update

Author: dirichy

.DS_Store

@@ -10,102 +10,102 @@
 \large
 \setlength{\baselineskip}{1.2em}
 \ifpreface
-  \input{../../../global/preface}
-  \newgeometry{left=2cm,right=2cm,top=2cm,bottom=2cm}
+	\input{../../../global/preface}
+	\newgeometry{left=2cm,right=2cm,top=2cm,bottom=2cm}
 \else
-  \newgeometry{left=2cm,right=2cm,top=2cm,bottom=2cm}
-  \maketitle
+	\newgeometry{left=2cm,right=2cm,top=2cm,bottom=2cm}
+	\maketitle
 \fi
 %from_here_to_type
 \newcommand{\A}{\mathbbm{A}}
 \begin{problem}
-  $P$ is an ideal of a unitary commutative ring $A$, then $P$ is prime ideal of $A\iff A/P$ is integral domain.
+$P$ is an ideal of a unitary commutative ring $A$, then $P$ is prime ideal of $A\iff A/P$ is integral domain.
 \end{problem}
 \(\)
 \begin{solution}
-  $\Rightarrow $:
+	$\Rightarrow $:
 
-  Since $A$ is a unitary commutative ring, so $A/P$ is unitary commutative ring, too. So we only need to prove $[ab]=[0]\Rightarrow [a]=[0]\xor [b]=[0]$.
-  Obviously $[ab]=0\iff ab\in P\iff a\in P\xor b\in P\iff [a]=[0]\xor [b]=[0]$.
+	Since $A$ is a unitary commutative ring, so $A/P$ is unitary commutative ring, too. So we only need to prove $[ab]=[0]\Rightarrow [a]=[0]\xor [b]=[0]$.
+	Obviously $[ab]=0\iff ab\in P\iff a\in P\xor b\in P\iff [a]=[0]\xor [b]=[0]$.
 
-  $\Leftarrow $:
+	$\Leftarrow $:
 
-  As the same, $ab\in P\iff [ab]=[0]\Rightarrow [a]=[0]\xor [b]=[0]\iff a\in P\xor b\in P$, so $P$ is prime ideal.
+	As the same, $ab\in P\iff [ab]=[0]\Rightarrow [a]=[0]\xor [b]=[0]\iff a\in P\xor b\in P$, so $P$ is prime ideal.
 \end{solution}
 
 \begin{problem}
-  $M$ is an ideal of a unitary commutative ring $A$, then $M$ is maximal ideal of $A \iff A/M$ is a field.
+$M$ is an ideal of a unitary commutative ring $A$, then $M$ is maximal ideal of $A \iff A/M$ is a field.
 \end{problem}
 \begin{solution}
-  $\Rightarrow $:
+	$\Rightarrow $:
 
-  Consider $[a]\in A/M\minus [0]$, we will prove it has a reverse.  Consider $N:=\{xm+ya:x,y\in A,m\in M\}$ is the minimum ideal of $A$ contains $M$ and $a$.
-  Since $[a]\neq [0]$ we know $a\notin M$, so $M\subsetneqq N$.
-  Noting $M$ is maximal, so $N=A$. That means $\exists x,y\in A,m\in M,xm+ya=1$. So $[xm+ya]=[1]$. Since $[xm]=[0]$ we get $[y][a]=1$, i.e., $[y]=[a]^{-1}$.
+	Consider $[a]\in A/M\minus [0]$, we will prove it has a reverse.  Consider $N:=\{xm+ya:x,y\in A,m\in M\}$ is the minimum ideal of $A$ contains $M$ and $a$.
+	Since $[a]\neq [0]$ we know $a\notin M$, so $M\subsetneqq N$.
+	Noting $M$ is maximal, so $N=A$. That means $\exists x,y\in A,m\in M,xm+ya=1$. So $[xm+ya]=[1]$. Since $[xm]=[0]$ we get $[y][a]=1$, i.e., $[y]=[a]^{-1}$.
 
