I'm reading the Gortz's Algebraic Geometry, Theorem 10.97 and stuck at final statement :
(Errata : For the last paragraph of the proof, we should replace all $\xi$ by $\eta$. Moreover, if $U$ is the non-empty open subset of $S$ which we are considering, the last equation holds for $v∈V :=Y \cap f^{−1}(U)$ (not all $Y$). )
Why the first equality in the underlined statement is true? As the first paragraph,
$$\operatorname{dim}_vX_{f(v)} = \operatorname{sup}_{Z\in I_v}\operatorname{dim}Z $$
, where $I_v$ is the set of irreducible components of $X_{f(v)}$ that contains $v$.
I'm struggling with this equality now and can't find breakthrough yet.
Can anyone help?
Furthur progress : Fix $v \in V:= Y \cap f^{-1}(U)$. Then since $Y\subseteq \bigcap_{i}X_i$, $v\in X_i \cap X_{f(v)}$ for all $i$.
For each $i$, let $Z_{i,v}$ be an irreducible component of $X_i \cap X_{f(v)}$ containing the $v$. Then by the final paragraph, $\operatorname{dim}Z_{i,v} = \operatorname{dim}(X_i \cap X_\eta)$ for all $i$.
Since irreducible subset of closed subset is irreducible in the ambient space(?), $Z_{i,v}$ is irreducible set in $X_{f(v)}$. So there is an irreducible component $Z'$ of $X_{f(v)}$ containing $Z_{i,v}$. Then, since $v\in Z_{i,v} \subseteq Z'$,
$$\operatorname{dim}Z_{i,v} \le \operatorname{dim}Z' \le \operatorname{sup}_{Z\in I_v}\operatorname{dim}Z $$
Since $i$ is arbitrary, $\operatorname{sup}_{i \in I}\operatorname{dim}Z_{i,v} \le \operatorname{sup}_{Z\in I_v} \operatorname{dim}Z$.
So, $\operatorname{sup}_{i \in I} \operatorname{dim}(X_i \cap X_{\eta}) \le \operatorname{dim}_v X_{f(v)}$.
And it remains to show the inverse inequality. Can we show that? If so, how?
Perhaps.. next is true? Fix $Z \in I_v$. Then
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$ v \in Z \cap X_i $ is an irreducible component of $X_i \cap X_{f(v)}$ ?
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$\operatorname{dim}Z = \operatorname{dim}(Z \cap X_i)$ ? For this statement, my first strategy is using next theorem (Gortz's Theorem 5.22.)
(From $Z \hookrightarrow X_{f(v)} \to \operatorname{Spec}(k(f(v)))$, the inclusion $Z\cap X_i \hookrightarrow Z$ is a morphism of $k(f(v))$-shcmes of locally finite type).
If $Z \cap X_i$ contains the generic point of $Z$, then by the theorem the equality of dimensions holds. And is it true?
Anyway, if these are true, then we may show the inverse inequality.
Can anyone help?
Best Answer
Edited: the original lemma was false as stated (consider the spectrum of a Noetherian local ring, without the maximal ideal), although the result was true in the situation where it was applied. This has been corrected.
There’s a hypothesis missing: we obviously need $v \in f^{-1}(S’) \cap Y$, where $S’ \subset S$ is the open subset from the previous paragraph. Then $X_{f(v)}$ is the reunion of the (finitely many) closed subsets $X_i \cap X_{f(v)}$, all of which contain $v$ and are pure of dimension that of $X_i \cap X_{\eta}$.
We conclude with the following (applied to the closed subsets $X_i \cap X_{f(v)} \subset X_{f(v)}$ and the point $v$) results:
Lemma 1: let $X$ be an irreducible scheme of finite type over a field, and $U \subset X$ be a nonempty open subset. Then $U$ and $X$ have the same dimension.
Proof: first note that we can assume $X$ reduced, hence $U,X$ integral. $X$ has a finite cover by affine open subsets $V$, and the dimension of $X$ is the maximum of these dimensions (the dimension of the topological space underlying a scheme is the maximal length of a sequence $(x_0,\ldots,x_n)$ where $x_{i+1}$ is a specialization of $x_i$). For the same reason, the dimension of $U$ is the maximum of the dimensions of the $U \cap V$, so we can reduce to the case where $X$ is affine. Then there is a principal open subset $V \subset U \subset X$ and $\dim{V} \leq \dim{U} \leq \dim{X}$, so we may assume that $X,V$ are affine. Since $X$ is integral, it is enough to show that the function field of $X$ determines $\dim{X}$ (since $X$ and $V$ have the same fraction field).
Now, if $d=\dim{X}$, then $X$ is finite over $\mathbb{A}^d$ (by Noether normalization), so that the fraction field of $X$ is a finite extension of $k(x_1,\ldots,x_d)$ so has transcendence degree $d$ over $k$.
Lemma: let $Z_1,\ldots,Z_r$ be closed subsets of a topological space $X$, which is the underlying space of a scheme of finite type over a field. Assume furthermore that every $Z_i$ is equidimensional of dimension $d_i$ and that $X=\bigcup_{i=1}^r{Z_i}$. Then if $x \in \bigcap_{i=1}^n{Z_i}$, $\dim_x{X}$ is the maximum $\delta$ of the $d_i$.
Proof: Let $x \in U$ be an open subset of $X$. Then the irreducible components of $Z_i \cap U$ (which is nonempty since it contains $x$) are exactly the nonempty $Z’ \cap U$, where $Z’ \subset Z_i$ is an irreducible component. But for such $Z’$, $Z’ \cap U$ is a dense open subset of $Z’$, so it is irreducible of dimension $d_i$ (because $Z’$ is the topological space underlying a scheme of finite type over a field), thus $Z_i \cap U$ is pure of dimension $d_i$. As $U$ is the reunion of the closed $Z_i \cap U$, $\dim{U}$ is thus $\delta$, QED.