677 lines
25 KiB
TeX
677 lines
25 KiB
TeX
Let $\mathfrak{l}$ be any field.
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\begin{definition}
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For a $\mathfrak{l}$-vector space $V$, let $\mathbb{P}(V)$ be
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the set of
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one-dimensional subspaces of $V$.
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Let $\mathbb{P}^n(\mathfrak{l}) \coloneqq
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\mathbb{P}(\mathfrak{l}^{n+1})$, the
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\vocab[Projective space]{$n$-dimensional projective space over $\mathfrak{l}$}.
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If $\mathfrak{l}$ is kept fixed, we will often write
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$\mathbb{P}^n$ for
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$\mathbb{P}^n(\mathfrak{l})$.
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When dealing with $\mathbb{P}^n$, the usual convention is to use
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$0$ as the
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index of the first coordinate.
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We denote the one-dimensional subspace generated by $(x_0,\ldots,x_n) \in
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\mathfrak{k}^{n+1} \setminus \{0\}$ by
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$[x_0,\ldots,x_n] \in \mathbb{P}^n$.
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If $x = [x_0,\ldots,x_n] \in \mathbb{P}^n$, the
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$(x_{i})_{i=0}^n$ are
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called
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\vocab{homogeneous coordinates} of $x$.
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At least one of the $x_{i}$ must be $\neq 0$.
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\end{definition}
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\begin{remark}
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There are points $[1,0],
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[0,1] \in \mathbb{P}^1$ but there
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is no point $[0,0]
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\in \mathbb{P}^1$.
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\end{remark}
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\begin{definition}[Infinite hyperplane]
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For $0 \le i \le n$ let $U_i \subseteq
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\mathbb{P}^n$ denote the set of $[x_0,\ldots,x_{n}]$ with
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$x_{i}\neq 0$.
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This is a correct definition since two different sets $[x_0,\ldots,x_{n}]$ and
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$[\xi_0,\ldots,\xi_n]$ of homogeneous coordinates for the
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same point $x \in
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\mathbb{P}^n$ differ by scaling with a $\lambda \in
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\mathfrak{l}^{\times}$,
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$x_i = \lambda \xi_i$.
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Since not all $x_i$ may be $0$, $\mathbb{P}^n =
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\bigcup_{i=0}^n U_i$.
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We identify $\mathbb{A}^n =
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\mathbb{A}^n(\mathfrak{l}) = \mathfrak{l}^n$
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with
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$U_0$ by identifying $(x_1,\ldots,x_n) \in \mathbb{A}^n$ with
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$[1,x_1,\ldots,x_n] \in \mathbb{P}^n$.
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Then $\mathbb{P}^1 = \mathbb{A}^1 \cup \{\infty\} $
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where $\infty=[0,1]$.
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More generally, when $n > 0$ $\mathbb{P}^n \setminus
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\mathbb{A}^n$ can be
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identified with $\mathbb{P}^{n-1}$ identifying
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$[0,x_1,\ldots,x_n] \in
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\mathbb{P}^n \setminus \mathbb{A}^n$ with
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$[x_1,\ldots,x_n] \in
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\mathbb{P}^{n-1}$.
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Thus $\mathbb{P}^n$ is $\mathbb{A}^n \cong
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\mathfrak{l}^n$ with a copy of
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$\mathbb{P}^{n-1}$ added as an \vocab{infinite hyperplane} .
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\end{definition}
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\subsubsection{Graded rings and homogeneous ideals}
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\begin{notation}
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Let $\mathbb{I} = \N$ or $\mathbb{I} = \Z$.
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\end{notation}
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\begin{definition}
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By an \vocab[Graded ring]{$\mathbb{I}$-graded ring} $A_\bullet$ we understand a
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ring $A$ with a collection $(A_d)_{d \in \mathbb{I}}$ of
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subgroups of the
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additive group $(A, +)$ such that $A_a \cdot A_b \subseteq
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A_{a + b}$ for $a,b
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\in \mathbb{I}$ and such that $A = \bigoplus_{d \in \mathbb{I}} A_d$ in
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the
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sense that every $r \in A$ has a unique decomposition $r =
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\sum_{d \in
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\mathbb{I}} r_d$ with $r_d \in A_d$ and but finitely many $r_d
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\neq 0$.
