612 lines
24 KiB
TeX
612 lines
24 KiB
TeX
Let $\mathfrak{l}$ be any field.
|
|
\begin{definition}
|
|
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 $\mathbb{P}^n$ for $\mathbb{P}^n(\mathfrak{l})$.
|
|
|
|
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
|
|
$[x_0,\ldots,x_n] \in \mathbb{P}^n$.
|
|
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 $[1,0], [0,1] \in \mathbb{P}^1$ but there
|
|
is no point $[0,0] \in \mathbb{P}^1$.
|
|
\end{remark}
|
|
\begin{definition}[Infinite hyperplane]
|
|
For $0 \le i \le n$ let $U_i \subseteq \mathbb{P}^n$
|
|
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 $\mathbb{P}^n$ 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}
|
|
\begin{notation}
|
|
Let $\mathbb{I} = \N$ or $\mathbb{I} = \Z$.
|
|
\end{notation}
|
|
\begin{definition}
|
|
By an \vocab[Graded ring]{$\mathbb{I}$-graded ring} $A_\bullet$
|
|
we understand a ring $A$ with a collection $(A_d)_{d \in \mathbb{I}}$ of
|
|
subgroups of the additive group $(A, +)$
|
|
such that $A_a \cdot A_b \subseteq A_{a + b}$ for $a,b \in \mathbb{I}$
|
|
and such that $A = \bigoplus_{d \in \mathbb{I}} A_d$ in
|
|
the sense that every $r \in A$ has a unique decomposition
|
|
$r = \sum_{d \in \mathbb{I}} r_d$ with $r_d \in A_d$ and but finitely many
|
|
$r_d \neq 0$.
|
|
|
|
We call the $r_d$ the \vocab{homogeneous components} of $r$.
|
|
|
|
An ideal $I \subseteq A$ is called \vocab{homogeneous} if
|
|
\[r \in I \implies \forall d \in \mathbb{I} ~ r_d \in I_d\]
|
|
where $I_d \coloneqq I \cap A_d$.
|
|
|
|
By a \vocab{graded ring} we understand an $\N$-graded ring.
|
|
In this case,
|
|
\[A_{+} \coloneqq \bigoplus_{d=1}^{\infty} A_d = \{r \in A | r_0 = 0\}\]
|
|
is called the \vocab{augmentation ideal} of $A$.
|
|
\end{definition}
|
|
\begin{remark}[Decomposition of $1$]
|
|
If $1 = \sum_{d \in \mathbb{I}} \varepsilon_d$
|
|
is the decomposition into homogeneous components,
|
|
then $\varepsilon_a = 1 \cdot \varepsilon_a
|
|
= \sum_{b \in \mathbb{I}} \varepsilon_a\varepsilon_b$
|
|
with $\varepsilon_a\varepsilon_b \in A_{a+b}$.
|
|
By the uniqueness of the decomposition into homogeneous components,
|
|
$\varepsilon_a \varepsilon_0 = \varepsilon_a$
|
|
and $b \neq 0 \implies \varepsilon_a \varepsilon_b = 0$.
|
|
Applying the last equation with $a = 0$ gives
|
|
$b\neq 0 \implies \varepsilon_b = \varepsilon_0 \varepsilon _b = 0$.
|
|
Thus $1 = \varepsilon_0 \in A_0$.
|
|
\end{remark}
|
|
\begin{remark}
|
|
The augmentation ideal of a graded ring is a homogeneous ideal.
|
|
\end{remark}
|
|
|
|
% Graded rings and homogeneous ideals (2)
|
|
|
|
\begin{proposition}
|
|
\footnote{This holds for both $\Z$-graded and $\N$-graded rings.}
|
|
\begin{itemize}
|
|
\item
|
|
A principal ideal generated by a homogeneous element is homogeneous.
|
|
\item
|
|
The operations $\sum, \bigcap, \sqrt{}$ preserve homogeneity.
|
|
\item
|
|
An ideal is homogeneous iff it can be generated by a family of homogeneous
|
|
elements.
