jrpie-notes/algebra
Archived
1
0
Fork 0
Code Issues 2 Pull requests Projects Releases Packages Wiki Activity

replace \bA and \bP

Browse source
This commit is contained in:
Maximilian Keßler 2022-02-16 02:04:04 +01:00
parent ba13af0c72
commit 181f0ab1bc
1 changed files with 63 additions and 63 deletions
Show stats Download patch file Download diff file Expand all files Collapse all files

126
2021_Algebra_I.tex
View file

@ -1916,26 +1916,26 @@ $\rad(A) = f A$ where $f = \prod_{i=1}^{n} p_i$.
\subsection{Projective spaces}
Let $\mathfrak{l}$ be any field.
\begin{definition}
For a $\mathfrak{l}$-vector space $V$, let $\bP(V)$ be the set of one-dimensional subspaces of $V$.
Let $\bP^n(\mathfrak{l}) \coloneqq \bP(\mathfrak{l}^{n+1})$, the \vocab[Projective space]{$n$-dimensional projective space over $\mathfrak{l}$}.
For a $\mathfrak{l}$-vector space $V$, let $\mathbb{P}(V)$ be the set of one-dimensional subspaces of $V$.
Let $\mathbb{P}^n(\mathfrak{l}) \coloneqq \mathbb{P}(\mathfrak{l}^{n+1})$, the \vocab[Projective space]{$n$-dimensional projective space over $\mathfrak{l}$}.
If $\mathfrak{l}$ is kept fixed, we will often write $\bP^n$ for $\bP^n(\mathfrak{l})$.
If $\mathfrak{l}$ is kept fixed, we will often write $\mathbb{P}^n$ for $\mathbb{P}^n(\mathfrak{l})$.
When dealing with $\bP^n$, the usual convention is to use $0$ as the index of the first coordinate.
When dealing with $\mathbb{P}^n$, the usual convention is to use $0$ as the index of the first coordinate.
We denote the one-dimensional subspace generated by $(x_0,\ldots,x_n) \in \mathfrak{k}^{n+1} \setminus \{0\}$ by $[x_0,\ldots,x_n] \in \bP^n$.
If $x = [x_0,\ldots,x_n] \in \bP^n$, the $(x_{i})_{i=0}^n$ are called \vocab{homogeneous coordinates} of $x$.
We denote the one-dimensional subspace generated by $(x_0,\ldots,x_n) \in \mathfrak{k}^{n+1} \setminus \{0\}$ by $[x_0,\ldots,x_n] \in \mathbb{P}^n$.
If $x = [x_0,\ldots,x_n] \in \mathbb{P}^n$, the $(x_{i})_{i=0}^n$ are called \vocab{homogeneous coordinates} of $x$.
At least one of the $x_{i}$ must be $\neq 0$.
\end{definition}
\begin{remark}
There are points $[1,0], [0,1] \in \bP^1$ but there is no point $[0,0] \in \bP^1$.
There are points $[1,0], [0,1] \in \mathbb{P}^1$ but there is no point $[0,0] \in \mathbb{P}^1$.
\end{remark}
\begin{definition}[Infinite hyperplane]
For $0 \le i \le n$ let $U_i \subseteq \bP^n$ denote the set of $[x_0,\ldots,x_{n}]$ with $x_{i}\neq 0$.
This is a correct definition since two different sets $[x_0,\ldots,x_{n}]$ and $[\xi_0,\ldots,\xi_n]$ of homogeneous coordinates for the same point $x \in \bP^n$ differ by scaling with a $\lambda \in \mathfrak{l}^{\times}$, $x_i = \lambda \xi_i$. Since not all $x_i$ may be $0$, $\bP^n = \bigcup_{i=0}^n U_i$. We identify $\bA^n = \bA^n(\mathfrak{l}) = \mathfrak{l}^n$ with $U_0$ by identifying $(x_1,\ldots,x_n) \in \bA^n$ with $[1,x_1,\ldots,x_n] \in \bP^n$.
Then $\bP^1 = \bA^1 \cup \{\infty\} $ where $\infty=[0,1]$. More generally, when $n > 0$ $\bP^n \setminus \bA^n$ can be identified with $\bP^{n-1}$ identifying $[0,x_1,\ldots,x_n] \in \bP^n \setminus \bA^n$ with $[x_1,\ldots,x_n] \in \bP^{n-1}$.
For $0 \le i \le n$ let $U_i \subseteq \mathbb{P}^n$ denote the set of $[x_0,\ldots,x_{n}]$ with $x_{i}\neq 0$.