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We call the $r_d$ the \vocab{homogeneous components} of $r$.
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An ideal $I \subseteq A$ is called \vocab{homogeneous} if $r \in I
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\implies
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\forall d \in \mathbb{I} ~ r_d \in I_d$ where $I_d \coloneqq I
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\cap A_d$.
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By a \vocab{graded ring} we understand an $\N$-graded ring.
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Tin this case, $A_{+} \coloneqq \bigoplus_{d=1}^{\infty}
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A_d = \{r \in A | r_0
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= 0\} $ is called the \vocab{augmentation ideal} of $A$.
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\end{definition}
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\begin{remark}[Decomposition of $1$]
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If $1 = \sum_{d \in \mathbb{I}} \varepsilon_d$ is the
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decomposition into homogeneous components, then $\varepsilon_a = 1 \cdot
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\varepsilon_a = \sum_{b \in \mathbb{I}} \varepsilon_a\varepsilon_b$ with
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$\varepsilon_a\varepsilon_b \in A_{a+b}$.
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By the uniqueness of the decomposition into homogeneous components,
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$\varepsilon_a \varepsilon_0 = \varepsilon_a$ and $b \neq 0 \implies
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\varepsilon_a \varepsilon_b = 0$.
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Applying the last equation with $a = 0$ gives $b\neq 0 \implies \varepsilon_b =
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\varepsilon_0 \varepsilon _b = 0$.
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Thus $1 = \varepsilon_0 \in A_0$.
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\end{remark}
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\begin{remark}
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The augmentation ideal of a graded ring is a homogeneous ideal.
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\end{remark}
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% Graded rings and homogeneous ideals (2)
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\begin{proposition}
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\footnote{This holds for both $\Z$-graded and $\N$-graded rings.}
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\begin{itemize}
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\item
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A principal ideal generated by a homogeneous element is homogeneous.
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\item
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The operations $\sum, \bigcap, \sqrt{}$ preserve homogeneity.
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\item
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An ideal is homogeneous iff it can be generated by a family of homogeneous
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elements.
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\end{itemize}
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\end{proposition}
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\begin{proof}
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Most assertions are trivial.
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We only show that $J$ homogeneous $\implies \sqrt{J} $
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homogeneous.
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Let $A$ be $\mathbb{I}$-graded, $f \in \sqrt{J} $ and
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$f = \sum_{d \in
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\mathbb{I}} f_d$ the decomposition.
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To show that all $f_d \in \sqrt{J} $, we use induction on $N_f
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\coloneqq \# \{d
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\in \mathbb{I} | f_d \neq 0\}$.
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$N_f = 0$ is trivial.
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Suppose $N_f > 0$ and $e \in \mathbb{I}$ is maximal with $f_e \neq
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0$.
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For $l \in \N$, the $le$-th homogeneous component of $f^l$ is $f_e^l$.
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Choosing $l$ large enough such that $f^l \in J$ and using the homogeneity of
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$J$, we find $f_e \in \sqrt{J}$.
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As $\sqrt{J} $ is an ideal, $\tilde f \coloneqq f - f_e \in
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\sqrt{J} $.
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As $N_{\tilde f} = N_f -1$, the induction assumption may be
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applied to $\tilde
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f$ and shows $f_d \in \sqrt{J} $ for $d \neq e$.
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\end{proof}
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\begin{fact}
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A homogeneous ideal is finitely generated iff it can be generated by finitely
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many of its homogeneous elements.
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In particular, this is always the case when $A$ is a Noetherian ring.
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\end{fact}
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\subsubsection{The Zariski topology on $\mathbb{P}^n$}
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\begin{notation}
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Recall that for $\alpha \in \N^{n+1}$ $|\alpha| =
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\sum_{i=0}^{n} \alpha_i$ and
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$x^\alpha = x_0^{\alpha_0} \cdot \ldots \cdot x_n^{\alpha_n}$.
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\end{notation}
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\begin{definition}[Homogeneous polynomials]
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Let $R$ be any ring and $f = \sum_{\alpha \in \N^{n+1}}
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f_\alpha X^{\alpha}\in R[X_0,\ldots,X_n]$.