|
|
\end{itemize}
|
|
\end{proposition}
|
|
\begin{proof}
|
|
Most assertions are trivial.
|
|
We only show that $J$ homogeneous $\implies \sqrt{J}$ homogeneous.
|
|
Let $A$ be $\mathbb{I}$-graded,
|
|
$f \in \sqrt{J} $ and
|
|
$f = \sum_{d \in \mathbb{I}} f_d$ the decomposition.
|
|
To show that all $f_d \in \sqrt{J} $,
|
|
we use induction on $N_f \coloneqq \# \{d \in \mathbb{I} | f_d \neq 0\}$.
|
|
$N_f = 0$ is trivial.
|
|
Suppose $N_f > 0$ and $e \in \mathbb{I}$ is maximal with $f_e \neq 0$.
|
|
For $l \in \N$, the $le$-th homogeneous component of $f^l$ is $f_e^l$.
|
|
Choosing $l$ large enough such that $f^l \in J$ and using the homogeneity of
|
|
$J$, we find $f_e \in \sqrt{J}$.
|
|
As $\sqrt{J} $ is an ideal,
|
|
$\tilde f \coloneqq f - f_e \in \sqrt{J} $.
|
|
As $N_{\tilde f} = N_f -1$,
|
|
the induction assumption may be applied to $\tilde f$
|
|
and shows $f_d \in \sqrt{J} $ for $d \neq e$.
|
|
\end{proof}
|
|
\begin{fact}
|
|
A homogeneous ideal is finitely generated
|
|
iff it can be generated by finitely many of its homogeneous elements.
|
|
In particular, this is always the case when $A$ is a Noetherian ring.
|
|
\end{fact}
|
|
|
|
|
|
\subsubsection{The Zariski topology on $\mathbb{P}^n$}
|
|
\begin{notation}
|
|
Recall that for
|
|
$\alpha \in \N^{n+1}$
|
|
\[|\alpha| = \sum_{i=0}^{n} \alpha_i \text{ and }
|
|
x^\alpha = x_0^{\alpha_0} \cdot \ldots \cdot x_n^{\alpha_n}.\]
|
|
\end{notation}
|
|
\begin{definition}[Homogeneous polynomials]
|
|
Let $R$ be any ring and
|
|
\[f = \sum_{\alpha \in \N^{n+1}} f_\alpha X^{\alpha}\in R[X_0,\ldots,X_n].\]
|
|
We say that $f$ is \vocab{homogeneous of degree $d$}
|
|
if
|
|
\[|\alpha| \neq d \implies f_\alpha = 0.\]
|
|
We denote the subset of homogeneous polynomials of degree $d$ by
|
|
\[R[X_0,\ldots,X_n]_d \subseteq R[X_0,\ldots,X_n].\]
|
|
\end{definition}
|
|
\begin{remark}
|
|
This definition gives $R$ the structure of a graded ring.
|
|
\end{remark}
|
|
\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 $\Vp(f) \coloneqq \{x \in \mathbb{P}^n | f(x) = 0\}$.
|
|
|
|
We call a subset $X \subseteq \mathbb{P}^n$ Zariski-closed if it
|
|
can be represented as
|
|
\[
|
|
X = \bigcap_{i=1}^k \Vp(f_i)
|
|
\]
|
|
where the $f_i \in A_{d_i}$ are homogeneous polynomials.
|
|
\end{definition}
|
|
\pagebreak
|
|
\begin{fact}
|
|
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
|
|
\[
|
|
\{(x_1,\ldots,x_n) \in \mathfrak{k}^n |
|
|
f_i(1,x_1,\ldots,x_n) = 0, 1 \le i \le k\}
|
|
\subseteq \mathfrak{k}^n.