This is a correct definition since two different sets $[x_0,\ldots,x_{n}]$ and $[\xi_0,\ldots,\xi_n]$ of homogeneous coordinates for the same point $x \in \mathbb{P}^n$ differ by scaling with a $\lambda \in \mathfrak{l}^{\times}$, $x_i = \lambda \xi_i$. Since not all $x_i$ may be $0$, $\mathbb{P}^n = \bigcup_{i=0}^n U_i$. We identify $\mathbb{A}^n = \mathbb{A}^n(\mathfrak{l}) = \mathfrak{l}^n$ with $U_0$ by identifying $(x_1,\ldots,x_n) \in \mathbb{A}^n$ with $[1,x_1,\ldots,x_n] \in \mathbb{P}^n$.
Then $\mathbb{P}^1 = \mathbb{A}^1 \cup \{\infty\} $ where $\infty=[0,1]$. More generally, when $n > 0$ $\mathbb{P}^n \setminus \mathbb{A}^n$ can be identified with $\mathbb{P}^{n-1}$ identifying $[0,x_1,\ldots,x_n] \in \mathbb{P}^n \setminus \mathbb{A}^n$ with $[x_1,\ldots,x_n] \in \mathbb{P}^{n-1}$.
Thus $\bP^n$ is $\bA^n \cong \mathfrak{l}^n$ with a copy of $\bP^{n-1}$ added as an \vocab{infinite hyperplane} .
Thus $\mathbb{P}^n$ is $\mathbb{A}^n \cong \mathfrak{l}^n$ with a copy of $\mathbb{P}^{n-1}$ added as an \vocab{infinite hyperplane} .
\end{definition}
\subsubsection{Graded rings and homogeneous ideals}
@ -1984,7 +1984,7 @@ Let $\mathfrak{l}$ be any field.
\end{fact}
\subsubsection{The Zariski topology on $\bP^n$}
\subsubsection{The Zariski topology on $\mathbb{P}^n$}
\begin{notation}
Recall that for $\alpha \in \N^{n+1}$ $|\alpha| = \sum_{i=0}^{n} \alpha_i$ and $x^\alpha = x_0^{\alpha_0} \cdot \ldots \cdot x_n^{\alpha_n}$.
\end{notation}
@ -1996,15 +1996,15 @@ Let $\mathfrak{l}$ be any field.
\begin{remark}
This definition gives $R$ the structure of a graded ring.
\end{remark}
\begin{definition}[Zariski topology on $\bP^n(\mathfrak{k})$]\label{ztoppn}
\begin{definition}[Zariski topology on $\mathbb{P}^n(\mathfrak{k})$]\label{ztoppn}
Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.\footnote{As always, $\mathfrak{k}$ is algebraically closed}
For $f \in A_d = \mathfrak{k}[X_0,\ldots,X_n]_d$, the validity of the equation $f(x_0,\ldots,x_{n}) = 0$ does not depend on the choice of homogeneous coordinates, as
\[
f(\lambda x_0,\ldots, \lambda x_n) 0 \lambda^d f(x_0,\ldots,x_n)
\]
Let $\Vp(f) \coloneqq \{x \in \bP^n | f(x) = 0\}$.
Let $\Vp(f) \coloneqq \{x \in \mathbb{P}^n | f(x) = 0\}$.
We call a subset $X \subseteq \bP^n$ Zariski-closed if it can be represented as
We call a subset $X \subseteq \mathbb{P}^n$ Zariski-closed if it can be represented as
\[
X = \bigcap_{i=1}^k \Vp(f_i)
\]
@ -2012,7 +2012,7 @@ Let $\mathfrak{l}$ be any field.
\end{definition}
\pagebreak
\begin{fact}
If $X = \bigcap_{i = 1}^k \Vp(f_i) \subseteq \bP^n$ is closed, then $Y = X \cap \bA^n$ can be identified with the closed subset
If $X = \bigcap_{i = 1}^k \Vp(f_i) \subseteq \mathbb{P}^n$ is closed, then $Y = X \cap \mathbb{A}^n$ can be identified with the closed subset
\[
\{(x_1,\ldots,x_n) \in \mathfrak{k}^n | f_i(1,x_1,\ldots,x_n) = 0, 1 \le i \le k\} \subseteq \mathfrak{k}^n
\]
@ -2020,17 +2020,17 @@ Let $\mathfrak{l}$ be any field.
