1.8 The Fibres of a Morphism

Definition

A morphism $f:X\to Y$ is dominating if its image is dense in $Y$, i.e., $Y=\overline{f(X)}$.

Proposition

If $f:X\to Y$ is any morphism, let $Z=\overline{f(X)}$. Then $Z$ is irreducible, the restricted morphism $f^{\prime }:X\to Z$ is dominating and $f^{\prime \ast }$ induces an injection

$$k(Z)\stackrel{f^{\prime \ast }}{\hookrightarrow }k(X).$$

\begin{proof} We first show that $Z=\overline{f(X)}$ is irreducible. Assume that $Z=W_1\cup W_2$ where $W_i$ are closed. Since $f$ is morphism, we know $f^{-1}(W_1)$ and $f^{-1}(W_2)$ are both closed. Note that $X=f^{-1}(W_1)\cup f^{-1}(W_2)$ by $W_1\cup W_2=Z$. Then either $X=f^{-1}(W_1)$ or $X=f^{-1}(W_2)$, because $X$ is irreducible. It deduces that either $Z=W_1$ or $Z=W_2$, and so $Z$ is irreducible.

Notice that $f^{\prime }(X)=f(X)$, and then $(\overline{f^{\prime }(X)}{)}_Z=(\overline{f(X)}{)}_X=Z$. Recall that we have proved that $f^{\prime }$ is a morphism in HW, thus $f^{\prime }$ is dominating. Now it remains to show $f^{\prime \ast }:k(Z)\hookrightarrow k(X)$ is injective.

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下面就没有看了。

For $U\subseteq Z$ affine, $f^{-1}(U)\subseteq X$ open in $X$. Here $f^{-1}(U)={\cup }_{i\in I}{V}_i$ where $V_i$ is affine and $|I|<\infty$. There exists $i$ such that $V_i\to U$ is a dominating morphism. Otherwise, $\overline{f(V_i)}⊊U$ for all $i$ and we claim that

$$\overline{f(f^{-1}(U))}\subseteq \cup \overline{f(V_i)}⊊U,$$

which induces a contradiction.

Now we prove that $\cup \overline{f(V_i)}⊊U$. For any two closed set $V(I),V(J)⊊U$, one have $(V(I)\cup V(J){)}^c=V(I{)}^c\cap V(J{)}^c\ne \varnothing$ as $U$ is irreducible.

Define $g=f{|}_V$, then $g$ is still dominating. It suffices to show $V\to U$ dominating is a morphism between affine varieties. $S=\Gamma (U)\to \Gamma (V)=R$ is injective, then $0\to I\to S\to R$ where $\ker (S\to R)=I$.

We claim that $f(V)\subseteq V(I)$. For $x_0\in V$ and $y_0=f(x_0)\in U$. For $m_{y_0}\subseteq S$ and $m_{x_0}\subseteq R$. For $\alpha \in (f^{\ast }{)}^{-1}(m_{x_0})$, $f^{\ast }(\alpha )\in m_{x_0}$ yields $f^{\ast }(\alpha )(x_0)=0$. Then $\alpha \circ f(x_0)=0$ and $\alpha (y_0)=0$. Then $\alpha \in m_{y_0}$. Now we have $(f^{\ast }{)}^{-1}(m_{x_0})\subseteq m_{y_0}$. In fact $(f^{\ast }{)}^{-1}(m_{x_0})$ is a maximal ideal, because

$$S/(f^{\ast }{)}^{-1}(m_{x_0})\hookrightarrow R/m_{x_0}=k$$

and so $S/(f^{\ast }{)}^{-1}(m_{x_0})$ is a field.

