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In this post here, the author asks how to show $X$ is irreducible if there exist a proper morphism $\pi:X \rightarrow Y$ between varieties where all the fibers of $f$ are irreducible and are equidimensional.

In one of the comments, a user posted a criterion for checking irreducibility by M. Mustaţă (link) where the assumption for properness of $\pi$ is dropped and only require $X$ to be equidimensional. However, the criterion was still only stated for algebraic varieties.

My question is, can the criterion be generalized for affine schemes of finite type over DVRs? In particular, suppose $R$ is an DVR with $\text{Spec}(R) = \{\eta,s\}$ where $\eta$ is the generic point and $s$ the special point and $$X = \text{Spec}(R[x_1,...,x_n]/(f_1,...,f_m)).$$ Let $\pi: X \rightarrow \text{Spec}(R)$ be the structure map. Suppose both the generic fiber $X_{\eta}$ and the special fiber $X_s$ of $X$ are irreducible and are of dimension $1$. Can we conclude that $X$ must also be irreducible and is of dimension $1$? (if the criterion in the notes applies then $X$ should automatically be equidimensional as the irreducible components of $X$ can only be $X_{\eta}$ or $X_{s}$ or $X_{\eta} \cup X_s$ according to the criterion?)

KSAKY
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1 Answers1

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Note that your hypothesis does not prevent $X$ from being the disjoint union of a generic fibre and a special fibre (and the conclusion obviously fails) – so we assume that $X$ is connected.

Let $F \subset X$ be the irreducible component containing the closed fibre $X_s$.

Suppose that $F$ contains a point $x \in X_{\eta}$. Then $F \cap X_{\eta}$ is a nonempty open subset of $F$, hence it is dense in $F$. In particular, the Zariski-closure of $X_{\eta}$ contains the Zariski-closure of $F \cap X_{\eta}$, hence contains $F$, hence contains $X_s$. Thus $X=\overline{X_{\eta}}$ is irreducible.

Thus, we assume that $F=X_s$. Since $X$ is connected, there is a closed point $x \in Y:=\overline{X_{\eta}}$ which lies in $X_s$.

Since $F=X_s$, $Y_s$ is a proper closed subset of $X_s$ (which is irreducible of dimension one), so that $\dim_x(Y_s) =0$.

Then, by Stacks, Lemma 02FZ, $\dim_{y}{Y_{\eta}}=0$, where $y \in X_{\eta}$ is the generic point. It follows that $\{y\} \subset X_{\eta}$ is open. This is not possible, since $X_{\eta}$ is of finite type over $Frac(R)$ and of dimension $1$.

Remarks:

  1. Note that the argument could have been the same if we assume that $X$ is equidimensional with irreducible fibres. I think a similar idea should work for reasonable bases (eg Noetherian?), but the notation gets more complicated (as in the link).

  2. the real place where something nontrivial happens is the use of Lemma 02FZ, which relies on Zariski’s Main Theorem.

Aphelli
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