\[
\{(x_1,\ldots,x_n) \in \mathfrak{k}^n | g_i(x_1,\ldots,x_n) = 0, 1 \le i \le k\}
\]
and can thus be identified with $X \cap \bA^n$ where $X \coloneqq \bigcap_{i=1}^k \Vp(f_i)$ is given by \[f_i(X_0,\ldots,X_n) \coloneqq X_0^{d_i} g_i(X_1 / X_0,\ldots, X_n / X_0), d_i \ge \deg(g_i)\]
Thus, the Zariski topology on $\mathfrak{k}^n$ can be identified with the topology induced by the Zariski topology on $\bA^n = U_0$, and the same holds for $U_i$ with $0 \le i \le n$.
and can thus be identified with $X \cap \mathbb{A}^n$ where $X \coloneqq \bigcap_{i=1}^k \Vp(f_i)$ is given by \[f_i(X_0,\ldots,X_n) \coloneqq X_0^{d_i} g_i(X_1 / X_0,\ldots, X_n / X_0), d_i \ge \deg(g_i)\]
Thus, the Zariski topology on $\mathfrak{k}^n$ can be identified with the topology induced by the Zariski topology on $\mathbb{A}^n = U_0$, and the same holds for $U_i$ with $0 \le i \le n$.
In this sense, the Zariski topology on $\bP^n$ can be thought of as gluing the Zariski topologies on the $U_i \cong \mathfrak{k}^n$.
In this sense, the Zariski topology on $\mathbb{P}^n$ can be thought of as gluing the Zariski topologies on the $U_i \cong \mathfrak{k}^n$.
\end{fact}
% The Zariski topology on P^n (2)
\begin{definition}
Let $I \subseteq A = \mathfrak{k}[X_0,\ldots,X_n]$ be a homogeneous ideal.
Let $\Vp(I) \coloneqq \{[x_0,\ldots,_n] \in \bP^n | \forall f \in I ~ f(x_0,\ldots,x_n) = 0\}$
Let $\Vp(I) \coloneqq \{[x_0,\ldots,_n] \in \mathbb{P}^n | \forall f \in I ~ f(x_0,\ldots,x_n) = 0\}$
As $I$ is homogeneous, it is sufficient to impose this condition for the homogeneous elements $f \in I$.
Because $A$ is Noetherian, $I$ can finitely generated by homogeneous elements $(f_i)_{i=1}^k$ and $\Vp(I)=\bigcap_{i=1}^k \Vp(f_i)$ as in \ref{ztoppn}.
Conversely, if the homogeneous $f_i$ are given, then $I = \langle f_1,\ldots,f_k \rangle_A$ is homogeneous.
@ -2053,10 +2053,10 @@ Let $\mathfrak{l}$ be any field.
By definition, a topological space is Noetherian $\iff$ all open subsets are quasi-compact.
\end{proof}
\begin{corollary}
The Zariski topology on $\bP^n$ is indeed a topology.
The induced topology on the open set $\bA^n = \bP^n \setminus \Vp(X_0) \cong \mathfrak{k}^n$ is the Zariski topology on $\mathfrak{k}^n$.
The same holds for all $U_i = \bP^n \setminus \Vp(X_i) \cong \mathfrak{k}^n$.
Moreover, the topological space $\bP^n$ is Noetherian.
The Zariski topology on $\mathbb{P}^n$ is indeed a topology.
The induced topology on the open set $\mathbb{A}^n = \mathbb{P}^n \setminus \Vp(X_0) \cong \mathfrak{k}^n$ is the Zariski topology on $\mathfrak{k}^n$.
The same holds for all $U_i = \mathbb{P}^n \setminus \Vp(X_i) \cong \mathfrak{k}^n$.
Moreover, the topological space $\mathbb{P}^n$ is Noetherian.
\end{corollary}
\subsection{Noetherianness of graded rings}
@ -2098,14 +2098,14 @@ Let $\mathfrak{l}$ be any field.
% Lecture 12
\subsection{The projective form of the Nullstellensatz and the closed subsets of $\bP^n$}
\subsection{The projective form of the Nullstellensatz and the closed subsets of $\mathbb{P}^n$}
Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
\begin{proposition}[Projective form of the Nullstellensatz]\label{hnsp}
If $I \subseteq A$ is a homogeneous ideal and $f \in A_d$ with $d>0$, then $\Vp(I) \subseteq \Vp(f) \iff f \in \sqrt{I}$.
\end{proposition}
\begin{proof}
$\impliedby$ is clear. Let $\Vp(I) \subseteq \Vp(f)$. If $x = (x_0,\ldots,x_n) \in \Va(I)$, then either $x = 0$ in which case $f(x) = 0$ since $d > 0$
or the point $[x_0,\ldots,x_n] \in \bP^n$ is well-defined and belongs to $\Vp(I) \subseteq \Vp(f)$, hence $f(x) = 0$.
or the point $[x_0,\ldots,x_n] \in \mathbb{P}^n$ is well-defined and belongs to $\Vp(I) \subseteq \Vp(f)$, hence $f(x) = 0$.