For $x_0\in V$ and $f(x_0)=y_0$, $m_{y_0}=(f^{\ast }{)}^{-1}(m_{x_0^{\prime }})\supseteq I$. Then $y_0\in V(I)$ yields that $f(V)\subseteq V(I)$ and $\overline{f(V)}\subseteq V(I)$. Since $\overline{f(V)}=U$, one have $I=(0)$ and so $S\to R$ is injective. \end{proof}

Theorem

Let $f:X\to Y$ be a dominating morphism of varieties and let $r=\dim X-\dim Y$. Let $W\subset Y$ be a closed irreducible subset and let $Z$ be a component of $f^{-1}(W)$ that dominates $W$. Then

$$\dim Z⩾\dim W+r$$

or

$$\mathrm{codim}(Z\text{ in }X)⩽\mathrm{codim}(W\text{ in }Y).$$

\begin{proof} (affine local base)

没有听。15:43

\end{proof}

Corollary

If $Z$ is a component of $f^{-1}(y)$, for some $y\in Y$, then $\dim Z⩾r$.

Nakayama?

Chevelly's theorem

Let $f:X\to Y$ be a dominating morphism of varieties and let $r=\dim X-\dim Y$. Then there exists a nonempty open set $U\subset Y$ such that:

• $U\subset f(X)$
• for all irreducible closed subsets $W\subset Y$ such that $W\cap U\ne \varnothing$, and for all components $Z$ of $f^{-1}(W)$ such that $Z\cap f^{-1}(U)\ne \varnothing$,

$$\dim Z=\dim W+r$$

or

$$\mathrm{codim}(Z\text{ in }X)=\mathrm{codim}(W\text{ in }Y)$$

\begin{proof} We can assume $Y$ is affine $X\stackrel{f}{\to }Y$ and $X=\cup X_i$ where $X_i$ are affine open. Since $Y=\cup \overline{f(X_i)}\supseteq \overline{f(X)}=Y$ and $Y$ is irreducible, there exists $i$ such that $\overline{f(X_i)}=Y$.

For i) it suffices to show the claim for $X_i\to Y$.

For ii)

---

We can assume $Y$ is affine $X\stackrel{f}{\to }Y$ and $X=\cup X_i$ where $X_i$ are affine open.

We claim $f:X_i\to Y$ is dominating. Otherwise $\overline{f(X_i)}⊊Z⊊Y$ and $f^{-1}(Z)⊊X$ closed. But $X_i\subseteq X$ is open, then $X_i⊊f^{-1}(Z)$ by $X$ irreducible. Then it suffices to show the claim for $f_i:X_i\to Y$.

To show “it suffices”:

• For i) done
• ii) If every $f_i$ satisfies the desired claim, then $Z\cap f^{-1}(U)=Z\cap (\cup f_i^{-1}(U))$. Note that $\dim (Z\cap f^{-1}(U))=\dim (Z\cap f^{-1}(U))$.

We can assume $X,Y$ are affine with $\Gamma (X)=R$, $\Gamma (Y)=S$.

Consider $X\stackrel{f}{\to }Y$ and $S\stackrel{f^{\ast }}{\hookrightarrow }R$. $k(Y)=\mathrm{Frac}(S)$. To compute $\mathrm{tr}.{\mathrm{deg}}_{k(S)}R{\otimes }_{S}k(S)=\mathrm{tr}.{\mathrm{deg}}_{k(S)}\mathrm{Frac}(R\otimes k(S))=\mathrm{tr}.{\mathrm{deg}}_{k(S)}k(R)=r$.

By Normalization, there exist transidental basis $x_1,\cdots ,x_r\in R{\otimes }_{S}k(S)$ over $k(S)$ such that $R{\otimes }_{S}k(S)$ is a finitely generated $k(S)[x_1,\cdots ,x_n]$-module, where $x_i=r_i\otimes s_i/t_i$. replacing $x_i$ by $r_i$, we can assume $x_i\in R$.

$R$ is a finitely generated $S[x_1,\cdots ,x_r]$-module. For $r\in R$, there exists a monic polynomial $X^n+p_{n-1}{X}^{n-1}+\cdots +p_0\in k(S)[x_1,\cdots ,x_n][X]$ such that $r$ is a root. There exists $g\in S$ such that $p_j\in S_{(g)}[x_1,\cdots ,x_r]$, and $r$ is a root of a monic poly with coeff in $S_{(g)}[x_1,\cdots ,x_r]$.

