some small changes

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Josia Pietsch 2023-07-22 21:52:48 +02:00
parent 80024d147a
commit f4bed5efb3
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GPG key ID: E70B571D66986A2D
8 changed files with 270 additions and 339 deletions

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@ -1,4 +1,5 @@
\documentclass[10pt,ngerman,a4paper, fancyfoot, git]{mkessler-script} \documentclass[english, fancyfoot% , git
]{mkessler-script}
\course{Algebra I} \course{Algebra I}
\lecturer{Prof.~Dr.~Jens Franke} \lecturer{Prof.~Dr.~Jens Franke}

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@ -1,7 +1,7 @@
\ProvidesPackage{algebra}[2022/02/10 - Style file for notes of Algebra I] \ProvidesPackage{algebra}[2022/02/10 - Style file for notes of Algebra I]
\RequirePackage{mkessler-math} \RequirePackage[english]{mkessler-math}
\RequirePackage{mkessler-refproof} \RequirePackage{mkessler-refproof}
\RequirePackage[number in = section]{fancythm} \RequirePackage[number in = section]{fancythm}
@ -11,7 +11,7 @@
\RequirePackage[utf8x]{inputenc} \RequirePackage[utf8x]{inputenc}
\RequirePackage{babel} \RequirePackage{babel}
\RequirePackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry} % \RequirePackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}
\RequirePackage[normalem]{ulem} \RequirePackage[normalem]{ulem}
\RequirePackage{pdflscape} \RequirePackage{pdflscape}