Thus $\Va(I) \subseteq \Va(f)$ and $f \in \sqrt{I}$ be the Nullstellensatz (\ref{hns3}).
\end{proof}
@ -2119,7 +2119,7 @@ Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
\begin{proposition}\label{bijproj}
There is a bijection
\begin{align}
f: \{I \subseteq A_+ | I \text{ homogeneous ideal}, I = \sqrt{I}\} &\longrightarrow \{X \subseteq \bP^n | X \text{ closed}\} \\
f: \{I \subseteq A_+ | I \text{ homogeneous ideal}, I = \sqrt{I}\} &\longrightarrow \{X \subseteq \mathbb{P}^n | X \text{ closed}\} \\
I &\longmapsto \Vp(I)\\
\langle \{f \in A_d | d > 0, X \subseteq \Vp(f)\} \rangle & \longmapsfrom X
\end{align}
@ -2127,7 +2127,7 @@ Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
\end{proposition}
\begin{proof}
From the projective form of the Nullstellensatz it follows that $f$ is injective and that $f^{-1}(\Vp\left( I \right)) = \sqrt{I} = I$.
If $X \subseteq \bP^n$ is closed, then $X = \Vp(J)$ for some homogeneous ideal $J \subseteq A$. \Wlog $J = \sqrt{J}$. If $J \not\subseteq A_+$, then $J = A$ (\ref{proja}), hence $X = \Vp(J) = \emptyset = \Vp(A_+)$.
If $X \subseteq \mathbb{P}^n$ is closed, then $X = \Vp(J)$ for some homogeneous ideal $J \subseteq A$. \Wlog $J = \sqrt{J}$. If $J \not\subseteq A_+$, then $J = A$ (\ref{proja}), hence $X = \Vp(J) = \emptyset = \Vp(A_+)$.
Thus we may assume $J \subseteq A_+$, and $f$ is surjective.
@ -2145,7 +2145,7 @@ Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
It is important that $I \subseteq A_{\color{red} +}$, since $\Vp(A) = \Vp(A_+) = \emptyset$ would be a counterexample.
\end{remark}
\begin{corollary}
$\bP^n$ is irreducible.
$\mathbb{P}^n$ is irreducible.
\end{corollary}
\begin{proof}
Apply \ref{bijproj} to $\{0\} \in \Proj(A_\bullet)$.
@ -2173,24 +2173,24 @@ Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
\[\{\fp \in \Spec R | \fp \text{ is a homogeneous ideal of } R_\bullet\} = \Proj(R_\bullet) \sqcup \{\fp \oplus R_+ | \fp \in \Spec R_0\}\]
\end{remark}
\subsection{Dimension of $\bP^n$}
\subsection{Dimension of $\mathbb{P}^n$}
\begin{proposition}
\begin{itemize}
\item $\bP^n$ is catenary.
\item $\dim(\bP^n) = n$. Moreover, $\codim(\{x\} ,\bP^n) = n$ for every $x \in \bP^n$.
\item If $X \subseteq \bP^n$ is irreducible and $x \in X$, then $\codim(\{x\}, X) = \dim(X) = n - \codim(X, \bP^n)$.
\item If $X \subseteq Y \subseteq \bP^n$ are irreducible subsets, then $\codim(X,Y) = \dim(Y) - \dim(X)$.
\item $\mathbb{P}^n$ is catenary.
\item $\dim(\mathbb{P}^n) = n$. Moreover, $\codim(\{x\} ,\mathbb{P}^n) = n$ for every $x \in \mathbb{P}^n$.
\item If $X \subseteq \mathbb{P}^n$ is irreducible and $x \in X$, then $\codim(\{x\}, X) = \dim(X) = n - \codim(X, \mathbb{P}^n)$.
\item If $X \subseteq Y \subseteq \mathbb{P}^n$ are irreducible subsets, then $\codim(X,Y) = \dim(Y) - \dim(X)$.
\end{itemize}
\end{proposition}
\begin{proof}
Let $X \subseteq \bP^n$ be irreducible. If $x \in X$, there is an integer $0 \le i \le n$ and $X \in U_i = \bP^n \setminus \Vp(X_i)$.
\Wlog $i = 0$. Then $\codim(X, \bP^n) = \codim(X \cap \bA^n, \bA^n)$ by the locality of Krull codimension (\ref{lockrullcodim}).