-  $\Leftarrow $:
+	$\Leftarrow $:
 
-  Consider $a\in A\minus M, N:=\{xp+ya:x,y\in A,p\in P\}$, we will prove $N=A$, which means $M$ is maximal.
-  Since $A/M$ is field, $\exists y\in A, [y]=[a]^{-1}$. That's means $ay-1\in M\subset N$. Noting $ay\in N$, so $1\in N$, thus $N=A$.
+	Consider $a\in A\setminus M, N:=\{xp+ya:x,y\in A,p\in P\}$, we will prove $N=A$, which means $M$ is maximal.
+	Since $A/M$ is field, $\exists y\in A, [y]=[a]^{-1}$. That's means $ay-1\in M\subset N$. Noting $ay\in N$, so $1\in N$, thus $N=A$.
 \end{solution}
 
 \begin{problem}
-  A ring $A$ is noetherian, $I\subset A$ is an ideal of $A$, then $A/I$ is noetherian.
+A ring $A$ is noetherian, $I\subset A$ is an ideal of $A$, then $A/I$ is noetherian.
 \end{problem}
 \begin{solution}
-  Consider an ideal $J\subset A/I$, let $M:=\{x\in A:[x]\in J\}$. Then $\forall a\in A,x\in M,[ax]=[a][x]\in J$, so $ax\in M$. $\forall a,b\in M,[a-b]=[a]-[b]\in J$, so $a-b\in M$. So $M$ is an ideal of $A$.
-  Since $A$ is noetherian, we can assume $M=(f_i,i=1,2,\cdots n)$. Now we will prove $J=([f_i],i=1,2,\cdots n)$.
-  Consider $[f]\in J$, from definition of $M$ we know $f\in M$, so $f=\sum_{i=1}^na_if_i,a_i\in A$, thus $[f]=\left[\sum_{i=1}^na_if_i\right]=\sum_{i=1}^n[a_i][f_i]$. So $J=([f_i],i=1,2,\cdots n)$.
+	Consider an ideal $J\subset A/I$, let $M:=\{x\in A:[x]\in J\}$. Then $\forall a\in A,x\in M,[ax]=[a][x]\in J$, so $ax\in M$. $\forall a,b\in M,[a-b]=[a]-[b]\in J$, so $a-b\in M$. So $M$ is an ideal of $A$.
+	Since $A$ is noetherian, we can assume $M=(f_i,i=1,2,\cdots n)$. Now we will prove $J=([f_i],i=1,2,\cdots n)$.
+	Consider $[f]\in J$, from definition of $M$ we know $f\in M$, so $f=\sum_{i=1}^na_if_i,a_i\in A$, thus $[f]=\left[\sum_{i=1}^na_if_i\right]=\sum_{i=1}^n[a_i][f_i]$. So $J=([f_i],i=1,2,\cdots n)$.
 \end{solution}
 
