I'm reading this paper of A. Vistoli and I have some questions about the discussion in page 5. This is the context (If you don't want to download the paper):

Let $A'$ be a noetherian local ring with maximal ideal $\mathfrak{m}_{A'}$ and residue field $\kappa=A'/\mathfrak{m}_{A'}$, and $\mathfrak{a}\subset A'$ an ideal such that $\mathfrak{m}_{A'}\mathfrak{a}=0$. We set $A=A'/\mathfrak{a}$.

Question 1: Is there a geometrical meaning for the condition $\mathfrak{m}_{A'}\mathfrak{a}=0$?

Let $M'$ be a flat scheme of finite type over $A'$ and let $X$ be a complete intersection closed subcheme of $M=M'|_{\operatorname{Spec} A}$, hence the ideal $\mathcal{I}$ of $X$ in $M$ is generated by a regular sequence $f_1,\ldots,f_r$ in $\mathcal{O}_M$. Assume also that $X$ is flat over $A$.

Now consider a lifting $X'$ of $X$, i.e., a subscheme $X'$ of $M'$ which is flat over $A'$ and such that $X=X'\cap M$. It was shown previously in the paper the following (Lemma 2.6):

The scheme $X'$ is flat over $A'$ if and only if the natural surjective map $\mathcal{I}'/\mathfrak{a}\mathcal{I}'\to \mathcal{I}$ is an isomorphism, where $\mathcal{I}'$ is the ideal of $X'$ in $M'$ (satisfying $\mathcal{I}'\mathcal{O}_M=\mathcal{I}$).

Choose liftings $f_1',\ldots,f_r'$ of $f_1,\ldots,f_r$ to $\mathcal{I}'$. He claims: from the equality $\mathcal{I}'/\mathfrak{a}\mathcal{I}'= \mathcal{I}$ and the fact that the ideal $\mathfrak{a}$ is nilpotent (since $\mathfrak{a}^2 \subset \mathfrak{m}_{A'}\mathfrak{a}=0$ ) we conclude that $f_1',\ldots,f_r'$ generate $\mathcal{I}'$.

Question 2: Where we use the fact that $\mathfrak{a}$ is nilpotent?

I ask because if we define $\mathcal{J}\subset \mathcal{I}'$ to be the ideal generated by the $f_1',\ldots,f_r'$, then for any $f\in \mathcal{I}'$ we can consider its image in $\mathcal{I}$ and consider a lifting $g\in \mathcal{J}$, hence we get from the equality $\mathcal{I}'/\mathfrak{a}\mathcal{I}'= \mathcal{I}$ that $f-g\in \mathfrak{a}\mathcal{I}'$. Therefore $\mathcal{I}'=\mathfrak{a}\mathcal{I}'+\mathcal{J}$ with $\mathfrak{a}\subset \mathfrak{m}_{A'}$, so by Nakayama's lemma we obtain $\mathcal{J}=\mathcal{I}'$.

We want to prove that $f_1',\ldots,f_r'$ is a regular sequence in $\mathcal{O}_{M'}$. For that he will consider the Koszul complex $\mathcal{K}_\bullet'$ of $f_1',\ldots,f_r'$. At this step I'm following the Lecture Notes of Joseph Le Potier (Chapter 3, Section 6.1, page 189) and the book of Fulton and Lang "Riemann-Roch Algebra" (Chapter 4, Section 2, pages 76-77).

We have the following Koszul resolution of $\mathcal{O}_{X'}$ (by flat sheaves over $A'$) associated to $f_1',\ldots,f_r'$:

$$0\to \bigwedge^r \mathcal{N}_{X'/M'}^\vee \to \bigwedge^{r-1} \mathcal{N}_{X'/M'}^\vee \ldots \to \bigwedge^1 \mathcal{N}_{X'/M'}^\vee \to \mathcal{O}_{M'}\to \mathcal{O}_{X'}\to 0 $$

Then $\mathcal{K}_\bullet = \mathcal{K}_\bullet' \otimes_{A'} A$ is the Koszul complex of $f_1,\ldots,f_r$.

