I have following question about so-called "principle of degeneration" in algebraic geometry (which in modern terms is an immediate consequence of Zariski's main theorem and goes in it's original form to Enriques) saying roughly that for a (wlog local) normal base ring $R$ with field of fractions $K:=\text{Frac}(R)$ a (geometrically) connected, pure $r$-dimensional projective subscheme $X_K \subset \mathbb{P}_K^n$ whose vanishing ideal $I(X)$ can be generated by polynomials in $R[X_0,..., X_n]$ specializes to a connected subscheme over a special point $s \in \operatorname{Spec}(R)$.
In modern terms it should be phrased as follows: that for any point $s \in \operatorname{Spec}(R)$ specializing the generic point of $R$ - so every point/prime $\mathfrak{p}_s \subset R$ - the fiber $X_s =X \otimes_R \kappa(s)$ over $\kappa(s)$ is connected. (note, that by assumption $X \subset \mathbb{P}_R^n$ (ie as scheme over $R$ make sense since we assumed that it's vanishing ideal $I(X)$ is generated by polynomials in $R[X_0,..., X_n]$)
Now the question: Wei-Liang Chow remarked in his paper On the connectedness theorem in algebraic geometry briefly (at first page) that even though it is hard to prove pure algebraically (indeed Zariski invented to show this rigorously whole apparatus of formal holomorphic functions) in contrast it is rather easy to show this in classical (=complex analytic) context by "transcendental methods".
Could somebody sketch the rough idea how to show this result on "Principle of Degeneration" with complex analytic methods what should be due to Chow not so hard? (indeed the notion of normality exists clearly also in analytic world, so the normal base ring $R$ makes also sense in complex analytic setting)
Why is this interesting (for me): I would like to develop better "geometric" intuition for nature of normality as necessary condition in this context.
In modern terms normality of a scheme/ring can be formulated in pure category theoretic terms as a universal property to be the "maximal" finite birational extension in the sense of if that if $X$ is normal then if $f:Y \to X$ is finite birational, then it's already an iso (modern argument: immediately by Stein factorization).
But I would like to go back in time a bit to unravel some black boxes to understand better the "inner mechanism" why it is necessary that the base ring is normal in the sense that it contains it's all "integral" elements.
This suggests, that somehow the whole claim could be somehow reduced to approval that a solution on a monomial equation with coefficients in $R$ which lives in $K$ already "lives" in $R$ (essentially that's what $R$ normal means).
Let consider following "toy situation": $R$ as before normal ring, $\mathfrak{p} \subset R$ a prime, and a geometrically connected $X_K =V_+(F) \subset \mathbb{P}_K^n$ is given by a simple homogeneous polynomial $F(X_0,..., X_n) \in R[X_0,..., X_n]$ (ie coeff in $R$!)
RMK on geometrically connectedness: One way to assume it is to assume that $X_K$ is connected and $K$ is separably closed. In analytic situation $K$ contains $\mathbb{C}$, let's do it, as I'm primary interested in "quick" analytic method Chow skipped in the linked paper.
And the question is if there is a direct argument based strongly on assumed normality of ring $R$ (resp it's localization at $\mathfrak{p}$) to see that $X_{\kappa(\mathfrak{p})} = V_+(\overline{F}) \subset \mathbb{P}_{\kappa(\mathfrak{p})}^n$ (here $F$ with coefficients reduced modulo $\mathfrak{p}$ (+localization) must be also connected.
Is there an ad "heuristic" argument why it is the case? (at leat, why it is plausible to expect ?)