Doing this for finite step, we can assume $R$ is a finitely generated $S_{(g)}[x_1,\cdots ,x_r]$-module. Then $R_{f^{-1}g}$ is a finitely generated $S_{(g)}[x_1,\cdots ,x_r]$-module.

We claim $f^{-1}(y_{(g)})=X_{(f^{\ast }g)}$. 「」 14:50

15:05 全是图，没法记了 好像最后也没证出来；回去自己看课本吧

\end{proof}

Definition

Let $X$ be a variety. A subset $A$ of $X$ is constructible if it is a finite union of locally-closed subsets of $X$.

Chevalley

Let $f:X\to Y$ be any morphism. Then the image of $f$ is a constructible set in $Y$. More generally, $f$ maps constructible sets in $X$ to constructible sets in $Y$.

\begin{proof} 15:44

Upper semi-continuity of dimension

Let $f:X\to Y$ be any morphism. For all $x\in X$, define

$$e(x)=\left\{\begin{array}{ll}\max (\dim Z)& \begin{array}{l}Z\text{ a component of }\\ f^{-1}(f(x))\text{ containing }x\end{array}\end{array}\right\}.$$

Then $e$ is upper semi-continuous, i.e., for all integers $n$

$$S_n(f)=\{x\in X\mid e(x)\ge n\}$$

is closed.

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Lemma

Let $X,Y$ be affine varieties with $\Gamma (X)=R$ and $\Gamma (Y)=S$. For a morphism $f:X\to Y$, define $f^{\ast }:S\to R$. Then for an ideal $I\subseteq S$, we have $f^{-1}(V(I))=V(f^{\ast }(I)\cdot R)$.

\begin{proof} Note that $y\in f^{-1}(V(I))$ $⟺$ $f(y)\in V(I)$ $⟺$ for any $\alpha \in I$, $\alpha (f(y))=0$ $⟺$ for any $\alpha \in I$, $(f^{\ast }\alpha )=0$ iff $\beta \in f^{\ast }I,\beta (y)=0$ iff $y\in V(f^{\ast }(I)R)$. \end{proof}

Remark. Note that $f(V(J))\ne V((f^{\ast }{)}^{-1}(J))$. Take $f:Y_f\hookrightarrow Y$ and $Y_f=V(0)$.

Take an affine variety $Y=\mathrm{Spec}(S)$, and take nonzero $f\in S$ with $f\ne 1$. Note that $S_f=R=S[1/f]$. Also take $X=Y_f=\{y\in Y:f(y)\ne 0\}$. Consider an embedding $i:X\hookrightarrow Y$, and it induces $i^{\ast }:S\to S_f=R$.

Going-up, Lying over, Going down

讲完going up那堆，证明了

\operatorname{codim}(Z \text { in } X)=\operatorname{codim}(W \text { in } Y)

Link to original

然后开始证明

Let $f:X\to Y$ be any morphism. Then the image of $f$ is a constructible set in $Y$. More generally, $f$ maps constructible sets in $X$ to constructible sets in $Y$.

Link to original

i)→ii) 对每一片考虑嵌入映射和f的复合，还是morphism

i） We can assume $f$ is dominating. Then by ^9y9p1b, there exists open $U$ such that $f(X)\supseteq U$ and $Y\setminus U=\cup Y_i$ with $Y_i$ irreducible and $\dim Y_i<\dim Y$. Then $f^{-1}(Y_i)=\cup X_i^{i_j}$ with $X_i^{i_j}$ irreducible. Consider $f_i^{i_j}:X_i^{i_j}\to Y_i$. By induction on dimension of $Y_i$, we can assume $f_i^{i_j}(X_i^{i_j})$ is constructible. Then $f(X)=\cup \cup f_i^{i_j}(X_i^{i_j})$ is constructible. \end{proof}

Then begin to prove

is closed.

