1.9 Complete Variety

结式 Resultant. For $f(x)=a_n{x}^n+\cdots +a_0$ and $g(x)=b_m{x}^m+,,,+b_m$, there exists $\mathrm{Res}(f,g)\in k[a_0,\cdots ,a_n,b_0,\cdots ,b_m]$ such that $f(x),g(x)$ have a common root iff $\mathrm{Res}(f,g)=0$.

Higher-dimensional case. (elimination theory).

Main theorem of elimination theory

Given $r$ polynomials, with coefficients in $k$ :

$$\begin{array}{c}{f}_1\left(x_0,\dots ,x_n;y_1,\dots ,y_m\right)\\ \dots \dots \dots \dots \dots \dots \dots \\ f_r\left(x_0,\dots ,x_n;y_1,\dots ,y_m\right)\end{array}$$

all of which are homogeneous in the variables $x_0,\dots ,x_n$, there is a second set of polynomials (with coefficients in $k$ ):

$$\begin{array}{c}{g}_1\left(y_1,\dots ,y_m\right)\\ \dots \dots \dots .\\ g_{\nu }\left(y_1,\dots ,y_m\right)\end{array}$$

such that for all $m$-tuples $\left(a_1,\dots ,a_m\right)$ in $k,g_i\left(a_1,\dots ,a_m\right)=0$, all $i$ if and only if there is a non-zero $(n+1)$-tuple $\left(b_0,\dots ,b_n\right)$ in $k$ such that $f_i\left(b_0,\dots ,b_m;a_1,\dots ,a_m\right)=0$, all $i$.

For any closed set $Z\subseteq {\mathbb{P}}^n\times {\mathbb{A}}^m$, $p_r(Z)\subseteq {\mathbb{A}}^m$ is closed, where

$$\begin{array}{rl}{p}_r:{\mathbb{P}}^n\times {\mathbb{A}}^m& \to {\mathbb{A}}^m\\ \left(\left[b_0:\cdots :b_n\right],\left(a_1,\dots ,a_m\right)\right)& \mapsto \left(a_1,\dots ,a_m\right)\end{array}$$

\begin{proof} $X=\cup X_i$ where $X_i$ are affine open set. If the statement holds for any $X_i$, so for $X$. Define $Y={\mathbb{P}}^n$.

Since $Z\subseteq Y\times X$ is closed, $Z\cap (Y\cap X_i)$ is closed in $Y\times X_i$. Then $p_r(Z\cap (Y\times X_i))=p_r(Z)\cap X_i$ is closed in $X_i$.

It remains to show $p_r(z)\cap X_i$ closed in $X_i$ yields $p_i(Z)$ closed. Notice that

$$X-p_r(Z)=(\cup X_i)\cap (p_r(Z{)}^c)=\cup (X_i\cap p_r(Z{)}^c)=\cup (X_i-p_r(Z))$$

is open.

Now we can assume $X$ is affine. There is an embedding $X\hookrightarrow {\mathbb{A}}^m$. Then $Z\subseteq Y\times X\hookrightarrow Y\times {\mathbb{A}}^m$. consider a commutative diagram 15:08

\end{proof}

Definition

We say $X$ is a complete variety, if for any $Y$ variety, $X\times Y\stackrel{p_r}{\to }Y$ is a closed map. (universal closed)

If $X$ is a complex manifold, then $X$ is universally closed iff $X$ is compact.

A compact complex manifold is universal closed.

Link to original

"→" If $X$ is universal closed, then $X$ is compact.

(exist compactification $\tilde{X}$ and $X\hookrightarrow \tilde{X}$ is an open dense subset)

Consider ${\Gamma }_j:\{(x,x)\in X\times \tilde{X}:x\in X\}$. If $\tilde{X}$ is Hausdorff, then ${\Gamma }_j$ is closed: because ${\Gamma }_j={\theta }^{-1}(\Delta )$ where $\theta :X\times \tilde{X}\to \tilde{X}\times \tilde{X}$.

Then $X=p_r({\Gamma }_j)$ is closed in $\tilde{X}$ and so $X=\overline{X}=\tilde{X}$. Hence $X$ is compact.

Proposition

Let $f:X\to Y$ be a morphism, with $X$ complete, then $f(X)$ is closed in $Y$ and is complete.

\begin{proof} As $Y$ is a variety, ${\Gamma }_f=\{(x,f(x))\in X\times Y\}\subseteq X\times Y$ closed. 15:29

Proposition

If $X$ and $Y$ are complete, then $X\times Y$ is complete.

\begin{proof} Since $X\times Y\times Z\to Y\times Z\to Z$ is closed, we have $(X\times Y)\times Z\to Z$ is closed. \end{proof}

Proposition

If $X$ is complete and $Y\subset X$ is a closed subvariety, then $Y$ is complete.

\begin{proof} 15:35 \end{proof}

Later, we will show ${\mathbb{P}}^n$ is complete. By ^tbgc8l any projective variety is complete.

Proposition

An affine variety $X$ is complete only if $\dim X=0$, i.e., $X$ consists in a single point.

\begin{proof}

15:44

\end{proof}

Nakayama lemma

Let $M$ be a finitely generated $R$-module, and let $I\subseteq R$ be an ideal of $R$. If $M=IM$, then there exists $f=1+r$ with $r\in I$ such that $f\cdot M=0$.

Remark. If we write $f=1-i$, then $fm=(1-i)m=0$ and so $m=im$ for any $m$. In wikipedia, this lemma has slogan as $M=IM$ yields there exists $i\in I$ such that $m=im$.