 \begin{problem}
-  \label{pro:4}
-  $K$ is a field, $A=K[x_1,x_2,\cdots x_n],\A^n_K=K^n$. For ideal $I$ of $A$ let $V(I):=\{p\in\A_K^n:f(p)=0,\forall f\in I\}$. Then $V(I_1)\cup V(I_2)=V(I_1\cap I_2)=V(I_1I_2)$
+\label{pro:4}
+$K$ is a field, $A=K[x_1,x_2,\cdots x_n],\A^n_K=K^n$. For ideal $I$ of $A$ let $V(I):=\{p\in\A_K^n:f(p)=0,\forall f\in I\}$. Then $V(I_1)\cup V(I_2)=V(I_1\cap I_2)=V(I_1I_2)$
 \end{problem}
 \begin{solution}
-  \begin{lemma}
-    \label{lem:1}
-    $I\subset J\Rightarrow V(I)\supset V(J)$.
-  \end{lemma}
-  \begin{proof}
-    Consider $p\in V(J)$, we get $\forall f\in J,f(p)=0$. Since $I\subset J$, so $\forall f\in I,f(p)=0$, i.e., $f\in V(I)$.
-  \end{proof}
-  From Lemma \Cref{lem:1} we know $V(I_1\cap I_2)\subset V(I_1I_2)$, so we only need to prove $V(I_1)\cup V(I_2)\subset V(I_1\cap I_2),V(I_1I_2)\subset V(I_1)\cup V(I_2)$.
-  \begin{itemize}
-    \item $V(I_1)\cup V(I_2)\subset V(I_1\cap I_2)$:
-      From \Cref{lem:1} It's obvious.
-    \item $V(I_1I_2)\subset V(I_1)\cup V(I_2)$:
-      Consider $p\in V(I_1I_2)$.
-      If $p\notin V(I_1)\cup V(I_2)$, then $\exists f_1\in I_1,f_2\in I_2, f_1(p)\neq 0,f_2(p)\neq 0$. Now consider $f=f_1f_2\in I_1I_2$, we get $f(p)=f_1(p)f_2(p)\neq 0$, so $p\notin V(I_1I_2)$, it's a contradiction.
-  \end{itemize}
+	\begin{lemma}
+		\label{lem:1}
+		$I\subset J\Rightarrow V(I)\supset V(J)$.
+	\end{lemma}
+	\begin{proof}
+		Consider $p\in V(J)$, we get $\forall f\in J,f(p)=0$. Since $I\subset J$, so $\forall f\in I,f(p)=0$, i.e., $f\in V(I)$.
+	\end{proof}
+	From Lemma \Cref{lem:1} we know $V(I_1\cap I_2)\subset V(I_1I_2)$, so we only need to prove $V(I_1)\cup V(I_2)\subset V(I_1\cap I_2),V(I_1I_2)\subset V(I_1)\cup V(I_2)$.
+	\begin{itemize}
+		\item $V(I_1)\cup V(I_2)\subset V(I_1\cap I_2)$:
+		      From \Cref{lem:1} It's obvious.
+		\item $V(I_1I_2)\subset V(I_1)\cup V(I_2)$:
+		      Consider $p\in V(I_1I_2)$.
+		      If $p\notin V(I_1)\cup V(I_2)$, then $\exists f_1\in I_1,f_2\in I_2, f_1(p)\neq 0,f_2(p)\neq 0$. Now consider $f=f_1f_2\in I_1I_2$, we get $f(p)=f_1(p)f_2(p)\neq 0$, so $p\notin V(I_1I_2)$, it's a contradiction.
+	\end{itemize}
 \end{solution}
 
 \begin{problem}
-  $K$ is an infinite field, then $\A_k^n$ is not Hausdorff.
+$K$ is an infinite field, then $\A_k^n$ is not Hausdorff.
 \end{problem}
 \begin{solution}
-  \begin{lemma}
-    \label{lem:2}
-    $K$ is an infinite field, $f\in K[x_1,x_2,\cdots x_n]\setminus \{0\}$, then exists $p\in \A_K^n,f(p)\neq 0$.
-  \end{lemma}
-  \begin{proof}
-    Use MI. When $n=0,K[x_1,x_2,\cdots x_n]=K$, so it's obvious. Assume for $n=k$ it's right, when goes to $k+1$:
+	\begin{lemma}
+		\label{lem:2}
+		$K$ is an infinite field, $f\in K[x_1,x_2,\cdots x_n]\setminus \{0\}$, then exists $p\in \A_K^n,f(p)\neq 0$.
+	\end{lemma}
+	\begin{proof}
+		Use MI. When $n=0,K[x_1,x_2,\cdots x_n]=K$, so it's obvious. Assume for $n=k$ it's right, when goes to $k+1$:
 