Let $X \subseteq \mathbb{P}^n$ be irreducible. If $x \in X$, there is an integer $0 \le i \le n$ and $X \in U_i = \mathbb{P}^n \setminus \Vp(X_i)$.
\Wlog $i = 0$. Then $\codim(X, \mathbb{P}^n) = \codim(X \cap \mathbb{A}^n, \mathbb{A}^n)$ by the locality of Krull codimension (\ref{lockrullcodim}).
Applying this with $X = \{x\}$ and our results about the affine case gives the second assertion.
If $Y$ and $Z$ are also irreducible with $X \subseteq Y \subseteq Z$, then $\codim(X,Y) = \codim(X \cap \bA^n, Y \cap \bA^n)$, $\codim(X,Z) = \codim(X \cap \bA^n, Z \cap \bA^n)$ and $\codim(Y,Z) = \codim(Y \cap \bA^n, Z \cap \bA^n)$.
If $Y$ and $Z$ are also irreducible with $X \subseteq Y \subseteq Z$, then $\codim(X,Y) = \codim(X \cap \mathbb{A}^n, Y \cap \mathbb{A}^n)$, $\codim(X,Z) = \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$ and $\codim(Y,Z) = \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$.
Thus
\begin{align}
\codim(X,Y) + \codim(Y,Z) &= \codim(X \cap \bA^n, Y \cap \bA^n) + \codim(Y \cap \bA^n, Z \cap \bA^n)\\
&= \codim(X \cap \bA^n, Z \cap \bA^n)\\
\codim(X,Y) + \codim(Y,Z) &= \codim(X \cap \mathbb{A}^n, Y \cap \mathbb{A}^n) + \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)\\
&= \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)\\
&= \codim(X, Z)
\end{align}
because $\mathfrak{k}^n$ is catenary and the first point follows.
@ -2199,7 +2199,7 @@ Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
\subsection{The cone $C(X)$}
\begin{definition}
If $X \subseteq \bP^n$ is closed, we define the \vocab{affine cone over $X$}
If $X \subseteq \mathbb{P}^n$ is closed, we define the \vocab{affine cone over $X$}
\[
C(X) = \{0\} \cup \{(x_0,\ldots,x_n) \in \mathfrak{k}^{n+1} \setminus \{0\} | [x_0,\ldots,x_n] \in X\}
\]
@ -2212,7 +2212,7 @@ Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
$\dim(C(X)) = \dim(X) + 1$ and
$\codim(C(X), \mathfrak{k}^{n+1}) = \codim(X, \bP^n)$
$\codim(C(X), \mathfrak{k}^{n+1}) = \codim(X, \mathbb{P}^n)$
\end{itemize}
\end{proposition}
\begin{proof}
@ -2220,42 +2220,42 @@ Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
Let $d = \dim(X)$ and
\[
X_0 \subsetneq \ldots \subsetneq X_d = X \subsetneq X_{d+1} \subsetneq \ldots \subsetneq X_n = \bP^n
X_0 \subsetneq \ldots \subsetneq X_d = X \subsetneq X_{d+1} \subsetneq \ldots \subsetneq X_n = \mathbb{P}^n
\]
be a chain of irreducible subsets of $\bP^n$. Then
be a chain of irreducible subsets of $\mathbb{P}^n$. Then
\[
\{0\} \subsetneq C(X_0) \subsetneq \ldots \subsetneq C(X_d) = C(X) \subsetneq \ldots \subsetneq C(X_n) = \mathfrak{k}^{n+1}
\]
is a chain of irreducible subsets of $\mathfrak{k}^{n+1}$. Hence $\dim(C(X)) \ge 1 + d$ and $\codim(C(X), \mathfrak{k}^{n+1}) \ge n-d$. Since $\dim(C(X)) + \codim(C(X), \mathfrak{k}^{n+1}) = \dim(\mathfrak{k}^{n+1}) = n+1$, the two inequalities must be equalities.
\end{proof}
\subsubsection{Application to hypersurfaces in $\bP^n$}
\subsubsection{Application to hypersurfaces in $\mathbb{P}^n$}
\begin{definition}[Hypersurface]
Let $n > 0$.
By a \vocab{hypersurface} in $\bP^n$ or $\bA^n$ we understand an irreducible closed subset of codimension $1$.
By a \vocab{hypersurface} in $\mathbb{P}^n$ or $\mathbb{A}^n$ we understand an irreducible closed subset of codimension $1$.
