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A is a Noetherian ring, B is an f.g. algebra over A, I is an ideal of A. let \$\hat B\$ be B's I-adic completion. Prove that \$\Omega^1_{\hat B/A}\$'s I-adic completion is isomorphic to \$\Omega^1_{B/A}\$'s I-adic completion.

This is an exercise from Liu's "Algebraic geometry and arithmetic curves", Exercise VI.1.3. It seems strange because according to Part(a) of that problem, there is an exact sequence involving these two objects. And if this is true, we must prove the first term in that exact sequence is actually zero under only Noetherian condition! I feel a bit puzzled, can anyone help me? Thanks!

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1 Answer 1

Yes $K_n=0$ in this situation ($A$ noetherian and $B$ essentially of finite type over $A$).

Hints: You need to know that $\Omega_{B/A}$ is a finitely generated $B$-module and you can use the canonical bijection 6.1.5 for both $B$ and $\hat{B}$ and with $M=\Omega_{B/A}\otimes_B \hat{B}$ to get a canonical map $\Omega_{\hat{B}/A}\to \Omega_{B/A}\otimes_B \hat{B}$...

EDIT: Let me detail a little more. Recall $\hat{B}$ is the formal completion of $B$ with respect to the $I$-adic topology for some ideal $I$ of $B$, and $B$ is essentially of finite type over $A$ (localization of a finitely generated $A$-algebra).

  1. Let $M$ be a finitely generated $\hat{B}$-module. Any $A$-derivation $B\to M$ induces a $A$-derivation $B/I^n \to M/I^nM$ for all $n\ge 1$. Passing to the limit, we get an $A$-derivation $\hat{B}\to \hat{M}=M$ whose restricton to $B$ is just the $A$-derviation we start with. Theorefore the natural (restriction) map ${\rm Der}_A(\hat{B}, M)\to {\rm Der}_A(B, M)$ has a section (it is therefore surjective).

  2. Applying the above result to $M= \Omega_{B/A}\otimes_B \hat{B}$, the canonical map $$Hom_{\hat{B}}(\Omega_{\hat{B}/A}, M)\to Hom_A(\Omega_{B/A}, M)$$ has a section $\sigma$. Consider $\psi: \Omega_{B/A}\to M$, $\omega\mapsto \omega\otimes 1$ and $\psi=\sigma(\phi) : \Omega_{\hat{B}/A}\to \Omega_{B/A}\otimes_B \hat{B}$. Now tensoring by $B/I^n=\hat{B}/I^n\hat{B}$, we see that the natural map $$\Omega_{B/A}\otimes_B B/I^n \to \Omega_{\hat{B}/A}\otimes_{\hat{B}} \hat{B}/I^n\hat{B}$$ is injective. This is an isomorphism by Part (a) of the exercise.

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That's pretty awesome. –  Harry Gindi Jan 24 '10 at 1:37
    
I imagine the notation $K^n$ is explained in the book? –  Mariano Suárez-Alvarez Jan 25 '10 at 0:56
    
Mariano: sorry I just reproduced the notation of Part(a) of the exercise. $K_n$ is the kernel of the natural map $\Omega_{B/A}\otimes B/I^n\to \Omega_{\hat{B}/A}\otimes \hat{B}/I^n\hat{B}$. Part (a) sayes that map is surjective and $K_{n+1}\to K_n$ too. –  Qing Liu Jan 25 '10 at 22:52
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