Base change preserving irreducibility?
问题内容
Let $S$ be a Dedekind scheme (using the less conventional definition: locally Noetherian, irreducible, all stalks normal and $\dim S \le 1$), $X$ an irreducible scheme, and $f: X \to S$ a dominant morphism of finite type. Consider a point $s \in S$, and let $T = \operatorname{Spec} \mathcal{O}_{S,s}$ with $g: T \to S$ the canonical map. Consider the base change $Y = X \times_S T$. For which points $s \in S$ is $Y$ irreducible, or is $Y$ always irreducible no matter the choice of $s$?
Also, are all of the assumptions here necessary? I, for one, am not sure whether $S$ Dedekind can simply be replaced by $S$ integral, and whether the finite type assumption is even needed.
回答 (2)
I think $Y$ is irreducible no matter what $s$ is. Indeed, $Y$ identifies as a topological space with the space of $x \in X$ such that $s$ is a specialization of $f(x)$, where the topology is induced by that of $X$.
So the generic point of $X$ belongs to $Y$ and lies in every non-empty open subspace of $Y$, i.e. is dense. Hence there is a point which is dense in $Y$, so $Y$ is irreducible.
In fact, we only need $X$ irreducible, $S$ integral, and $f:X\to S$ dominant. Let $T=\operatorname{Spec}\mathcal O_{S,s}$. Choose an affine open neighbourhood $V=\operatorname{Spec}R\subseteq S$ of $s$, say $s\leftrightarrow\mathfrak p$. Since $S$ is integral, $R$ is an integral domain, and $T=\operatorname{Spec}R_{\mathfrak p}$. Also
$$Y=X\times_ST\cong f^{-1}(V)\times_VT.$$
Since $f^{-1}(V)$ is a nonempty open subset of the irreducible scheme $X$, it is irreducible. Take an affine open cover
$$f^{-1}(V)=\bigcup_iU_i,\qquad U_i=\operatorname{Spec}A_i.$$
Let $\xi$ be the generic point of $X$. Every $U_i$ contains $\xi$, so $A_i$ has a unique minimal prime $\mathfrak q_i$, corresponding to $\xi$. Let
$$\phi_i:R\to A_i$$
be the map induced by $f\vert{}_{U_i}$. Since $f$ is dominant, $f(\xi)$ is the generic point of $S$, hence
$$\phi_i^{-1}(\mathfrak q_i)=(0).$$
Thus
$$\mathfrak q_i\cap\phi_i(R\setminus\mathfrak p)=\varnothing.$$
Now
$$U_i\times_VT=\operatorname{Spec}(A_i\otimes_RR_{\mathfrak p})=\operatorname{Spec}\big((R\setminus\mathfrak p)^{-1}A_i\big).$$
The prime $\mathfrak q_i$ survives the localization. Since it is the unique minimal prime of $A_i$, every prime of $(R\setminus\mathfrak p)^{-1}A_i$ contains $(R\setminus\mathfrak p)^{-1}\mathfrak q_i$. Hence this is the unique minimal prime, so $U_i\times_VT$ is irreducible and nonempty. Finally, for any $(i,j)$, choose an affine open $W\subseteq U_i\cap U_j$ containing $\xi$. By the same argument, $W\times_VT\neq\varnothing$. Since $W\subseteq U_i,U_j$, its base change is contained in both $U_i\times_VT$ and $U_j\times_VT$. Thus these opens pairwise meet. Therefore
$$Y=\bigcup_i(U_i\times_VT)$$
is a union of irreducible opens which pairwise meet, and hence $Y$ is irreducible. Thus
$$\boxed{X\times_S\operatorname{Spec}\mathcal O_{S,s}\text{ is irreducible for every }s\in S.}$$