I am trying to solve problem 3 from Appendix B from Milnor's “Singular Points of Complex Hypersurfaces” (p. 115). This shows that Milnor's topological definition of intersection multiplicity agrees with another more algebraic one.
Let $f = (f_1, \dots, f_n)$ be holomorphic functions with an isolated zero at the origin in $\mathbb C^n$. Milnor (p. 59) defines the intersection multiplicity $\mu$ of $f_1, \dots, f_n$ at the origin as the degree of the map
$$g : S^{2n-1} \to S^{2n-1}, \qquad g(x) = \dfrac {f(tx)} {\Vert f(tx) \Vert}$$
for small enough $t > 0$. Milnor also shows (p. 113) that, for “most” values $p$ near the origin, the fiber $f^{-1}(p)$ has exactly $\mu$ points.
Actually, I am not trying to solve the problem from scratch, but rather heavily reusing a proof in Ebeling's book “Functions of Several Complex Variables and Their Singularities” (pp. 150-151), which, if it goes through, then it solves the problem.
By hypothesis, $f_1, \dots, f_n$ form a system of parameters of the local ring $B = \mathbb C \{ x_1, \dots, x_n \}$. Then $B / \langle f_1, \dots, f_n \rangle$ is a finite-dimensional $\mathbb C$-vector space, say, of dimension $d$. The goal is to show that $d = \mu$.
Let $A = \mathbb C \{ f_1, \dots, f_n \}$. By the Weierstrass preparation theorem for modules (p. 97), $B$ is a finite $A$-module, minimally generated by $d$ elements. By miracle flatness (p. 109, or also Stacks Project 10.128), $B$ is freely generated by these $d$ elements. In particular, $B$ is an integral extension of $A$.
Let $K = \mathrm{Frac}(A)$ and $L = \mathrm{Frac}(B)$. Then $L = K(x_1, \dots, x_n)$ is an algebraic extension of $K$ of degree $d$. By the primitive element, we may write $L = K(x_0)$, where $x_0$ is a $K$-linear combination of $x_1, \dots, x_n$. Then $x_0$'s minimal polynomial $g_0 \in K[T]$ has degree $d$. Multiplying $x_0$ times the least common denominator of $g_0$'s coefficients, we obtain another element $y_1 \in L$ such that $L = K(y_1)$, but whose minimal polynomial is in $A[T]$. This implies that $y_1 \in B$. In fact, $y_1$ is in $B$'s maximal ideal.
So far so good, but then I do not understand the next step. Ebeling claims (p. 151) that, after a change of coordinates, we may assume $x_1 = y_1$. Equivalently, there exist $y_2, \dots, y_n \in B$ such that $y_1, \dots, y_n$ is a system of parameters that generates the whole maximal ideal $\langle x_1, \dots, x_n \rangle$. Equivalently, $y_1$ has order $1$. Why is this the case?
Moving on to the rest of the proof.
Let $g \in A[T]$ be $y_1$'s minimal polynomial. Using Hensel's lemma (p. 73), one can show (p. 101) that $g$'s non-leading coefficients are in $A$'s maximal ideal.
The inclusion ring map $A \to A[y_1]$ corresponds geometrically to the restriction of
$$\pi : U \times \mathbb C \longrightarrow U, \qquad (f_1, \dots, f_n, y_1) \longmapsto (f_1, \dots, f_n)$$
to the hypersurface $g = 0$. Away from the subset of $U$ where the discriminant $\Delta g$ vanishes, the fibers $\pi^{-1}(p)$ have exactly $d$ points on this hypersurface.
IMO, Ebeling handwaves the last part, so I am going to try to “unhandwave” it. The inclusion ring map $A[y_1] \to B$ is a normalization map, because $A[y_1]$ and $B$ have the same field of fractions, and $B$ is integrally closed. Moreover, one constructs the normalization $B$ by adding $y_2, \dots, y_n$, so, if we invert the product $h = y_2 \cdots y_n$, then the localization $A[y_1]_h \to B_h$ is actually an isomorphism. In other words, geometrically, the restriction of
$$\iota : V \longrightarrow U \times \mathbb C, \qquad (y_1, \dots, y_n) \longmapsto (f_1, \dots, f_n, y_1)$$
is “almost an isomorphic embedding” into the hypersurface $g = 0$, and it actually becomes an isomorphism after we remove the points where $h = 0$ from both the domain and the codomain. Is the highlighted idea right?
Finally, the composition $\pi \circ \iota : V \to U$ agrees with the original map $f$, and it has fibers of size $d$ on an open subset of $U$ whose closure contains the origin. Therefore, $d$ agrees with Milnor's intersection multiplicity $\mu$.
Change of strategy:
Instead of considering the ring extension $A \to B$ in one go, I will set $A_j = A[x_1, \dots, x_j]$ and consider the tower of extensions
$$A = A_0 \longrightarrow A_1 \longrightarrow A_2 \longrightarrow \dots \longrightarrow A_n = B$$
Similarly, I will set $K_j = \mathrm{Frac}(A_j) = K(x_1, \dots, x_j)$ and consider the tower of extensions
$$K = K_0 \longrightarrow K_1 \longrightarrow K_2 \longrightarrow \dots \longrightarrow K_n = L$$
By construction, $K_j = K_{j-1}(x_j)$, hence $x_j$'s minimal polynomial $g_j \in K_{j-1}[T]$ has degree $d_j = [K_j : K_{j-1}]$. Geometrically, the inclusion $A_{j-1} \to A_j$ corresponds to the restriction of
$$(f_1, \dots, f_n, x_1, \dots, x_j) \longmapsto (f_1, \dots, f_n, x_1, \dots, x_{j-1})$$
to the zero locus of $g_1 = \dots g_j = 0$. Away from the subset of the codomain where the discriminant $\Delta g_j$ vanishes, the fiber over each point has exactly $d_j$ points. Therefore, the general fiber of
$$(f_1, \dots, f_n, x_1, \dots, x_n) \longmapsto (f_1, \dots, f_n)$$
has exactly $\mu = d_1 \cdots d_n$ points. Finally, the map
$$(x_1, \dots, x_n) \longmapsto (f_1, \dots, f_n, x_1, \dots, x_n)$$
is an isomorphism, because it is a parametrization of the graph of $f(x)$, so we are done. Does this idea work?