-    Consider $h\in K(x_1,x_2,\cdots x_k)[x_{k+1}],h(x_{k+1}):=f(x_1,x_2,\cdots x_k,x_{k+1})$ is a non-zero polynomial so it has finite root. So exists $a\in K,h(a)\neq 0$. So $g:=f(x_1,x_2,\cdots x_k,a)\in K[x_1,x_2,\cdots x_k]\neq 0$.
-    By induction hypothesis we get $\exists b\in \A_K^k,g(b)\neq 0$. Let $p:=(b,a)\in \A_K^{k+1}$, then $f(p)\neq 0$.
-  \end{proof}
-  In fact, it's not only not Hausdorff, it's kind of ''absolutly not Hausdorff'' because every pair of point can't be seperated.
-  Consider two point $p\neq q,p,q\in \A_K^n$.
-  Assume $p=(p_1,p_2,\cdots p_n),q=(q_1,q_2,\cdots q_n),p_1\neq q_1$. Assume two open set $V(I_1)^c,V(I_2)^c$ can seperate $p,q$, then $V(I_1)\cup V(I_2)=\A_K^n$.
-  From \Cref{pro:4} we know $V(I_1I_2)=\A_K^n$. So $\forall f\in V(I_1I_2),\forall p\in \A_K^n,f(p)=0$.
-  Then from Lemma\Cref{lem:2} we can get $f=0$. So $I_1I_2=\{0\}$.
-  Since $p\notin V(I_1),q\notin V(I_2)$, we know $I_1,I_2\neq \{0\}$. So $\exists f_1\in I_1\neq 0,\exists f_2\in I_2\neq 0$, and thus $f=f_1f_2\in I_1I_2\neq 0$, contradiction!
-  \(\phi\)
-  So there is not a pair of points can be seperated by two disjoint open set.
+		Consider $h\in K(x_1,x_2,\cdots x_k)[x_{k+1}],h(x_{k+1}):=f(x_1,x_2,\cdots x_k,x_{k+1})$ is a non-zero polynomial so it has finite root. So exists $a\in K,h(a)\neq 0$. So $g:=f(x_1,x_2,\cdots x_k,a)\in K[x_1,x_2,\cdots x_k]\neq 0$.
+		By induction hypothesis we get $\exists b\in \A_K^k,g(b)\neq 0$. Let $p:=(b,a)\in \A_K^{k+1}$, then $f(p)\neq 0$.
+	\end{proof}
+	In fact, it's not only not Hausdorff, it's kind of ''absolutly not Hausdorff'' because every pair of point can't be seperated.
+	Consider two point $p\neq q,p,q\in \A_K^n$.
+	Assume $p=(p_1,p_2,\cdots p_n),q=(q_1,q_2,\cdots q_n),p_1\neq q_1$. Assume two open set $V(I_1)^c,V(I_2)^c$ can seperate $p,q$, then $V(I_1)\cup V(I_2)=\A_K^n$.
+	From \Cref{pro:4} we know $V(I_1I_2)=\A_K^n$. So $\forall f\in V(I_1I_2),\forall p\in \A_K^n,f(p)=0$.
+	Then from Lemma\Cref{lem:2} we can get $f=0$. So $I_1I_2=\{0\}$.
+	Since $p\notin V(I_1),q\notin V(I_2)$, we know $I_1,I_2\neq \{0\}$. So $\exists f_1\in I_1\neq 0,\exists f_2\in I_2\neq 0$, and thus $f=f_1f_2\in I_1I_2\neq 0$, contradiction!
+	\(\phi\)
+	So there is not a pair of points can be seperated by two disjoint open set.
 \end{solution}
 \begin{figure}
-  \centering
-  \includegraphics{~/Pictures/2-3.jpg}%![aaa](~/Pictures/2-3.jpg)
+	\centering
+	\includegraphics{~/Pictures/2-3.jpg}%![aaa](~/Pictures/2-3.jpg)
 
