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So I have a normal surface (ie, all components are dimension 2) $X$ which is smooth and affine over a Dedekind domain $R$ (so $X$ is an affine scheme). Suppose $R'$ is integral (possibly not dedekind) and $Spec(R')\rightarrow Spec(R)$ is finite flat (but possibly ramified).

Does $X_{R'}$ (the base change of $X$ to $Spec(R')$) have the property that connected components are irreducible?

Note that $X$ has this property since it's normal. If $K,K'$ are the fraction fields of $R,R'$ respectively, then $X_K$ also has this property since it's also normal, and since $X_{K'}$ is smooth over $X_K$ and normality is smooth-local, $X_{K'}$ is also normal and has this property.

The intuition is this:

  1. By There is a bijection between irreducible components of the generic fiber and irreducible components passing through it. together with the fact that $X_{R'}\rightarrow Spec(R')$ is smooth (hence open), we know that the irreducible components of $X_{R'}$ are in bijection with the irreducible components of $X_{K'}$, which are its connected components.

  2. Suppose two irreducible components $C_1,C_2\subset X_{R'}$ intersect. If they intersect in a curve, then there are two possibilities: The intersection is fibral (ie, lies in a fiber of the map $X_{R'}\rightarrow Spec(R')$). In this case, that fiber should be either singular (nonreduced?), probably(?) contradicting the smoothness assumption of $X\rightarrow Spec(R)$ (and hence $X_{R'}\rightarrow Spec(R')$), or higher dimension than the other fibers, contradicting flatness. If the intersection is not fibral, then the generic point of the intersection should map to the generic point of $Spec(R')$, and hence this should correspond to two irreducible components of $X_{K'}$ intersecting, which is impossible by normality.

  3. If $C_1,C_2$ intersect at a point, then that point should be a singular point in its fiber, which again probably(?) contradicts smoothness.

Does this make sense? If it doesn't, does anyone have a counterexample?

EDIT: Here are some really simple examples:

$X = Spec k[x,y,z]/(xy)$, so $X$ is the union of two planes intersecting in the $z$-axis. Then over $R = Spec k[x]$, we see that the fiber above $x = 0$ is 2-dimensional, whereas the fiber above $x = a$ $(a\ne 0)$ is only 1-dimensional. This contradicts flatness. If $R = Spec k[x-y]$, then the fiber at $x-y = 0$ is Spec $k[x,z]/(x^2)$, which is nonreduced, contradicting smoothness.

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  • Normality ascends along smooth extensions, therefore $X_{R^\prime}$ is normal, which, as you state, seems to imply the property you are asking for (I did not check the latter). – Hagen Knaf May 06 '15 at 18:35
  • @Hagen I only know that $X_R$ is smooth over $R$. It's not necessarily the case that $X_{R'}$ is smooth over $X_R$. – oxeimon May 06 '15 at 19:01
  • @Hagen oh you mean $X_{R'}$ is smooth over $R'$ which is normal, so $X_{R'}$ is. Ugh, I meant to say that $R'$ is not necessarily dedekind. – oxeimon May 06 '15 at 20:13
  • So then my argument doesn't work. My guess is that your conjecture is true. I'm however struggling with the case of a fibral curve in the intersection of two components. Why should it necessarily be singular? Couldn't we just take two normal surfaces having isomorphic components in their repective closed fibres and glue them together along that componnents? – Hagen Knaf May 07 '15 at 03:53
  • @Hagen The thought is that if they intersect in a fibral curve, then the fiber should have a component of "multiplicity 2" in some sense, probably corresponding to the fiber being nonreduced. Here's an example: – oxeimon May 07 '15 at 17:45
  • @Hagen I wrote the example in the OP – oxeimon May 07 '15 at 17:53

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