Link to original

\begin{proof} We can assume $f$ is dominating. Otherwise take $Z=\overline{f(X)}\subseteq Y$ and replace $Y$ by $Z$, which has no influence on $e$.

For $r=\dim X-\dim Y$, by ^9y9p1b there exists $U\subseteq f(X)\subseteq Y$ such that $\dim Z=r$ for any irreducible component of $f^{-1}(y)$ with $y\in U$.

i) $n⩽r$. (before $W\subseteq Y$ irreducible and $Z\subseteq f^{-1}(W)$ irreducible. Then $\overline{f(Z)}=W$ yields $\dim Z⩾\dim W+r$) Then $S_n=X$: For any $x\in X$, $W=f(x)$ with $\dim Z⩾r$, $Z$ component of $f^{-1}(f(x))$.

ii) $n>r$. Then $S_n\subseteq X\setminus f^{-1}(U)$. Consider $X\setminus f^{-1}(U)=\cup f^{-1}(Y_i)$ and $f_i^{ij}:X_i^{ij}\to Y_i$, where $\cup X_i^{ij}=f^{-1}(Y_i)$.

$S_n(f_i^{ij})$ is closed by induction hypothesis. It suffices to show $S_n(f)=\cup S_n(f_i^{ij})$. https://g.co/gemini/share/4567ec170e40