 \end{figure}
 \end{document}

AlgebraicGeometry/.DS_Store

@@ -11,11 +11,11 @@
 \large
 \setlength{\baselineskip}{1.2em}
 \ifpreface
-  \input{../../../global/preface}
-  \newgeometry{left=2cm,right=2cm,top=2cm,bottom=2cm}
+\input{../../../global/preface}
+\newgeometry{left=2cm,right=2cm,top=2cm,bottom=2cm}
 \else
-  \newgeometry{left=2cm,right=2cm,top=2cm,bottom=2cm}
-  \maketitle
+\newgeometry{left=2cm,right=2cm,top=2cm,bottom=2cm}
+\maketitle
 \fi
 %from_here_to_type
 \begin{problem}
@@ -34,9 +34,6 @@
   So \(\ker \theta = \ide(V)\). And easily \(\theta\) is surjective, so we get
   \(m_p=M_p/\ker \theta=M_p / \ide(V)\).
 \end{solution}
-\begin{enumerate}
-  \item
-\end{enumerate}
 
 \begin{problem}
   Prove that \(M_p/M_p^2+\ide(V) \cong m_p/m_p^2\).
@@ -73,5 +70,6 @@
 
   Obviously \(\tau\) is injective, so it's isomorphic.
 \end{solution}
-
 \end{document}
+
+

