Nakayama's lemma implies isomorphism It was originally related to this. I think it worthy to ask separately.
Known is: $X=\mathrm{Spec}(S),\ Y=\mathrm{Spec}(R)$ are affine, $f$ is finite morphism, actually the fiber has only one point at $y$ which is $\mathrm{Spec}(\mathbb C)$. If needed we can also think they are just closed subsets of $\mathbb {A}^n,\mathbb {A}^m$. Now consider the fiber product:
$$\require{AMScd}
\begin{CD}
Z @>>> X\\
@V{f'}VV @VV{f}V \\
\mathrm{Spec}(\mathscr{O}_{Y,y}) @>>> Y
\end{CD}$$
We want to conclude that $Z\to \mathrm{Spec}(\mathscr{O}_{Y,y})$ isomorphism.
We can translate into commutative algebra by reversing the arrows:
$$\require{AMScd}
\begin{CD}
R @>>> R_{\mathfrak{p}}\\
@V{g}VV @VV{g'}V \\
S @>>> S\otimes_{R}R_{\mathfrak{p}}
\end{CD}$$
We know that the $g,g'$ arrows are finite. $B:=S\otimes_{R}R_{\mathfrak{p}}$ is a finite $R$-algebra. 
The author in proposition 2.6 just claimed that $B$ is also local and then by Nakayama, they are isomorphism? I do not understand these arguments.
 A: You may assume from the outset that $Y=\text{Spec}(R)$ for $(R,{\mathfrak m})$ local and that $X=\text{Spec}(S)$ with $(S,{\mathfrak n})$ local and finitely generated over $R$. The assumption on the fibre translates as $S/{\mathfrak m}\cong{\mathbb C}$, and you intend to conclude that $R\to S$ is an isomorphism.
In this generality, however, this isn't true: Pick for example $R={\mathbb R}\to {\mathbb C}=S$. 
In your original question, though, you seem to be working over $\text{Spec}({\mathbb C})$, so let's assume $R,S$ are $k$-algebras for some field $k$ and $k\stackrel{\cong}{\to} S/{\mathfrak n}$. Since $k\to S/{\mathfrak n}$ factors as $k\to R/{\mathfrak m}\to S/{\mathfrak m}S\to S/{\mathfrak n}$, we get that $R/{\mathfrak m}\cong S/{\mathfrak m} S\cong k$. We still want to prove that $R\to S$ is an isomorphism.
However, it's still not true. Pick for example $R = k[x]/(x^2)\twoheadrightarrow k=S$. 
The missing property here was already mentioned in How to show a morphism is etale?: Flatness. So, let's assume in addition that $S$ is a flat $R$ module. Then we're in the game: A finitely generated, flat module over a Noetherian ring is projective, and finitely generated projective modules over local Noetherian rings are free. Then, ignoring ring structures, $R\to S$ is a morphism between finitely generated free $R$-modules which becomes an isomorphism under $-\otimes_R R/{\mathfrak m}$. Nakayama then implies that $R\to S$ is an isomorphism of $R$-modules, hence of rings.
If you want to avoid the general above statements about flat, projective and free modules, here's a more direct argument, this time starting with Nakayama: The morphism of finitely-generated $R$-modules $R\to S$ is an isomorphism (in particular surjective) after applying $-\otimes_R R/{\mathfrak m}$, so Nakayama implies that already $R\to S$ is surjective. Hence $S=R/I$ for some ideal $I\lhd R$. It's therefore sufficient to show that $R\to R/I$ is flat if and only if $I=0$: For this, consider the exact sequence of $R$-modules $0\to I\to R\to R/I\to 0$. Applying $-\otimes_R R/I$ gives $I/I^2\to R/I\stackrel{\cong}{\to} R/I\to 0$, so if $R/I$ is flat over $R$, we get $I/I^2 = 0$, hence $I = I^2$, and a forteriori $I = {\mathfrak m}I$. Viewing $I$ as a finitely generated $R$-module, Nakayama implies $I=0$ as desired.