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We say that $f$ is \vocab{homogeneous of degree $d$} if $|\alpha| \neq d
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\implies f_\alpha = 0$ .
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We denote the subset of homogeneous polynomials of degree $d$ by
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$R[X_0,\ldots,X_n]_d \subseteq R[X_0,\ldots,X_n]$.
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\end{definition}
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\begin{remark}
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This definition gives $R$ the structure of a graded ring.
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\end{remark}
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\begin{definition}[Zariski topology on $\mathbb{P}^n(\mathfrak{k})$]
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\label{ztoppn}
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Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
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\footnote{As always, $\mathfrak{k}$ is algebraically closed}
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For $f \in A_d = \mathfrak{k}[X_0,\ldots,X_n]_d$, the validity of the equation
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$f(x_0,\ldots,x_{n}) = 0$ does not depend on the choice of homogeneous
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coordinates, as
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\[
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f(\lambda x_0,\ldots, \lambda x_n) 0 \lambda^d
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f(x_0,\ldots,x_n)
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\]
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Let $\Vp(f) \coloneqq \{x \in \mathbb{P}^n | f(x)
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= 0\}$.
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We call a subset $X \subseteq \mathbb{P}^n$ Zariski-closed if it
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can be
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represented as
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\[
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X = \bigcap_{i=1}^k \Vp(f_i)
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\]
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where the $f_i \in A_{d_i}$
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are homogeneous polynomials.
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\end{definition}
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\pagebreak
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\begin{fact}
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If $X = \bigcap_{i = 1}^k \Vp(f_i) \subseteq
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\mathbb{P}^n$ is closed, then $Y
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= X \cap \mathbb{A}^n$ can be identified with the closed subset
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\[
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\{(x_1,\ldots,x_n) \in \mathfrak{k}^n | f_i(1,x_1,\ldots,x_n) = 0, 1
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\le i \le
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k\} \subseteq \mathfrak{k}^n
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\]
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Conversely, if $Y \subseteq \mathfrak{k}^n$ is
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closed it has the form
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\[
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\{(x_1,\ldots,x_n) \in \mathfrak{k}^n |
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g_i(x_1,\ldots,x_n) = 0, 1 \le i \le k\}
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\]
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and can thus be identified with $X
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\cap \mathbb{A}^n$ where $X \coloneqq \bigcap_{i=1}^k
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\Vp(f_i)$ is given by
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\[
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f_i(X_0,\ldots,X_n) \coloneqq X_0^{d_i} g_i(X_1 / X_0,\ldots, X_n / X_0), d_i
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\ge \deg(g_i)
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\]
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Thus, the Zariski topology on $\mathfrak{k}^n$ can be
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identified with the topology induced by the Zariski topology on
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$\mathbb{A}^n =
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U_0$, and the same holds for $U_i$ with $0 \le i \le n$.
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In this sense, the Zariski topology on $\mathbb{P}^n$ can be
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thought of as
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gluing the Zariski topologies on the $U_i \cong \mathfrak{k}^n$.
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\end{fact}
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% The Zariski topology on P^n (2)
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\begin{definition}
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Let $I \subseteq A = \mathfrak{k}[X_0,\ldots,X_n]$ be a homogeneous ideal.
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Let $\Vp(I) \coloneqq
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\{[x_0,\ldots,_n] \in \mathbb{P}^n |
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\forall f \in I ~
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f(x_0,\ldots,x_n) = 0\}$ As $I$ is homogeneous, it is sufficient to impose this
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condition for the homogeneous elements $f \in I$.
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Because $A$ is Noetherian, $I$ can finitely generated by homogeneous elements
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$(f_i)_{i=1}^k$ and
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$\Vp(I)=\bigcap_{i=1}^k \Vp(f_i)$ as
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in
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\ref{ztoppn}.
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Conversely, if the homogeneous $f_i$ are given, then $I = \langle
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f_1,\ldots,f_k \rangle_A$ is homogeneous.
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\end{definition}
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\begin{remark}
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Note that $V(A) = V(A_+) = \emptyset$.