AlgebraicGeometry/1/.DS_Store

@@ -0,0 +1 @@
+RiemannGeometry

AlgebraicGeometry/1/solution/AlgebraicGeometry1.tex

@@ -0,0 +1 @@
+1

AlgebraicGeometry/10/solution/AlgebraicGeometry10.pdf

@@ -0,0 +1,144 @@
+%arara: xelatex
+%!Mode:: "TeX:UTF-8"
+%!TEX TS-program = xelatex
+\documentclass{ctexart}
+\newif\ifpreface
+\prefacetrue
+\input{../../../global/all}
+\DeclareMathOperator{\sgn}{sgn}
+\begin{document}
+\large
+\setlength{\baselineskip}{1.2em}
+\ifpreface
+\input{../../../global/preface}
+\else
+\maketitle
+\fi
+\newgeometry{left=2cm,right=2cm,top=2cm,bottom=2cm}
+\everymath{\displaystyle}
+\newlength\inlineHeight
+\newlength\inlineWidth
+\long\def\(#1\){%
+  \settoheight{\inlineHeight}{$#1$}%
+  \settowidth{\inlineWidth}{$#1$}%
+  \ifdim \inlineWidth > 0.5\textwidth%
+  $$#1$$%
+  \else%
+  \ifdim \inlineHeight > 1.5em%
+  $$#1$$%
+  \else%
+  $#1$%
+  \fi%
+  \fi%
+}
+%from_here_to_type
+\begin{problem}\label{pro:1.1}
+  Assume \(\mathcal{A}_0=\{(U_\alpha,\phi_\alpha):\alpha \in I\}\) is a \(C^r\)-compatible coordinate covery of a \(m\)-dimensional manifold \(M\), let
+  \(\mathcal{A}:= \{(U,\phi): (U,\phi) \text{ is chart of } M, \AND \forall (V,\psi)\in \mathcal{A}_0,(U,\phi)\text{ is compatible with }(V,\psi)\}\).
+  Then \(\mathcal{A}\) is unique \(C^r\)-differential structure on \(M\) contains \(\mathcal{A}_0\).
+\end{problem}
+\begin{solution}
+
+  First, easily \(\mathcal{A}_0 \subset \mathcal{A}\) by definition of \(\mathcal{A}_0\).
+  Now we should prove \(\mathcal{A}\) is differential structure on \(M\). Let \((U,\phi),(V,\psi) \in \mathcal{A}\).
+  If \((U,\phi)\in \mathcal{A}_0\), then by definition of \(\mathcal{A}_0\) we know \((U,\phi)\) is compatible to \((V,\psi)\).
+  If \((U,\phi),(V,\psi) \notin \mathcal{A}_0\), then consider \(U\cap V\). If \(U\cap V=\varnothing\), then \((U,\phi)\) is compatible with \((V,\psi)\).
+  Now assume \(W=U \cap V\neq \varnothing\). Consider \(\gamma:=\psi \circ \phi^{-1} :\phi(W) \to \psi(W)\).
+  For any \(x \in W\), since \(\mathcal{A}_0\) is covery of \(M\), we know \(\exists (T,\tau) \in \mathcal{A}_0,x \in T\).
+  Then by the definition of \(\mathcal{A}\), we know \((T,\tau)\) is compatible locally on \(x\) with both \((U,\phi)\) and \((V,\psi)\).
+  So \((U,\phi)\) is locally compatible with \((V,\psi)\) on \(x\). Since \(x\) is arbitrary, we know \((U,\phi)\) is compatible with \((V,\psi)\).
+  So \(\mathcal{A}\) is differential structure of \(M\).
+
+  Now we assume \(\mathcal{B}\) is another differential structure of \(M\) contains \(\mathcal{A}_0\).
+  Since \(\mathcal{B}\) is compatible, we get \(\mathcal{B} \subset \mathcal{A}\).
+  Since \(\mathcal{B}\) is maximal, we get \(\mathcal{B}=\mathcal{A}\).
+  So \(\mathcal{A}\) is unique.
+\end{solution}
+\begin{problem}\label{pro:1.2}
+  Assume \((U,\phi;x^i),(V,\psi;y^i),(W,\chi;z^i)\) are three local coordinate on an \(m\)-dimensional smooth manifold \(M\), and \(W \cap V \cap U \neq \varnothing\).
+  Prove that on \(\phi(U \cap V \cap W)\), we have:
+  \(\left( \frac{\partial z^i}{\partial x^j}\right)=\left(\frac{\partial z^i}{\partial y^k}\right)\left(\frac{\partial y^k}{\partial x^j}\right)\).
+\end{problem}
+\begin{solution}
+  For fixed \(1 \leq i,j \leq m\), we have \(\frac{\partial z^i}{\partial x^j}=\sum_{k=1}^{m} \frac{\partial z^i}{\partial y^k} \frac{\partial y^k}{\partial x^j}\).