Gemini写了一个证明，我还没有读：

证明补全：

1. 归约: 如你所述，可以假设 $f$ 是主导态射（用 $Y$ 替换 $\overline{f(X)}$ 不影响 $e(x)$）。
2. Chevalley 定理: 设 $r=\dim X-\dim Y$。根据 Chevalley 定理 (Theorem 3)，存在 $Y$ 的一个非空开集 $U\subseteq f(X)$，使得对于所有 $y\in U$，纤维 $f^{-1}(y)$ 的所有不可约分支 $Z$ 都满足 $\dim Z=r$。因此，对于所有 $x\in f^{-1}(U)$，都有 $e(x)=r$。
3. 情况 i) $n\le r$:
   • 你正确地指出此时 $S_n(f)=X$。
   • 理由：根据纤维维数的下界定理 (Theorem 2)，对于任意 $y\in Y$，其纤维 $f^{-1}(y)$ 的任意不可约分支 $Z$ 都有 $\dim Z\ge r$。因此，对于任意 $x\in X$，$e(x)\ge r$。
   • 如果 $n\le r$，那么对于所有 $x\in X$ 都有 $e(x)\ge r\ge n$，所以 $S_n(f)=X$，$X$ 当然是闭集。
4. 情况 ii) $n>r$:
   • $S_n(f)$ 包含在闭集中: 你正确地指出 $S_n(f)\subseteq X\setminus f^{-1}(U)$。因为如果 $x\in f^{-1}(U)$，则 $e(x)=r<n$，所以 $x\notin S_n(f)$。$X\setminus f^{-1}(U)=f^{-1}(Y\setminus U)$ 是 $X$ 中的闭集。
   • 归纳设置: 令 $Y^{\prime }=Y\setminus U$。$Y^{\prime }$ 是 $Y$ 的真闭子集，其所有不可约分支 $Y_i$ 的维数都严格小于 $\dim Y$ ($\dim Y_i<\dim Y$)。
   • 令 $X^{\prime }=f^{-1}(Y^{\prime })=X\setminus f^{-1}(U)$。将 $X^{\prime }$ 分解为有限个不可约闭集 $X_{ij}$（其中 $X_{ij}$ 是 $X_i=f^{-1}(Y_i)$ 的不可约分支）。
   • 考虑限制态射 $f_{ij}:X_{ij}\to Y_i$。由于 $\dim Y_i<\dim Y$，我们可以对 $f_{ij}$ 应用归纳假设：集合 $S_n(f_{ij})=\{x\in X_{ij}\mid e_{ij}(x)\ge n\}$ 在 $X_{ij}$ 中是闭集。（这里 $e_{ij}(x)$ 是用态射 $f_{ij}$ 定义的纤维维数函数）。
   • 关键等式: 你卡住的地方是证明 $S_n(f)={\bigcup }_{i,j}{S}_n(f_{ij})$。我们来证明这个等式。
     • 证明 $S_n(f)\subseteq \cup S_n(f_{ij})$:
       • 设 $x\in S_n(f)$。因为 $n>r$，我们已知 $x\in X^{\prime }=\cup X_{ij}$。所以 $x$ 必然属于某个 $X_{ij}$。
       • $e(x)=\max \{\dim Z\mid Z\text{ 是 }{f}^{-1}(f(x))\text{ 包含 }x\text{ 的分支}\}$。
       • $e_{ij}(x)=\max \{\dim Z^{\prime }\mid Z^{\prime }\text{ 是 }{f}_{ij}^{-1}(f(x))=f^{-1}(f(x))\cap X_{ij}\text{ 包含 }x\text{ 的分支}\}$。
       • 设 $Z$ 是 $f^{-1}(f(x))$ 的一个包含 $x$ 且 $\dim Z=e(x)$ 的分支。因为 $x\in X_{ij}$ 且 $f(x)\in Y_i\subseteq Y^{\prime }$，所以 $Z\subseteq f^{-1}(Y_i)=X_i={\cup }_k{X}_{ik}$。$Z$ 作为一个不可约闭集，必须完全包含在 $X_i$ 的某个包含 $x$ 的不可约分支 $X_{ik}$ 中。
       • 如果 $k=j$，即 $Z\subseteq X_{ij}$，那么 $Z$ 也是 $f^{-1}(f(x))\cap X_{ij}$ 的一个不可约闭子集。事实上，$Z$ 必须是 $f^{-1}(f(x))\cap X_{ij}$ 的一个分支。（否则 $Z$ 会被 $f^{-1}(f(x))\cap X_{ij}$ 的一个更大的分支 $Z^{\prime }$ 真包含，而 $Z^{\prime }$ 也是 $f^{-1}(f(x))$ 的一个不可约闭子集，这意味着 $Z$ 不是 $f^{-1}(f(x))$ 的分支，矛盾。）
       • 因此 $\dim Z\le e_{ij}(x)$。这表明 $e(x)\le {\max }_{k\text{ s.t. }x\in X_{ik}}{e}_{ik}(x)$。
       • （反过来 $e(x)\ge e_{ij}(x)$ 也成立，因为 $f^{-1}(f(x))\cap X_{ij}$ 的分支也是 $f^{-1}(f(x))$ 的不可约闭子集）。
       • 所以 $e(x)={\max }_{k\text{ s.t. }x\in X_{ik}}{e}_{ik}(x)$。
       • 因此，如果 $e(x)\ge n$，那么必然存在某个 $j$ 使得 $x\in X_{ij}$ 且 $e_{ij}(x)\ge n$，即 $x\in S_n(f_{ij})$。
     • 证明 $\cup S_n(f_{ij})\subseteq S_n(f)$:
       • 设 $x\in S_n(f_{ij})$ 对于某个 $i,j$。这意味着 $x\in X_{ij}$ 并且 $e_{ij}(x)\ge n$。
       • $e_{ij}(x)$ 是 $f^{-1}(f(x))\cap X_{ij}$ 的某个包含 $x$ 的分支 $Z^{\prime }$ 的维数。
       • $Z^{\prime }$ 也是 $f^{-1}(f(x))$ 的一个包含 $x$ 的不可约闭子集。它必然包含在 $f^{-1}(f(x))$ 的某个包含 $x$ 的分支 $Z$ 中。
       • 所以 $\dim Z^{\prime }\le \dim Z\le e(x)$。
       • 因此 $e(x)\ge e_{ij}(x)\ge n$。故 $x\in S_n(f)$。
     • 等式成立: $S_n(f)={\bigcup }_{i,j}{S}_n(f_{ij})$。
   • 得出结论:
     • 根据归纳假设，$S_n(f_{ij})$ 在 $X_{ij}$ 中是闭集。
     • 因为 $X_{ij}$ 是 $X$ 中的闭集，所以 $S_n(f_{ij})$ 在 $X$ 中也是闭集。
     • $S_n(f)$ 是有限个闭集 $S_n(f_{ij})$ 的并集。
     • 有限个闭集的并集是闭集。
   • 最终结论: 因此 $S_n(f)$ 在 $X$ 中是闭集。