|
|
\]
|
|
Conversely, if $Y \subseteq \mathfrak{k}^n$ is closed it has the form
|
|
\[
|
|
\{(x_1,\ldots,x_n) \in \mathfrak{k}^n |
|
|
g_i(x_1,\ldots,x_n) = 0, 1 \le i \le k\}
|
|
\]
|
|
and can thus be identified with $X \cap \mathbb{A}^n$
|
|
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 $\mathbb{P}^n$ 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 $\Vp(I) \coloneqq \{[x_0,\ldots,_n] \in \mathbb{P}^n | \forall f \in I ~%
|
|
f(x_0,\ldots,x_n) = 0\}$
|
|
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.
|
|
\end{definition}
|
|
\begin{remark}
|
|
Note that $V(A) = V(A_+) = \emptyset$.
|
|
\end{remark}
|
|
\begin{fact}
|
|
For homogeneous ideals in $A$ and $m \in \N$, we have:
|
|
\begin{itemize}
|
|
\item
|
|
$\Vp(\sum_{\lambda \in \Lambda} I_\lambda) = \bigcap_{\lambda \in \Lambda} \Vp(I_\lambda)$.
|
|
\item
|
|
$\Vp(\bigcap_{k=1}^m I_k) = \Vp(\prod_{k=1}^{m} I_k) = \bigcup_{k=1}^m \Vp(I_k)$.
|
|
\item
|
|
$\Vp(\sqrt{I}) = \Vp(I)$.
|
|
\end{itemize}
|
|
\end{fact}
|
|
\begin{fact}
|
|
If $X = \bigcup_{\lambda \in \Lambda} U_\lambda$ is an open covering
|
|
of a topological space then $X$ is Noetherian
|
|
iff there is a finite subcovering and all $U_\lambda$ are Noetherian.
|
|
\end{fact}
|
|
\begin{proof}
|
|
By definition, a topological space is Noetherian
|
|
$\iff$ all open subsets are quasi-compact.
|
|
\end{proof}
|
|
\begin{corollary}
|
|
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}
|
|
\begin{proposition}
|
|
For a graded ring $R_{\bullet}$,
|
|
the following conditions are equivalent:
|
|
\begin{enumerate}[A]
|
|
\item
|
|
$R$ is Noetherian.
|
|
\item
|
|
Every homogeneous ideal of $R_{\bullet}$ is finitely generated.
|
|
\item
|
|
Every chain $I_0\subseteq I_1 \subseteq \ldots$ of
|
|
homogeneous ideals terminates.
|
|
\item
|
|
Every set $\mathfrak{M} \neq \emptyset$ of homogeneous ideals
|
|
has a $\subseteq$-maximal element.
|
|
\item
|
|
$R_0$ is Noetherian and the ideal $R_+$ is finitely generated.
|
|
\item
|
|
$R_0$ is Noetherian and $R / R_0$ is of finite type.
|
|
\end{enumerate}
|
|
\end{proposition}
|
|
\begin{proof}
|
|
\noindent\textbf{A $\implies$ B,C,D} trivial.
|
|
|
|
\noindent\textbf{B $\iff$ C $\iff$ D} similar to the proof about Noetherianness.
|
|
|
|
\noindent\textbf{B $\land$ C $\implies $E}
|
|
B implies that $R_+$ is finitely generated.
|
|
Since $I \oplus R_+$ is homogeneous for any homogeneous ideal $I \subseteq
|
|
R_0$, C implies the Noetherianness of $R_0$.
|
|
|
|
\noindent\textbf{E $\implies$ F}
|
|
Let $R_+$ be generated by $f_i \in R_{d_i}, d_i > 0$ as
|
|
an ideal.
|
|
\begin{claim}
|
|
The $R_0$-subalgebra $\tilde R$ of $R$ generated by the $f_i$ equals $R$.
|
|
\end{claim}
|
|
\begin{subproof}
|
|
It is sufficient to show that every homogeneous $f \in R_d$
|
|
belongs to $\tilde R$.
|
|
We use induction on $d$.
|
|
The case of $d = 0$ is trivial.
|
|
Let $d > 0$ and $R_e \subseteq \tilde R$ for all $e < d$.