+  So easily to get that \(\left( \frac{\partial z^i}{\partial x^j}\right)=\left(\frac{\partial z^i}{\partial y^k}\right)\left(\frac{\partial y^k}{\partial x^j}\right)\).
+
+  We let \((W,\chi;z^i)=(U,\phi;x^i)\), then we get:
+  \(I_m=\left(\frac{\partial x^i}{\partial y^k}\right)\left(\frac{\partial y^k}{\partial x^j}\right)\).
+  So both terms on the right side are invertible, thus non-singular.
+\end{solution}
+\begin{problem}\label{pro:1.3}
+  Assume \(M\) is orientable and connected, prove that \(M\) has exactly two different oriention.
+\end{problem}
+% \begin{lemma}\label{lem:connected_manifold_is_path_connected}
+%   Assume \(M\) is a connected manifold, then it's path-connected.
+% \end{lemma}
+% \begin{proof}
+%   For \(x,y \in M\), we define \(x \sim y\) if and only if there is a path from \(x\) to \(y\).
+%   Easily \(\sim\) is equivalent relation.
+%   For \(x \in M\), we define \(U:=\{y \in M: x \sim y\}\). Obviously \(x \in U \neq \varnothing\),
+%   so we only need to prove \(U\) is both open and closed.
+%   Assume \(y \in U\), since \(M\) is manifold, we obtain \(\exists V\) is homeomorphic to \(\mathbb{R}^n\), thus path-connected, and \(y \in V\).
+%   So \(\forall z \in V,z \sim y\), so \(z \sim x\). So \(V \subset U\). So \(U\) is open.
+%   For \(y \notin U\), samely \(\exists V\) is open and path-connected and \(y \in V\).
+%   If \(z \in V\) and \(z \sim x\) then we will get \(y \sim x\), contradiction!
+%   So \(\forall z \in V,z \not \sim x\). So \(V \subset U^c\), so \(U^c\) is open, thus \(U\) is closed.
+%   Since \(M\) is connected and \(U\) is both open and closed and \(U \neq \varnothing\), we obtain \(U = M\).
+%   So \(M\) is path-connected.
+% \end{proof}
+
+% \begin{lemma}\label{lem:1.3}
+%   Assume \(\mathcal{A}\) is an oriention of \(M\). Then for every local corrdinate \((V;y^i)\), one of \((V;y^i)\) and \((V;z^i)\) is in \(\mathcal{A}\),
+%   where \(z^i=
+%     \begin{cases}
+%       y^i & i>1\\
+%       -y^i & i=1
+%   \end{cases}\).
+% \end{lemma}
+% \begin{proof}
+%   Since Jacobi determinant is continious, and Jacobi determinant of transition map is non-zero, so it's always positive or negative.
+%   Let \(\mathcal{B}:=\{(U;x^i) \in \mathcal{A}:U \cap V \neq \varnothing, J_{x,y}>0\}\) and \(\mathcal{C}:=\{(U;x^i) \in \mathcal{A}:U \cap V \neq \varnothing, J_{x,y}<0\}\).
+%   If \(\mathcal{C}\) is empty then \((V;y^i)\in \mathcal{A}\) by maximalism of \(\mathcal{A}\).
+%   If \(\mathcal{B}\) is empty then \((V;z^i)\in \mathcal{A}\) because \(J_{x,y}=-J_{x,z}\).
+%   So we only need to prove it's impossible that \(\mathcal{B}\) and \(\mathcal{C}\) are both non-empty.
+%
+%   Assume \((U;x^i) \in \mathcal{B}\) and \((T;z^i) \in \mathcal{C}\). By definition of \(\mathcal{B},\mathcal{C}\) we get \(U \cap V \neq \varnothing,T \cap V \neq \varnothing\).
+%   Let \(p \in U \cap V,q \in T \cap V\). Since \((V;y^i)\) is homeomorphic to a subset of \(\mathbb{R}^n\), we know there is a path from \(p\) to \(q\).
+%   Easily this path is compact, so \(\exists p_0=p,p_1,p_2,\cdots,p_k=q\) on this path and \(\{(U_i;y_i^j):i=0,\cdots,k\} \subset \mathcal{A}\) is local coordinate of \(p_i\), and this path is in \(\bigcup_{t=0}^{k} U_t\).
+% \end{proof}
+
+\begin{solution}
