Mori's “projective manifolds with ample tangent bundles”, Proposition 3 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.

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 "$+M^n$".