Now he claims: we have a homology spectral sequence

$$E_{pq}^2 = \operatorname{Tor}_p^{A'}(\operatorname{H}_q(\mathcal{K}_\bullet'),A) \Rightarrow \operatorname{H}_{p+q}(\mathcal{K}_\bullet)=\left\{ \begin{array}{ll} 0 & \mbox{if }p+q>0 \\ \mathcal{O}_X & \mbox{if }p+q=0 \end{array} \right. $$

I guess that this is a particular case of the Grothendieck spectral sequence (see for instance the book of Lang "Algebra", Chapter 20, Section 9, Theorem 9.6, page 821), but in its analog version for right exact functors:

Theorem (Grothendieck spectral sequence) Let $\mathcal{C}, \mathcal{D}, \mathcal{E}$ be categories such that $\mathcal{C}$ and $\mathcal{D}$ have enough projective objects. Let $$ F:\mathcal{C} \to \mathcal{D} \;\;\text{and}\;\;G:\mathcal{D}\to \mathcal{E}$$ be covariant right exact functors such that if $P$ is a projective object in $\mathcal{C}$, then $$L_rG(F(P))=0,\;\forall r>0. $$ Then, for every object $A$ in $\mathcal{C}$ there is a spectral sequence with $$E_{p,q}^2 = L_pG (L_qF(A)) \Rightarrow L_{p+q}G\circ F(A) $$

Question 3: Is the following right?

In our case, I guess that We should apply the theorem to $\mathcal{C}=\mathbf{\mbox{Comp}}_{A'}$ (the category of chain complexes of $A'$-modules), $ \mathcal{D} = \mathcal{E} =\mathbf{\mbox{Mod}}_{A'}$ (the category of $A'$-modules) and the functors $$F: \mathbf{\mbox{Comp}}_{A'} \to \mathbf{\mbox{Mod}}_{A'} $$ $$ C_\bullet \longmapsto H_0(C_\bullet), $$ $$G:\mathbf{\mbox{Mod}}_{A'} \to \mathbf{\mbox{Mod}}_{A'} $$ $$ N \longmapsto N\otimes_{A'} A $$

Here the left derivated functors are $L_rG(N)=\operatorname{Tor}_r^{A'}(N,A)$ and $L_rF(C_\bullet)=H_r(C_\bullet)$, so they satisfy the hypotheses of the theorem, since projective objects are acyclic with respect to the left derived functors.

By applying the theorem to the Koszul resolution of $\mathcal{O}_{X'}$ we obtain the claimed spectral sequence. Using this spectral sequence it can be deduced (see the paper for more explanation) that $H_q(\mathcal{K}_\bullet')=0$ for all $q>0$, thus $f_1',\ldots,f_r'$ is a regular sequence.

Finally, we analyse the converse situation. Namely, given liftings $f_1',\ldots,f_r'$ of $f_1,\ldots,f_r$ to $\mathcal{O}_{M'}$, we define $X'$ via $\mathcal{O}_{X'}=\mathcal{O}_{M'}/(f_1',\ldots,f_r')$.

We want to prove that $X'$ is a lifting of $X$ to $M'$. So by the previous discussion we need to show that the surjective map $\mathcal{I}'/\mathfrak{a}\mathcal{I}'\to \mathcal{I}$ is injective. Then we consider $\sum_i a_i'f_i' \in \mathcal{I}'$ such that its image $\sum_i a_i f_i$ in $\mathcal{I}$ is 0.

He claims that since $f_1,\ldots,f_r$ is a regular sequence we can write $(a_1,\ldots,a_r)\in \mathcal{O}_M^r$ as a linear combination of "standard relations" of the form

$$(0,\ldots,0,\underbrace{f_j}_{i\operatorname{-th\;place}},0,\ldots,0,\underbrace{-f_i}_{j\operatorname{-th\;place}},0,\ldots,0). $$

Question 4: How do we get these relations and how to use them to obtain the last equality of sums in the following paragraph?

Then, by lifting them we obtain relations

$$(0,\ldots,0,\underbrace{f_j'}_{i\operatorname{-th\;place}},0,\ldots,0,\underbrace{-f_i'}_{j\operatorname{-th\;place}},0,\ldots,0). $$

among the $f_i'$. Then $(a_1',\ldots,a_r')$ can be written as a relation among the $f_i'$, plus an element $(b_1',\ldots,b_r')\in (\mathfrak{a}\mathcal{O}_{M'})^r$, so that $\sum_i a_i'f_i'=\sum_i b_i'f_i' \in \mathfrak{a}\mathcal{I}'$. Proving in this way that the liftings of $X$ are obtained locally by lifting equations for $X$.

Sorry about the extension. I wanted to be clear and self-contained as possible. Thank you very much for your help in advance.

  • $\begingroup$ Dear Pedro, what is the definition of $A$? $\endgroup$
    – Mark Grant
    Mar 20, 2014 at 7:05
  • $\begingroup$ I'm sorry, I didn't write it: We set $A=A'/\mathfrak{a}$. Thanks for your comment, I will edit. $\endgroup$ Mar 20, 2014 at 7:56
  • $\begingroup$ Dear Pedro, not a problem. I'm afraid this is not really my area, but your question appears to be well-written, and I hope an expert can give you an answer. $\endgroup$
    – Mark Grant
    Mar 20, 2014 at 10:53


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