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\end{remark}
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\begin{fact}
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For homogeneous ideals in $A$ and $m \in \N$, we have:
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\begin{itemize}
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\item
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$\Vp(\sum_{\lambda \in \Lambda} I_\lambda) = \bigcap_{\lambda \in \Lambda}
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\Vp(I_\lambda)$
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\item
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$\Vp(\bigcap_{k=1}^m I_k) = \Vp(\prod_{k=1}^{m} I_k) =
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\bigcup_{k=1}^m \Vp(I_k)$
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\item
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$\Vp(\sqrt{I}) = \Vp(I)$
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\end{itemize}
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\end{fact}
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\begin{fact}
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If $X = \bigcup_{\lambda \in \Lambda} U_\lambda$ is an
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open covering of a topological space then $X$ is Noetherian iff there is a
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finite subcovering and all $U_\lambda$ are Noetherian.
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\end{fact}
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\begin{proof}
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By definition, a topological space is Noetherian $\iff$ all open subsets are
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quasi-compact.
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\end{proof}
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\begin{corollary}
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The Zariski topology on $\mathbb{P}^n$ is indeed a topology.
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The induced topology on the open set $\mathbb{A}^n =
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\mathbb{P}^n \setminus
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\Vp(X_0) \cong \mathfrak{k}^n$ is the Zariski
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topology on $\mathfrak{k}^n$.
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The same holds for all $U_i = \mathbb{P}^n \setminus
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\Vp(X_i) \cong
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\mathfrak{k}^n$.
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Moreover, the topological space $\mathbb{P}^n$ is Noetherian.
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\end{corollary}
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\subsection{Noetherianness of graded rings}
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\begin{proposition}
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For a graded ring $R_{\bullet}$, the following conditions
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are equivalent:
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\begin{enumerate}[A]
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\item
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$R$ is Noetherian.
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\item
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Every homogeneous ideal of $R_{\bullet}$ is finitely
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generated.
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\item
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Every chain $I_0\subseteq I_1 \subseteq \ldots$ of homogeneous ideals
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terminates.
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\item
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Every set $\mathfrak{M} \neq \emptyset$ of homogeneous ideals has a
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$\subseteq$-maximal element.
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\item
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$R_0$ is Noetherian and the ideal $R_+$ is finitely generated.
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\item
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$R_0$ is Noetherian and $R / R_0$ is of finite type.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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\noindent\textbf{A $\implies$ B,C,D} trivial.
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\noindent\textbf{B $\iff$ C $\iff$ D} similar to the proof about Noetherianness.
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\noindent\textbf{B $\land$ C $\implies $E}
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B implies that $R_+$ is finitely generated.
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Since $I \oplus R_+$ is homogeneous for any homogeneous ideal $I \subseteq
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R_0$, C implies the Noetherianness of $R_0$.
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\noindent\textbf{E $\implies$ F}
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Let $R_+$ be generated by $f_i \in R_{d_i}, d_i > 0$ as
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an ideal.
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\begin{claim}
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The $R_0$-subalgebra $\tilde R$ of $R$ generated by the $f_i$ equals $R$.
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\end{claim}
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\begin{subproof}
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It is sufficient to show that every homogeneous $f \in R_d$ belongs to $\tilde
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R$.
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We use induction on $d$.
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The case of $d = 0$ is trivial.
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Let $d > 0$ and $R_e \subseteq \tilde R$ for all $e < d$.
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as $f \in R_+$, $f = \sum_{i=1}^{k} g_if_i$.
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Let $f_a = \sum_{i=1}^{k} g_{i, a-d_i} f_i$, where
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$g_i = \sum_{b=0}^{\infty}
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g_{i,b}$ is the decomposition into homogeneous
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components.
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Then $f = \sum_{a=0}^{\infty} f_a$ is the decomposition of $f$ into
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homogeneous
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components, hence $a \neq d \implies f_a = 0 $.
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Thus we may assume $g_i \in R_{d-d_i}$.
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As $d_i > 0$, the induction assumption may now be applied to $g_i$, hence $g_i
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\in \tilde R$, hence $f \in \tilde R$.