\end{definition}
\begin{corollary}
If $P \in A_d$ is a prime element, then $H = \Vp(P)$ is a hypersurface in $\bP^n$ and every hypersurface $H$ in $\bP^n$ can be obtained in this way.
If $P \in A_d$ is a prime element, then $H = \Vp(P)$ is a hypersurface in $\mathbb{P}^n$ and every hypersurface $H$ in $\mathbb{P}^n$ can be obtained in this way.
\end{corollary}
\begin{proof}
If $H = \Vp(P)$ then $C(H) = \Va(P)$ is a hypersurface in $\mathfrak{k}^{n+1}$ by \ref{irredcodimone}. By \ref{conedim}, $H$ is irreducible and of codimension $1$.
Conversely, let $H$ be a hypersurface in $\bP^n$. By \ref{conedim}, $C(H)$ is a hypersurface in $\mathfrak{k}^{n+1}$, hence $C(H) = \Vp(P)$ for some prime element $P \in A$ (again by \ref{irredcodimone}).
Conversely, let $H$ be a hypersurface in $\mathbb{P}^n$. By \ref{conedim}, $C(H)$ is a hypersurface in $\mathfrak{k}^{n+1}$, hence $C(H) = \Vp(P)$ for some prime element $P \in A$ (again by \ref{irredcodimone}).
We have $H = \Vp(\fp)$ for some $\fp \in \Proj(A)$ and $C(H) = \Va(\fp)$. By the bijection between closed subsets of $\mathfrak{k}^{n+1}$ and ideals $I = \sqrt{I} \subseteq A$ (\ref{antimonbij}), $\fp = P \cdot A$.
Let $P = \sum_{k=0}^{d}P_k$ with $P_d \neq 0$ be the decomposition into homogeneous components.
If $P_e $ with $e < d$ was $\neq 0$, it could not be a multiple of $P$ contradicting the homogeneity of $\fp = P \cdot A$. Thus, $P$ is homogeneous of degree $d$.
\end{proof}
\begin{definition}
A hypersurface $H \subseteq \bP^n$ has \vocab{degree $d$} if $H = \Vp(P)$ where $P \in A_d$ is an irreducible polynomial.
A hypersurface $H \subseteq \mathbb{P}^n$ has \vocab{degree $d$} if $H = \Vp(P)$ where $P \in A_d$ is an irreducible polynomial.
\end{definition}
\subsubsection{Application to intersections in $\bP^n$ and Bezout's theorem}
\subsubsection{Application to intersections in $\mathbb{P}^n$ and Bezout's theorem}
\begin{corollary}
Let $A \subseteq \bP^n$ and $B \subseteq \bP^n$ be irreducible subsets of dimensions $a$ and $b$. If $a+ b \ge n$, then $A \cap B \neq \emptyset$ and every irreducible component of $A \cap B$ as dimension $\ge a + b - n$.
Let $A \subseteq \mathbb{P}^n$ and $B \subseteq \mathbb{P}^n$ be irreducible subsets of dimensions $a$ and $b$. If $a+ b \ge n$, then $A \cap B \neq \emptyset$ and every irreducible component of $A \cap B$ as dimension $\ge a + b - n$.
\end{corollary}
\begin{remark}
This shows that $\bP^n$ indeed fulfilled the goal of allowing for nicer results of algebraic geometry because ``solutions at infinity'' to systems of algebraic equations are present in $\bP^n$
This shows that $\mathbb{P}^n$ indeed fulfilled the goal of allowing for nicer results of algebraic geometry because ``solutions at infinity'' to systems of algebraic equations are present in $\mathbb{P}^n$
(see \ref{affineproblem}).
\end{remark}
@ -2266,7 +2266,7 @@ If $P_e $ with $e < d$ was $\neq 0$, it could not be a multiple of $P$ contradic
If $A \cap B = \emptyset$, then $C(A) \cap C(B) = \{0\}$ with $\{0\} $ as an irreducible component, contradicting the lower bound $a + b + 1 - n > 0$ for the dimension of irreducible components of $C(A) \cap C(B)$ (again \ref{codimintersection}).
\end{proof}
\begin{remark}[Bezout's theorem]
If $A \neq B$ are hypersurfaces of degree $a$ and $b$ in $\bP^2$, then $A \cap B$ has $ab$ points counted by (suitably defined) multiplicity.
If $A \neq B$ are hypersurfaces of degree $a$ and $b$ in $\mathbb{P}^2$, then $A \cap B$ has $ab$ points counted by (suitably defined) multiplicity.
\end{remark}
@ -2408,8 +2408,8 @@ is an isomorphism.
