T^i_{B/k}$ and $T^i_{B'/k}$ are isomorphic for $i=1,2$ as modules over the completions of $B$ and $B'$.

A good answer would only use the definitions and basic properties of these functors, and/or would give a reference where this is worked out in some amount of detail, i.e. not where it is just stated as a fact.

Thanks!

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What do you mean by the "deformation theory modules" ? Do you mean the first homology groups of the cotangent complex ? (as described in Lichtenbaum-Schlessinger, perhaps) – Damian Rössler Aug 11 '12 at 7:58
Yes. I didn't think there were other T^i modules out there, but in case there were, I wanted to clarify that I meant the ones discussed in deformation theory, and first defined in the paper of L-S. Would you like me to edit the question? – HNuer Aug 12 '12 at 1:53

Here is a stab at an answer, but it is incomplete.

Let $S:={\rm Spec}\, B$ and let $\widehat{S}:={\rm Spec}\,\widehat{B}$ be the completion of $B$ along its maximal ideal $m$. Let $\phi:\widehat{S}\to S$ be the natural morphism (which is faithfully flat). The composition of morphisms $\widehat{S}\to S\to{\rm Spec}\,k$ gives rise to a triangle of cotangent complexes and hence to an exact sequence $$\dots\to T^1_{\phi}\to T^1_{\widehat{S}/k}\to \phi^*T^1_{S/k}\to T^2_{\phi}\to T^2_{\widehat{S}/k}\to \phi^*T^2_{S/k}\to T^3_\phi\to\dots {\rm (*)}$$ so what you need to show is that $T^i_\phi$ vanishes for $i=1,2,3$. You would then get isomorphisms $T^i_{\widehat{S}/k}\to \phi^*T^i_{S/k}$ and since you can repeat this for $B'$ instead of $B$, you would get the required isomorphism.

Let $L_\phi$ be the cotangent complex of $\widehat{S}/S$. This complex $L_\phi$ is concentrated in degree $0$. A quick way to see this is to notice that

• the formation of the cotangent complex is compatible with direct limits of rings (see Quillen, "Cohomology of commutative rings", eq. (4.11))

• the ring $B$ is excellent because it is essentially of finite type over a field (Grothendieck) and thus the fibres of $\phi$ are geometrically regular;

• a (deep...) result of Popescu (see for instance Th. 1.3 in "Approximations of versal deformations", by B. Conrad and AJ de Jong) then implies that $\widehat{B}$ is a direct limit of smooth $B$-algebras, and for the latter the cotangent complex is clearly concentrated at $0$.

By considering the sequence analogous to (*) for the modules $T_i$ instead of $T^i$, this proves the analog of your assertion for the $T_i$. Furthermore, we see that $T^i_\phi={\rm Ext}^i(L_\phi,\widehat{B})={\rm Ext}^i(\Omega_\phi,\widehat{B})$.

So the issue is to show that ${\rm Ext}^i(\Omega_\phi,\widehat{B})=0$ for $i>0$. Maybe your hypothesis on the fact that the singularity of $S$ is isolated at the closed point plays a role here.

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For the last part, one has a much more general statement: $\mathrm{RHom}(L_\phi,N) = 0$ for any $m$-adically complete $\widehat{B}$-module $N$. Indeed, by completeness (and commuting $\mathrm{R}\lim$ with $\mathrm{RHom}$ appropriately), it suffices to show the claim for $N$ killed by some power $m^n$ of $m$. By adjunction, it suffices to show that $L_\phi \otimes_{\widehat{B}} \widehat{B}/m^n \simeq 0$, but this follows from the base change for the cotangent complex of $\phi$ together with fact that $\phi \mod m^n$ is an isomorphism. This also shows that excellence is not necessary. – anon Sep 19 '12 at 0:22
Thank you very much for this comment. So the answer to Hnuer's question is positive in general (without any assumption on singularities). Could you expand on your parenthetical "... and commuting with RLim and RHom appropriately..." ? – Damian Rössler Sep 19 '12 at 23:33
I mean the derived generalisation of the defining property of the limit: $\mathrm{Hom}(M,\lim_k N_k) \simeq \lim_k \mathrm{Hom}(M,N_k)$. Fixing an $M$ gives two isomorphic left-exact functors out of an abelian category of projective systems, so deriving the isomorphism gives an isomorphism $\mathrm{RHom}(M,\mathrm{R}\lim -) \simeq \mathrm{R}\lim \mathrm{RHom}(M,-)$ of derived functors on projective systems. – anon Sep 20 '12 at 2:05