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)?

  • $\begingroup$ If you can say a bit more about what parts of the statement are confusing for you, you are probably more likely to receive some good answers (though unfortunately, I am unable to help you beyond this advice) $\endgroup$ – BaronVT Jan 27 '14 at 16:32
  • $\begingroup$ I believe it is the Noether normalization theorem. $\endgroup$ – Youngsu Jan 27 '14 at 16:45

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.

| cite | improve this answer | |
  • $\begingroup$ thanks for the nice answer; could you also comment a bit on why this implies that we have a finite (say $d$) branched covering, which in turn is used to constrain integration on small disks around points, roughly speaking as $\int_{\Delta \cap X} vol \leq d \cdot \int_{\Delta' \in \mathbb C^n} vol$? $\endgroup$ – jj_p Sep 12 '14 at 16:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.