\]
Hence $\phi = F\defon{X}$.
\end{proof}
\subsubsection{The structure sheaf on closed subsets of $\bP^n$}
Let $X \subseteq \bP^n$ be closed and $R_\bullet = \mathfrak{k}[X_0,\ldots,X_n]$ with its usual grading.
\subsubsection{The structure sheaf on closed subsets of $\mathbb{P}^n$}
Let $X \subseteq \mathbb{P}^n$ be closed and $R_\bullet = \mathfrak{k}[X_0,\ldots,X_n]$ with its usual grading.
\begin{definition}\label{structuresheafpn}
For open $U \subseteq X$, let $\mathcal{O}_X(U)$ be the set of functions $U \xrightarrow{\phi} \mathfrak{k}$ such that for every $x \in U$, there are an open subset $W \subseteq U$, a natural number $d$ and $f,g \in R_d$ such that $W \cap \Vp(g) = \emptyset$ and $\phi(y) = \frac{f(y_0,\ldots,y_n)}{g(y_0,\ldots,y_n)}$ for $y = [y_0,\ldots,y_n] \in W$.
@ -2417,10 +2417,10 @@ Let $X \subseteq \bP^n$ be closed and $R_\bullet = \mathfrak{k}[X_0,\ldots,X_n]$
\begin{remark}
This is a subsheaf of rings of the sheaf of $\mathfrak{k}$-valued functions on $X$.
Under the identification $\bA^n =\mathfrak{k}^n$ with $\bP^n \setminus \Vp(X_0)$, one has $\mathcal{O}_X \defon{X \setminus \Vp(X_0)} = \mathcal{O}_{X \cap \bA^n}$ as subsheaves of the sheaf of $\mathfrak{k}$-valued functions, where the second sheaf is a sheaf on a closed subset of $\mathfrak{k}^n$:
Under the identification $\mathbb{A}^n =\mathfrak{k}^n$ with $\mathbb{P}^n \setminus \Vp(X_0)$, one has $\mathcal{O}_X \defon{X \setminus \Vp(X_0)} = \mathcal{O}_{X \cap \mathbb{A}^n}$ as subsheaves of the sheaf of $\mathfrak{k}$-valued functions, where the second sheaf is a sheaf on a closed subset of $\mathfrak{k}^n$:
Indeed, if $W$ is as in the definition then $\phi([1,y_1,\ldots,y_n]) = \frac{f(1,y_1,\ldots,y_n)}{g(1,y_1,\ldots,y_n)}$ for $[1,y_1,\ldots,y_n] \in W$.
Conversely if $\phi([1,y_1,\ldots,y_n]) = \frac{f(y_1,\ldots,y_n)}{g(y_1,\ldots,y_n)}$ on an open subset $W $ of $X \cap \bA^n$ then
Conversely if $\phi([1,y_1,\ldots,y_n]) = \frac{f(y_1,\ldots,y_n)}{g(y_1,\ldots,y_n)}$ on an open subset $W $ of $X \cap \mathbb{A}^n$ then
$\phi([y_0,\ldots,y_n]) = \frac{F(y_0,\ldots,y_n)}{G(y_0,\ldots,y_n)}$ on $W$ where $F(X_0,\ldots,X_n) \coloneqq X_0^d f(\frac{X_1}{X_0}, \ldots, \frac{X_n}{X_0})$ and $G(X_0,\ldots,X_n) = X_0^d g(\frac{X_1}{X_0},\ldots, \frac{X_n}{X_0})$ with a sufficiently large $d \in \N$.
\end{remark}
\begin{remark}
@ -2523,8 +2523,8 @@ The following is somewhat harder than in the affine case:
\begin{example}
\begin{itemize}
\item If $(X, \mathcal{O}_X)$ is a variety and $U \subseteq X$ open, then $(U, \mathcal{O}_X\defon{U})$ is a variety (called an \vocab{open subvariety} of $X$), and the embedding $U \to X$ is a morphism of varieties.
\item If $X$ is a closed subset of $\mathfrak{k}^n$ or $\bP^n$, then $(X, \mathcal{O}_X)$ is a variety, where $\mathcal{O}_X$ is the structure sheaf on $X$ (\ref{structuresheafkn}, reps. \ref{structuresheafpn}).
A variety is called \vocab[Variety!affine]{affine} (resp. \vocab[Variety!projective]{projective}) if it is isomorphic to a variety of this form, with $X $ closed in $\mathfrak{k}^n$ (resp. $\bP^n$).
