See http://www.jstor.org/stable/1971241?seq=6#page_scan_tab_contents, pages 597 and 598.

At the bottom of page 597, we get the following diagram:

$\begin{array}{ccccccccc} A/I & \xrightarrow{nat.} & A/(I+M^n) \\ & {\alpha}\searrow & \uparrow nat. \\ & & A/(MI+M^n+(r_1,...,r_a)) \end{array}$

where the $r_i$ are in $I+M^n$. $A$ is a formal power series ring over $k$ and $M$ its maximal ideal. $I \subset M^2, I\cap M^n = M(I\cap M^{n-1}) \subset MI$ (for some n, following from the Artin-Rees lemma). I am not sure how much context is required, so for now I will refer to the link for more. Do feel free to ask.

The crucial claim, on which the entire argument seems to rely, is that "since $I \subset M^2$, $\alpha$ is surjective by the above diagram." The following part of the proof is then a mere formality.

Why is $\alpha$ surjective? This seems not at all as obvious as the statement of the claim makes it sound. I believe thus that I am completely missing something critical.

For what it is worth, I am not yet familiar enough with obstruction theory. Maybe this is where the crux lies (see the linked paper, on the same page)? If so, I would appreciate some pointers on where to educate myself about this.

  • $\begingroup$ Is $$\alpha$ some lift of $A$-modules, making the diagram commutative? $\endgroup$ – Mohan May 23 '17 at 0:24
  • $\begingroup$ Yes. It come from an extension of scheme morphisms obtained in the paragraph above. $\endgroup$ – numberjedi May 23 '17 at 1:17
  • $\begingroup$ Do you know Nakayama's lemma? $\endgroup$ – Mohan May 23 '17 at 1:30
  • $\begingroup$ Yes. Not yet seeing how to apply it though. Let me take a look... $\endgroup$ – numberjedi May 23 '17 at 1:42
  • $\begingroup$ I don't see it, especially what the fact $I \subset M^2$ has to do with it. Note that we do not know that the $r_i$ generate $I$; in fact this is the goal of the proof. $\endgroup$ – numberjedi May 23 '17 at 1:54

So finally here it is. Let $B:= (MI+M^n+(r_1+\cdots +r_a))$ and write $I_B$ for $IB$. For $a \in B, \sigma\circ\pi(a) = a+\bar a,$where $\bar a \in I_B$, $\pi$ is the natural map from $B$ to $B/I_B = A/(I+M^n)$, and $\sigma$ is a section from $B/I_B$ to $B$ obtained by restricting $A/I\xrightarrow{\alpha}B$ to $A/(I+M^n)$.

Now let $a,b \in \mathfrak m_B \subset B$. Then $\sigma\circ\pi(ab) \in ab + \mathfrak m_BI_B +I_B^2.$ In particular, $I_B\subset \mathfrak m_B^2 \subset \mathfrak m_BI_B$. Now, at this point one could invoke Nakayama's lemma to point out that $I_B = 0$, but the containment itself is actually enough for us to see that $\alpha$ is surjective ($MI=0$ in $B$), and $I$ is generated by at most $a$ elements.

Note that in the proof we can actually save us some hassle by not using "$+M^n$".


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