|
|
As $f \in R_+$, $f = \sum_{i=1}^{k} g_if_i$.
|
|
Let $f_a = \sum_{i=1}^{k} g_{i,
|
|
a-d_i} f_i$, where $g_i = \sum_{b=0}^{\infty} g_{i,b}$
|
|
is the decomposition into homogeneous components.
|
|
Then $f = \sum_{a=0}^{\infty} f_a$ is the decomposition of $f$ into
|
|
homogeneous components,
|
|
hence $a \neq d \implies f_a = 0 $.
|
|
Thus we may assume $g_i \in R_{d-d_i}$.
|
|
As $d_i > 0$,
|
|
the induction assumption may now be applied to $g_i$,
|
|
hence $g_i \in \tilde R$,
|
|
hence $f \in \tilde R$.
|
|
\end{subproof}
|
|
|
|
\noindent\textbf{F $\implies$ A}
|
|
Hilbert's Basissatz (\ref{basissatz})
|
|
|
|
\end{proof}
|
|
|
|
|
|
|
|
|
|
\subsection{The projective form of the Nullstellensatz and the closed subsets
|
|
of $\mathbb{P}^n$} Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
|
|
% Lecture 12
|
|
\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 $[x_0,\ldots,x_n] \in \mathbb{P}^n$ 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}
|
|
|
|
\begin{definition}%
|
|
\footnote{This definition is not too important, the characterization in the following remark suffices.}
|
|
For a graded ring $R_\bullet$,
|
|
let $\Proj(R_\bullet)$ be the set of $\fp \in \Spec R$
|
|
such that $\fp$ is a homogeneous ideal and $\fp \not\supseteq R_+$.
|
|
\end{definition}
|
|
\begin{remark}
|
|
\label{proja}
|
|
As the elements of $A_0 \setminus \{0\}$ are units in $A$
|
|
it follows that for every homogeneous ideal $I$
|
|
we have $I \subseteq A_+$ or $I = A$.
|
|
In particular,
|
|
$\Proj(A_\bullet) = \{\fp \in \Spec A \setminus A_+ | \fp \text{ is homogeneous}\} $.
|
|
\end{remark}
|
|
\begin{proposition}
|
|
\label{bijproj}
|
|
There is a bijection
|
|
\begin{align*}
|
|
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*}
|
|
Under this bijection,
|
|
the irreducible subsets correspond to the elements of
|
|
$\Proj(A_\bullet)$.
|
|
\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 $X \subseteq \mathbb{P}^n$ is closed,
|
|
then $X = \Vp(J)$ for some homogeneous ideal $J \subseteq A$.
|
|
Without loss of generality loss of generality $J = \sqrt{J}$.
|
|
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.
|
|
|
|
Suppose $\fp \in \Proj(A_\bullet)$.
|
|
Then $\fp \neq A_+$ hence $X = \Vp(\fp) \neq \emptyset$ by the
|
|
proven part of the proposition.
|
|
Assume $X = X_1 \cup X_2$ is a decomposition into proper closed subsets,
|
|
where $X_k = \Vp(I_k)$ for some $I_k \subseteq A_+, I_k = \sqrt{I_k}$.
|
|
Since $X_k$ is a proper subset of $X$, there is $f_k \in I_k \setminus \fp$.
|
|
We have $\Vp(f_1f_2) \supseteq \Vp(f_k) \supseteq \Vp(I_k)$
|
|
hence $\Vp(f_1f_2) \supseteq \Vp(I_1) \cup \Vp(I_2) = X = \Vp(\fp)$
|
|
and it follows that $f_1f_2\in \sqrt{\fp} = \fp \lightning$.
|
|
|
|
Assume $X = \Vp(\fp)$ is irreducible,
|
|
where $\fp = \sqrt{\fp} \in A_+$ is homogeneous.
|
|
The $\fp \neq A_+$ as $X = \emptyset$ otherwise.
|
|
Assume that $f_1f_2 \in \fp$ but $f_i \not\in A_{d_i} \setminus \fp$.