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main theory of elimination theory

${\mathbb{P}}_k^n$ is a complete variety.

\begin{proof} For a given variety $Y$ and $p:{\mathbb{P}}^n\times Y\to Y$, we aim to show $p$ is closed. Assume $Y={\cup }_{i=1}^{\ell }{U}_i$ with open affine subvarieties $U_i$, if ${\mathbb{P}}_k^n\times U_i\to U_i$ is closed, then $p$ is closed. Hence, we can assume that $Y$ is an affine variety and $\Gamma (Y)=R$.

For closed $Z\subseteq {\mathbb{P}}^n\times Y$, ${\mathbb{P}}^n=\cup U_i$ with $\Gamma (U_i)=k[x_0/x_i,\cdots ,x_n/x_i]$. Then $Z\subseteq {\mathbb{P}}^n\times Y=\cup (U_i\times Y)$, where $\Gamma (U_i\times Y)=k[x_0/x_i,\cdots ,x_n/x_0]{\otimes }_{k}R=R[x_0/x_1,\cdots ,x_n/x_0]$. Let $S=R[x_0,\cdots ,x_n]$, and let $S_m=$ degree $m$ pieces. Define $A_m=\{f\in S_m:(x_0/x_i,\cdots ,x_n/x_i)\in I(U_i\times Y)\cap Z,\forall i\}$ and $A={\sum }_{m\in \mathbb{N}}{A}_m$.

Lemma

Fix $i$ and $g\in I((U_i\times Y)\cap Z)$, there exists $f\in A_m$ for some $m$ such that $f(x_0/x_1,\cdots ,x_n/x_i)=g$.

\begin{proof} Since $g\in \Gamma (U_i\times Y)=R[x_0/x_i,\cdots ,x_n/x_i]$, there exists $N$ such that $x_i^N\cdot g$ is a homogenous polynomial of degree $N$, and $f(x_0/x_i,\cdots ,x_n/x_i)=g$. However, $f\in A_m$ not necessarily holds, that is, $f(x_0/x_j,\cdots ,x_n/x_j)$ may not contained in $I(U_j\times Y\cap Z)$ for all $j$.

Notices that $(U_j\times Y)-(U_i\cap U_j)\times Y\subseteq \{x_i=0\}\times Y$. Then for any $j$, $x_{i}f(x_0/x_j,\cdots ,x_n/x_j)\in I(U_j\times Y\cap Z)$, which is what we desired. \end{proof}

Now we go back to the proof of the theorem.

To show $p(Z)\subseteq Y$ is closed, it suffices to show for any $y\notin p(Z)$, there exists an open subvariety $y\in V$ such that $V\cap p(Z)=\varnothing$.

Define $m_y\subseteq R$ as the maximal ideal of $y$. Recall that for $p:U_i\times Y\to Y$, there is $I(p^{-1}(y))=I(U_i\times \{y\})=m_y\cdot R[x_0/x_i,\cdots ,x_n/x_i]$. If $V\cap p(Z)=\varnothing$ holds, then

$$I(U_i\times \{y\})+I(V_i\times Y\cap Z)=R[x_0/x_i,\cdots ,x_n/x_i]$$

and so $I(U_i\times \{y\})+m_{y}R[x_0/x_i,\cdots ,x_n/x_i]=R[x_0/x_i,\cdots ,x_n/x_i]$. Thus, there exists $g\in I((U_i\times Y)\cap Z)$, $m_{\alpha }\in m_y$ and $g_{\alpha }\in R[x_0/x_i,\cdots ,x_n/x_i]$ such that $1=g+{\sum }_{\alpha }{m}_{\alpha }{g}_{\alpha }$. There exists $N\gg 1$ such that $x_i^{N}g\in A_N$ and $x_i^N=x_i^{N}g+\sum m_{\alpha }(x_i^N{g}_{\alpha })\in A_N+m_y\cdot S_N$.

Let $h$ be a monomial of $\mathrm{deg}h=d⩾(n+1)\cdot N$. There exists $j$ such that $x_j^N\mid h$ and $h\in A_d+m_y\cdot S_d$. In other words, if $d⩾(n+1)\cdot N$, $S_d\subseteq A_d+m_y\cdot S_d$.

Note that $S_d/A_d$ is an $R$-module and $m_y\subseteq R$ is a maximal ideal. Since $m_y(S_d/A_d)=A_d+m_y\cdot S_d/A_d\supseteq S_d/A_d$, by ^0ws17n, there exists $f\in 1+m_y$ such that $f\cdot S_d/A_d=0$, i.e., $f\cdot S_d\subseteq A_d$.

Let $V=Y_f=\{y:f(y)\ne 0\}$, and we claim that $V\cap p(Z)=\varnothing$.

It suffices to show for any $z\in Z$, there is $f(p(z))=0$. Since $z\in U_j\times Y$ for some $j$, we have $z=([x_0:\cdots :x_n],y)$ and $x_j\ne 0$. Recall that $f\cdot S_d\subseteq A_d$, there is $f{x}_j^d\in A_d$ and so $f{x}_j^d/x_k^d$ is zero on $(U_k\times Y)\cap Z$. In particular, take $k=j$ and $f(p(z))=0$ for all $z\in U_j\times Y$. Now we finish the proof. \end{proof}

Corollary

Every projective variety is complete.

\begin{proof} For a closed variety $X\subseteq {\mathbb{P}}_k^n$, $X\times Y$ is a closed set of ${\mathbb{P}}_k^n\times Y$. For any closed $Z\subseteq X\times Y$, we have $p_X(Z)=p_{{\mathbb{P}}_k^n}(X)\subseteq Y$ is closed in $Y$ and so $X$ is complete. \end{proof}