+  Since \(M\) is orientable, we can assume that \(\mathcal{B} \subset \mathcal{A}\) is an oriention of \(M\), where \(\mathcal{A}\) is all local coordinate of \(M\).
+  Now consider \(\mathcal{C}:=\{(U;-x^i):(U;x^i) \in \mathcal{B}\}\). Easily to check that \(\mathcal{C}\) is an oriention of \(M\), too.
+  And obviously \(\mathcal{B} \cap \mathcal{C} = \varnothing\), thus \(\mathcal{B} \neq \mathcal{C}\). So there is two oriention. Now we need to prove there is no other oriention.
+
+  Assume \(\mathcal{D}\) is an oriention of \(M\). We will define a function \(\sgn:M \to \{1,-1\}\) by
+  \(\sgn(p)=1 \iff \exists (U_0,x_0^i) \in \mathcal{B},\exists (V_0,y_0^i) \in \mathcal{D},p \in U_0 \cap V_0,J_{x_0,y_0}(p)>0\).
+  We will prove that \(\sgn(p)=1 \implies \forall (U,x^i) \in \mathcal{B},\forall (V;y^i) \in \mathcal{D},p \in U \cap V \implies J_{x,y}(p)>0\).
+  It's easy because \(J_{x,y}=J_{x,x_0}J_{x_0,y_0}J_{y_0,y}\), and \(J_{x,x_0}>0,J_{y,y_0}>0\) by definition of oriention.
+  So \(\sgn(p)=-1 \iff \exists (U_0,x_0^i) \in \mathcal{C},\exists (V_0,y_0^i) \in \mathcal{D},p \in U_0 \cap V_0,J_{x_0,y_0}(p)>0 \)
+  \(\iff \forall (U,x^i) \in \mathcal{B},\forall (V;y^i) \in \mathcal{D},p \in U \cap V \implies J_{x,y}(p)>0 \).
+  Noting that if \(\sgn(p)=1\), we have \(\forall q \in U_0 \cap V_0,\sgn(q)=1\), and so is \(\sgn(p)=-1\).
+  So \(\sgn\) is continuous. So \(\sgn\) is constant because \(M\) is connected.
+  Easy to check that \(\sgn(p)=1 \iff \mathcal{B}=\mathcal{D}\), and \(\sgn(p)=-1 \iff \mathcal{C}=\mathcal{D}\).
+  So there is only two oriention of \(M\).
+\end{solution}
+
+\begin{problem}\label{pro:1.4}
+  Let \(S^n(a):=\{(x^{(1)},\cdots,x^{(n + 1)} )\in \mathbb{R}_1^{n+1} :\sum_{k=1}^{n+1} (x^{(k)})^2=a^2\}\) be the ball in \(\mathbb{R}^{n+1}\) with radius \(a>0\).
+  Let \(S:=(0,\cdots,-a),N:=(0,\cdots,a)\) be the South Pole and North Pole respectively.
+  Let \(U_+=S^n(a)\setminus\{S\},U_-=S^n(a)\setminus\{N\}\).
+  Let \(\phi_+:U_+ \to \mathbb{R}^n,\phi_-:U_- \to \mathbb{R}^n\).
+  \((\xi^{(1)},\cdots,\xi^{(n)})=\phi_+(x^{(1)},\cdots,x^{(n+1)}):=\left(\frac{ax^{(1)}}{a+x^{(n+1)}},\cdots,\frac{ax^{(n)}}{a+x^{(n+1)}}\right)\).
+  \((\eta^{(1)},\cdots,\eta^{(n)})=\phi_+(x^{(1)},\cdots,x^{(n+1)}):=\left(\frac{ax^{(1)}}{a-x^{(n+1)}},\cdots,\frac{ax^{(n)}}{a-x^{(n+1)}}\right)\).
+  Calculate the inverses of \(\phi_+\) and \(\phi_-\), thus prove \(\{(U_+,\phi_+),(U_-,\phi_-)\}\) gives a smooth structuction of \(S^n(a)\).
+\end{problem}
+%TODO:Calculate this out.
+\begin{solution}
+  Noting \(\xi^{(i)}=\frac{a x^{(i)}}{a+x^{(n+1)}}\), so \(x^{(i)}=\frac{\xi^{(i)}(a+x^{(n+1)})}{a}\).
+  And \(\sum_{k=1}^{n+1} (x^{(k)})^2=a^2\), so \(\sum_{k=1}^{n} \frac{(\xi^{(i)})^2 (a+x^{(n+1)})^2}{a^2} + (x^{(n+1)})^2 = a^2\).
+\end{solution}
+\begin{problem}\label{pro:1.5}
+\end{problem}
+\end{document}
+

AlgebraicGeometry/10/solution/AlgebraicGeometry10.tex

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+%!Mode:: "TeX:UTF-8"
+%!TEX TS-program = xelatex
+\documentclass{ctexart}
+\newif\ifpreface
+\prefacetrue
+\input{../../../global/all}
+\begin{document}
+\large
+\setlength{\baselineskip}{1.2em}
+\ifpreface
+    \input{../../../global/preface}
+\else
+\maketitle
+\fi
+\newgeometry{left=2cm,right=2cm,top=2cm,bottom=2cm}
+%from_here_to_type
+
+\end{document}