|
|
Then $X \not \subseteq \Vp(f_i)$ by the projective
|
|
Nullstellensatz when $d_i > 0$
|
|
and because $\Vp(1) = \emptyset$ when $d_i = 0$.
|
|
Thus $X = (X \cap \Vp\left( f_1 \right)) \cup (X \cap \Vp(f_2))$ is a
|
|
proper decomposition $\lightning$.
|
|
By lemma \ref{homprime}, $\fp$ is a prime ideal.
|
|
\end{proof}
|
|
\begin{remark}
|
|
It is important that $I \subseteq A_{\color{red} +}$,
|
|
since $\Vp(A) = \Vp(A_+) = \emptyset$ would be a counterexample.
|
|
\end{remark}
|
|
\begin{corollary}
|
|
$\mathbb{P}^n$ is irreducible.
|
|
\end{corollary}
|
|
\begin{proof}
|
|
Apply \ref{bijproj} to $\{0\} \in \Proj(A_\bullet)$.
|
|
\end{proof}
|
|
|
|
\subsection{Some remarks on homogeneous prime ideals}
|
|
\begin{lemma}
|
|
\label{homprime}
|
|
Let $R_\bullet$ be an $\mathbb{I}$ graded ring
|
|
($\mathbb{I} = \N$ or $\mathbb{I} = \Z$).
|
|
A homogeneous ideal $I \subseteq R$ is a prime ideal
|
|
iff $1 \not\in I$ and for all homogeneous elements $f, g \in R$
|
|
\[fg \in I \implies f \in I \lor g \in I.\]
|
|
\end{lemma}
|
|
\begin{proof}
|
|
$\implies$ is trivial.
|
|
It suffices to show that for arbitrary $f,g \in R$
|
|
we have that $fg \in I \implies f \in I \lor g \in I$.
|
|
Let $f = \sum_{d \in \mathbb{I}} f_d, g = \sum_{d \in \mathbb{I}} g_d $ be the
|
|
decompositions into homogeneous components.
|
|
If $f \not\in I$ and $g \not\in I$ there are $d,e \in I$ with $f_d \in I$,
|
|
$g_e \in I$, and they may assumed to be maximal with this property.
|
|
As $I$ is homogeneous and $fg \in I$, we have $(fg)_{d+e} \in I$ but
|
|
\[
|
|
(fg)_{d+e} = f_dg_e + \sum_{\delta = 1}^{\infty} (f_{d + \delta} g_{e - \delta}
|
|
+ f_{d - \delta} g_{e + \delta})
|
|
\]
|
|
where $f_dg_e \not\in I$ by our assumption
|
|
on $I$ and all other summands on the right hand side are $\in I$
|
|
(as $f_{d+ \delta} \in I$ and $g_{e + \delta} \in I$ by the maximality
|
|
of $d$ and $e$),
|
|
a contradiction.
|
|
\end{proof}
|
|
|
|
\begin{remark}
|
|
If $R_\bullet$ is $\N$-graded and $\fp \in \Spec R_0$, then
|
|
\[\fp \oplus R_+ = \{r \in R | r_0 \in \fp\}\]
|
|
is a homogeneous prime ideal of $R$.
|
|
\begin{IEEEeqnarray*}{rCl}
|
|
&&\{\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{IEEEeqnarray*}
|
|
\end{remark}
|
|
|
|
\pagebreak
|
|
\subsection{Dimension of $\mathbb{P}^n$}
|
|
\begin{proposition}\,
|
|
\begin{itemize}
|
|
\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 $X \subseteq \mathbb{P}^n$ be irreducible.
|
|
If $x \in X$, there is an integer $0 \le i \le n$ and
|
|
$X \in U_i = \mathbb{P}^n \setminus \Vp(X_i)$.
|
|
Without loss of generality loss of generality $i = 0$.
|
|
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
|
|
$\codim(X,Y) = \codim(X \cap \mathbb{A}^n, Y \cap \mathbb{A}^n)$,
|
|
$\codim(X,Z) = \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$
|
|
and $\codim(Y,Z) = \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$.