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\end{subproof}
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\noindent\textbf{F $\implies$ A}
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Hilbert's Basissatz (
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\ref{basissatz})
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\end{proof}
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\subsection{The projective form of the Nullstellensatz and the closed subsets
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of $\mathbb{P}^n$} Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
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% Lecture 12
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\begin{proposition}[Projective form of the Nullstellensatz]
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\label{hnsp}
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If $I \subseteq A$ is a homogeneous ideal and $f \in A_d$ with $d>0$, then
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$\Vp(I) \subseteq \Vp(f) \iff f \in
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\sqrt{I}$.
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\end{proposition}
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\begin{proof}
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$\impliedby$ is clear.
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Let $\Vp(I) \subseteq \Vp(f)$.
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If $x = (x_0,\ldots,x_n) \in \Va(I)$, then either $x = 0$ in
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which case $f(x) =
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0$ since $d > 0$ or the point $[x_0,\ldots,x_n] \in
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\mathbb{P}^n$ is
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well-defined and belongs to $\Vp(I) \subseteq
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\Vp(f)$, hence $f(x) = 0$.
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Thus $\Va(I) \subseteq \Va(f)$ and $f \in
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\sqrt{I}$ be the Nullstellensatz
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(
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\ref{hns3}).
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\end{proof}
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\begin{definition}
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\footnote{This definition is not too important, the characterization in the following remark suffices.}.
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For a graded ring $R_\bullet$, let $\Proj(R_\bullet)$ be the set of
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$\fp \in
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\Spec R$ such that $\fp$ is a homogeneous ideal and $\fp \not\supseteq R_+$.
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\end{definition}
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\begin{remark}
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\label{proja}
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As the elements of $A_0 \setminus \{0\}$ are units in $A$ it follows that for
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every homogeneous ideal $I$ we have $I \subseteq A_+$ or $I = A$.
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In particular, $\Proj(A_\bullet) = \{\fp \in \Spec A \setminus A_+ |
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\fp
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\text{ is homogeneous}\} $.
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\end{remark}
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\begin{proposition}
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\label{bijproj}
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There is a bijection
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\begin{align*}
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f: \{I \subseteq A_+ | I \text{ homogeneous
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ideal}, I = \sqrt{I}\} & \longrightarrow \{X \subseteq \mathbb{P}^n | X \text{
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closed}\} \\ I & \longmapsto \Vp(I)\\ \langle \{f \in A_d | d > 0, X
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\subseteq \Vp(f)\} \rangle & \longmapsfrom X
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\end{align*}
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Under this bijection,
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the irreducible subsets correspond to the elements of
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$\Proj(A_\bullet)$.
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\end{proposition}
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\begin{proof}
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From the projective form of the Nullstellensatz it follows that $f$ is
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injective and that $f^{-1}(\Vp\left( I \right))
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= \sqrt{I} = I$.
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If $X \subseteq \mathbb{P}^n$ is closed, then $X =
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\Vp(J)$ for some homogeneous
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ideal $J \subseteq A$.
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Without loss of generality loss of generality $J = \sqrt{J}$.
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If $J \not\subseteq A_+$, then $J = A$ (
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\ref{proja}), hence $X =
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\Vp(J) =
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\emptyset = \Vp(A_+)$.
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Thus we may assume $J \subseteq A_+$, and $f$ is surjective.
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Suppose $\fp \in \Proj(A_\bullet)$.
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Then $\fp \neq A_+$ hence $X = \Vp(\fp) \neq \emptyset$ by the
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proven part of
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the proposition.
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Assume $X = X_1 \cup X_2$ is a decomposition into proper closed subsets, where
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$X_k = \Vp(I_k)$ 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
|
|
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 $f,g \in R fg \in I \implies f \in I
|
|
\lor g \in I$.
|
|
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 $R$.
|
|
\[
|
|
\{\fp \in \Spec R | \fp \text{ is a homogeneous ideal of }
|
|
R_\bullet\} = \Proj(R_\bullet) \sqcup \{\fp \oplus R_+ | \fp \in \Spec
|
|
R_0\}
|
|
\]
|
|
\end{remark}
|
|
|
|
\subsection{Dimension of $\mathbb{P}^n$}
|
|
\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{align*}
|
|
\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.
|
|
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!
|