regarding finite integral ring extension My question regards understanding (and possibly a source for proof) of the following, cited in the book Complex Geometry by Huybrechts (Theorem 1.1.30.) 
(Also, it is there stated that this is a particular form of a more general theorem: does anyone know which theorem is?)
Theorem
Let $X \in \mathbb{C}^n$ be an irreducible analytic germ
defined by a prime ideal
$\mathfrak{p}\subset \mathcal{O}_{\mathbb{C}^n,0}.$
Then one can find a coordinate system
$$(z_1 , \ldots , z_{n-d},z_{n-d+1} ,\ldots , z_n )$$
such that the projection
$(z_1 , \ldots , z_n) \mapsto (z_{n-d+1} ,\ldots , z_n )$
induces a surjective map (of germs!)
$\pi:X \to \mathbb{C}^d$
and such that the induced ring homomorphism
$ \mathcal{O}_{\mathbb{C}^d,0} \to  \mathcal{O}_{\mathbb{C}^n,0}/\mathfrak{p}$
is a finite integral ring extension.

What does it mean that a ring homomorphism
$ \mathcal{O}_{\mathbb{C}^d,0} \to  \mathcal{O}_{\mathbb{C}^n,0}/\mathfrak{p}$
is a finite integral ring extension (starting from a definition of finite integral ring extension)?
 A: The standard notion of integral ring extension is for subrings: an inclusion of rings $R \subseteq S$ is said to be integral if every element $x \in S$ satisfies a monic polynomial with coefficients in $R$, i.e. there exist $a_1, ..., a_{n-1} \in R$ such that $x^n + a_{n-1}x^{n-1} + \ldots + a_0 = 0$. One can extend this to general maps by saying that a ring homomorphism $\varphi : R \to S$ is integral if the inclusion $\varphi(R) \subseteq S$ is integral. In the case that $R$ is a field, this reduces down to the notion of being algebraic over a field, as taught in abstract algebra courses.
The word finite in "finite integral extension" should refer to the fact that the map $\mathcal{O}_{\mathbb{C}^d,0} \to \mathcal{O}_{\mathbb{C}^n,0}/\mathfrak{p}$ makes $\mathcal{O}_{\mathbb{C}^n,0}/\mathfrak{p}$ into a finite $\mathcal{O}_{\mathbb{C}^d,0}$-module. However, any map $\varphi : R \to S$ which makes $S$ a finite $R$-module is automatically an integral extension, so depending on convention, specifying integrality is not needed here (although "finite ring extension" typically means something else - perhaps "module-finite ring extension" is the most accurate). 
As mentioned in the comments, the theorem you mention seems to be an application of the Noether normalization theorem. In the algebraic context, one (weak) version states that if $R$ is a finitely generated algebra over a field $k$, then there exist algebraically independent (over $k$) elements $y_1, ..., y_d \in R$ such that the extension $k[y_1, ..., y_d] \subseteq R$ is integral. Since in this case $R$ is of finite type over $k$, this is equivalent to saying that $R$ is a finite $k[y_1,...,y_d]$-module.