\item If $X$ is a closed subset of $\mathfrak{k}^n$ or $\mathbb{P}^n$, then $(X, \mathcal{O}_X)$ is a variety, where $\mathcal{O}_X$ is the structure sheaf on $X$ (\ref{structuresheafkn}, reps. \ref{structuresheafpn}).
A variety is called \vocab[Variety!affine]{affine} (resp. \vocab[Variety!projective]{projective}) if it is isomorphic to a variety of this form, with $X $ closed in $\mathfrak{k}^n$ (resp. $\mathbb{P}^n$).
A variety which is isomorphic to and open subvariety of $X$ is called \vocab[Variety!quasi-affine]{quasi-affine} (resp. \vocab[Variety!quasi-projective]{quasi-projective}).
\item If $X = V(X^2 - Y^3) \subseteq \mathfrak{k}^2$ then $\mathfrak{k} \xrightarrow{t \mapsto (t^3,t^2)} X$ is a morphism which is a homeomorphism of topological spaces but not an isomorphism of varieties.
% TODO
@ -2786,20 +2786,20 @@ If $X$ is a set, then $\cB \subseteq \mathcal{P}(X)$ is a base for some topology
\end{proof}
\subsubsection{Intersection multiplicities and Bezout's theorem}
\begin{definition}
Let $R = \mathfrak{k}[X_0,X_1,X_2]$ equipped with its usual grading and let $x \in \bP^{2}$.
Let $R = \mathfrak{k}[X_0,X_1,X_2]$ equipped with its usual grading and let $x \in \mathbb{P}^{2}$.
Let $G \in R_g, H \in R_h$ be homogeneous polynomials with $x \in V(G) \cap V(h)$.
Let $\ell\in R_1$ such that $\ell(x) \neq 0$. Then $x \in U = \bP^2 \setminus V(\ell)$ and the rational functions $\gamma = \ell^{-g}G, \eta = \ell^{-h}H$ are elements of $\mathcal{O}_{\bP^2}(U)$.
Let $I_x(G,H) \subseteq \mathcal{O}_{\bP^2,x}$ denote the ideal generated by $\gamma_x$ and $\eta_x$.
Let $\ell\in R_1$ such that $\ell(x) \neq 0$. Then $x \in U = \mathbb{P}^2 \setminus V(\ell)$ and the rational functions $\gamma = \ell^{-g}G, \eta = \ell^{-h}H$ are elements of $\mathcal{O}_{\mathbb{P}^2}(U)$.
Let $I_x(G,H) \subseteq \mathcal{O}_{\mathbb{P}^2,x}$ denote the ideal generated by $\gamma_x$ and $\eta_x$.
\noindent The dimension $\dim_{\mathfrak{k}}(\mathcal{O}_{X,x} / I_x(G,H)) \text{\reflectbox{$\coloneqq$}} i_x(G,H)$ is called the \vocab{intersection multiplicity} of $G$ and $H$ at $x$.
\end{definition}
\begin{remark}
If $\tilde \ell \in R_1$ also satisfies $\tilde \ell(x) \neq 0$, then the image of $\tilde \ell / \ell$ under $\mathcal{O}_{\bP^2}(U) \to \mathcal{O}_{\bP^2,x}$ is a unit, showing that the image of $\tilde \gamma = \tilde \ell^{-g} G$ in $\mathcal{O}_{\bP^2,x}$ is multiplicatively equivalent to $\gamma_x$, and similarly for $\eta_x$.
If $\tilde \ell \in R_1$ also satisfies $\tilde \ell(x) \neq 0$, then the image of $\tilde \ell / \ell$ under $\mathcal{O}_{\mathbb{P}^2}(U) \to \mathcal{O}_{\mathbb{P}^2,x}$ is a unit, showing that the image of $\tilde \gamma = \tilde \ell^{-g} G$ in $\mathcal{O}_{\mathbb{P}^2,x}$ is multiplicatively equivalent to $\gamma_x$, and similarly for $\eta_x$.
Thus $I_x(G,H)$ does not depend on the choice of $\ell \in R_1$ with $\ell(x) \neq 0$.
\end{remark}
\begin{theorem}[Bezout's theorem]
In the above situation, assume that $V(H)$ and $V(G)$ intersect properly in the sense that $V(G) \cap V(H) \subseteq \bP^2$ has no irreducible component of dimension $\ge 1$.
In the above situation, assume that $V(H)$ and $V(G)$ intersect properly in the sense that $V(G) \cap V(H) \subseteq \mathbb{P}^2$ has no irreducible component of dimension $\ge 1$.
Then
\[
\sum_{x \in V(G) \cap V(H)} i_x(G,H) = gh