\end{proof}

这个证明的关键在于利用归纳法，将问题从 $Y$ 归约到维数更低的闭子集 $Y_i$，并证明了维数函数 $e(x)$ 和限制映射的维数函数 $e_{ij}(x)$ 之间的关系，从而可以将闭集性质从 $S_n(f_{ij})$ “并”回 $S_n(f)$。

If it holds, then $S_n(f)$ is a union of closed set and so it is also closed. \end{proof}

Counterexample. 然后又讲了一个定理 18“09

有一个反例

blow up

A3 * A1 中一条线（1维） blow up后扣掉某个纤维（2维）

投影到A3上，只有一个点是1维的，剩下2维的

Definition

$f:X\to Y$ is a birational morphism if $k(Y)\stackrel{f^{\ast }}{\to }k(X)$ is an isomorphism.

Theorem

$f:X\to Y$ is birational if there exists $U\subseteq Y$ such that $f^{-1}(U)\stackrel{\simeq }{\to }U$ is an isomorphism.

\begin{proof} Recall that $k(f^{-1}(U))\simeq k(U)\simeq k(Y)$.

Replacing $Y$ be an affine open set. We can assume $Y$ is affine, and assume $\Gamma (Y)=S$. Let $U\subseteq X$ be an affine open set with $\Gamma (U)=R$ and $\overline{f(X-U)}=W$.

$Z$ is a component of $W$, has $\dim Z<\dim X$. Note that $\dim Y=\mathrm{tr}\mathrm{deg}k(Y)=\mathrm{tr}\mathrm{deg}k(X)=\dim X$ and $\dim (X-U)<\dim X$. Then we have

$$\dim Z⩽\dim \overline{f(X-U)}⩽\dim (X-U)<\dim X$$

Pick $g\in S=\Gamma (Y)$ such that $g=0$ on $W$ but $g\ne 0$. We have $W⊊Y$ and $I(W)\ne (0)$ by $g\in I(W)\setminus \{0\}$. Then $f^{-1}(Y_{(g)})\subseteq U$. Consider

$$U\to Y,S\to R,f^{-1}(V(I))=V(I\cdot R)\text{\hspace{1em}(*)}$$

by last class. then $f^{-1}(Y_{(g)})=U_{(g)}$.

Then $\{g=0\}\supseteq \overline{f(X-U)}$ yields $f^{-1}(\{g=0\})\supseteq X-U$.

By computation, we have $f^{-1}(Y_{(g)})=U-V(gR)=U_{(g)}$ and then $U_{(g)}$ is affine.

$k(U_{(g)})=k(Y_{(g)})$. $\Gamma (U_{(g)})$ is finitely generated by $x_1,\cdots ,x_m$ as a $k$-algebra. Assume $x_1=y_1/h,\cdots ,x_m=y_m/h$ with $y_i,h\in \Gamma (Y_{(g)})$.

There is an isomorphism $U_{(g,h)}\to Y_{(g,h)}$, because $\Gamma (Y_{g,h})\hookrightarrow \Gamma (U_{g,h})$ is an isomorphism where $\Gamma (U_{(h)})\simeq \tilde{R}$ and $\Gamma (U_{(g,h)})={\tilde{R}}_{(g)}$.

We firstly find an affine variety of $X$ containing $f^{-1}(U)$. Then shrink $U_{(g)}$ to an subset which is isomorphic to $f^{-1}(U)$.