|
|
Thus
|
|
\begin{IEEEeqnarray*}{rCl}
|
|
\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{IEEEeqnarray*}
|
|
because $\mathfrak{k}^n$ is catenary and the first point follows.
|
|
The remaining assertions can easily be derived from the first two.
|
|
\end{proof}
|
|
|
|
\subsection{The cone $C(X)$}
|
|
\begin{definition}
|
|
If $X \subseteq \mathbb{P}^n$ is closed, we define the
|
|
\vocab{affine cone over $X$}
|
|
\[
|
|
C(X) = \{0\} \cup \{(x_0,\ldots,x_n) \in \mathfrak{k}^{n+1} \setminus \{0\} | [x_0,\ldots,x_n] \in X\}
|
|
\]
|
|
If $X = \Vp(I)$ where $I \subseteq A_+ = \mathfrak{k}[X_0,\ldots,X_n]_+$
|
|
is homogeneous, then $C(X) = \Va(I)$.
|
|
\end{definition}
|
|
\begin{proposition}\,
|
|
\label{conedim}
|
|
\begin{itemize}
|
|
\item
|
|
$C(X)$ is irreducible iff $X$ is irreducible or $X = \emptyset$.
|
|
\item
|
|
If $X$ is irreducible, then
|
|
\[\dim(C(X)) = \dim(X) + 1\] and
|
|
\[\codim(C(X), \mathfrak{k}^{n+1}) = \codim(X, \mathbb{P}^n).\]
|
|
\end{itemize}
|
|
\end{proposition}
|
|
\begin{proof}
|
|
The first assertion follows from \ref{bijproj} and \ref{bijiredprim}
|
|
(bijection of irreducible subsets and prime ideals in the projective
|
|
and affine case).
|
|
|
|
Let $d = \dim(X)$ and
|
|
\[
|
|
X_0 \subsetneq \ldots \subsetneq X_d = X
|
|
\subsetneq X_{d+1} \subsetneq \ldots \subsetneq X_n = \mathbb{P}^n
|
|
\]
|
|
be a chain of irreducible subsets of $\mathbb{P}^n$.
|
|
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 $\mathbb{P}^n$}
|
|
\begin{definition}[Hypersurface]
|
|
Let $n > 0$.
|
|
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 $\mathbb{P}^n$
|
|
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 $\mathbb{P}^n$.
|
|
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 $H \subseteq \mathbb{P}^n$ has \vocab{degree $d$}
|
|
if $H = \Vp(P)$,
|
|
where $P \in A_d$ is an irreducible polynomial.
|
|
\end{definition}
|
|
|
|
\subsubsection{Application to intersections in $\mathbb{P}^n$
|
|
and Bezout's theorem}
|
|
\begin{corollary}
|
|
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 $A \cap B$
|
|
has dimension $\ge a + b - n$.
|
|
\end{corollary}
|
|
|
|
\begin{remark}
|
|
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}
|
|
|
|
\begin{proof}
|
|
The lower bound on the dimension of irreducible components of $A \cap B$ is
|
|
easily derived from the similar affine result
|
|
(corollary of the principal ideal theorem, \ref{codimintersection}).
|
|
|
|
From the definition of the affine cone it follows that
|
|
$C(A \cap B) = C(A) \cap C(B)$.
|
|
We have $\dim(C(A)) = a+1$ and $\dim(C(B)) = b + 1$ by \ref{conedim}.
|
|
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 $\mathbb{P}^2$,
|
|
then $A \cap B$ has $ab$ points counted by (suitably defined) multiplicity.
|
|
\end{remark}
|
|
|
|
|
|
%TODO Proof of "Dimension of P^n"
|
|
% SLIDE APPLICATION TO HYPERSURFACES IN $\P^n$
|
|
%ERROR: C(H) = V_A(P)
|
|
%If n = 0, P = 0, V_P(P) = \emptyset is a problem!
|