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@ -122,11 +122,11 @@
\begin{definition}[Generated (sub)algebra, algebra of finite type] \begin{definition}[Generated (sub)algebra, algebra of finite type]
Let $(A, \alpha)$ be an $R$-algebra. Let $(A, \alpha)$ be an $R$-algebra.
\begin{align} \begin{align*}
\alpha: R[X_1,\ldots,X_m] & \longrightarrow A[X_1,\ldots,X_m] \\ \alpha: R[X_1,\ldots,X_m] & \longrightarrow A[X_1,\ldots,X_m] \\
P = \sum_{\beta \in \N^m} p_\beta X^{\beta} & \longmapsto \sum_{\beta \in \N^m} \alpha(p_\beta) P = \sum_{\beta \in \N^m} p_\beta X^{\beta} & \longmapsto \sum_{\beta \in \N^m} \alpha(p_\beta)
X^{\beta} X^{\beta}
\end{align} \end{align*}
is a ring homomorphism. is a ring homomorphism.
We will sometimes write $P(a_1,\ldots,a_m)$ instead of We will sometimes write $P(a_1,\ldots,a_m)$ instead of
$(\alpha(P))(a_1,\ldots,a_m)$. $(\alpha(P))(a_1,\ldots,a_m)$.
@ -285,10 +285,10 @@ commutative rings with $1$.
and finite over $R$. and finite over $R$.
Let $(b_i)_{i=1}^{n}$ generate $B$ as an Let $(b_i)_{i=1}^{n}$ generate $B$ as an
$R$-module. $R$-module.
\begin{align} \begin{align*}
q: R^n & \longrightarrow B \\ q: R^n & \longrightarrow B \\
(r_1,\ldots,r_n) & \longmapsto \sum_{i=1}^{n} r_i b_i (r_1,\ldots,r_n) & \longmapsto \sum_{i=1}^{n} r_i b_i
\end{align} \end{align*}
is surjective. is surjective.
Thus there are $\rho_{i} = \left( r_{i,j} \right)_{j=1}^n \in R^n$ Thus there are $\rho_{i} = \left( r_{i,j} \right)_{j=1}^n \in R^n$
such that such that
@ -478,10 +478,10 @@ commutative rings with $1$.
Then there are $a = (a_i)_{i=1}^{n} \in A$ which Then there are $a = (a_i)_{i=1}^{n} \in A$ which
are algebraically independent are algebraically independent
over $K$, i.e. the ring homomorphism over $K$, i.e. the ring homomorphism
\begin{align} \begin{align*}
\ev_a: K[X_1,\ldots,X_n] & \ev_a: K[X_1,\ldots,X_n] &
\longrightarrow A \\ P & \longmapsto P(a_1,\ldots,a_n) \longrightarrow A \\ P & \longmapsto P(a_1,\ldots,a_n)
\end{align} \end{align*}
is is
injective. injective.
$n$ and the $a_i$ can be chosen such that $A$ is finite over the image of $n$ and the $a_i$ can be chosen such that $A$ is finite over the image of
@ -496,53 +496,41 @@ commutative rings with $1$.
Let $\tilde{A}$ be the $K$-subalgebra generated by the $a_i$. Let $\tilde{A}$ be the $K$-subalgebra generated by the $a_i$.
If suffices to show that the $a_i$ are algebraically independent. If suffices to show that the $a_i$ are algebraically independent.
Since $A$ is of finite type over $K$ and thus over $\tilde{A}$, Since $A$ is of finite type over $K$ and thus over $\tilde{A}$,
by fact by fact \ref{ftaiimplf} (integral and finite type $\implies$ finite),
\ref{ftaiimplf} (integral and finite type $\implies$ finite) $A$ is finite over $A$ is finite over $\tilde{A}$.
$\tilde{A}$. Thus we only need to show that the $a_i$ are algebraically independent
Thus we only need to show that the $a_i$ are algebraically independent over over $K$.
$K$. Assume there is $P \in K[X_1,\ldots,X_n] \setminus \{0\}$
Assume there is $P \in K[X_1,\ldots,X_n] \setminus \{0\} $ such such that $P(a_1,\ldots,a_n) = 0$.
that
$P(a_1,\ldots,a_n) = 0$.
Let $P = \sum_{\alpha \in \N^n} p_\alpha X^{\alpha}$ and Let $P = \sum_{\alpha \in \N^n} p_\alpha X^{\alpha}$ and
$S = \{ \alpha \in $S = \{ \alpha \in \N^n | p_\alpha \neq 0\}$.
\N^n | p_\alpha \neq 0\}$. For $\vec{k} = (k_i)_{i=1}^{n} \in \N^n$ and $\alpha \in \N^n$ we define
For $\vec{k} = (k_i)_{i=1}^{n} $w_{\vec{k}}(\alpha) \coloneqq \sum_{i=1}^{n} k_i\alpha_i$.
\in \N^n$ and $\alpha \in \N^n$ we define
$w_{\vec{k}}(\alpha) \coloneqq \sum_{i=1}^{n}
k_i\alpha_i$.
By By \ref{nntechlemma} it is possible to choose $\vec{k} \in \N^n$
\ref{nntechlemma} it is possible to choose $\vec{k} such that $k_1 = 1$ and for $\alpha \neq \beta \in S$ we have
\in \N^n$ such that $k_1 $w_{\vec{k}}(\alpha) \neq w_{\vec{k}}(\beta)$.
= 1$ and for $\alpha \neq \beta \in S$ we have
$w_{\vec{k}}(\alpha) \neq
w_{\vec{k}}(\beta)$.
Define $b_i \coloneqq a_{i+1} - Define $b_i \coloneqq a_{i+1} - a^{k_{i+1}}_1$ for $1 \le i < n$.
a^{k_{i+1}}_1$ for $1 \le i < n$.
\begin{claim} \begin{claim}
$A$ is integral over the subalgebra $B$ generated by the $b_i$. $A$ is integral over the subalgebra $B$ generated by the $b_i$.
\end{claim} \end{claim}
\begin{subproof} \begin{subproof}
By the transitivity of integrality, it is sufficient to show that the $a_i$ are By the transitivity of integrality,
integral over $B$. it is sufficient to show that the $a_i$ are integral over $B$.
For $i > 1$ we have $a_i = b_{i-1} + a_1^{k_i}$. For $i > 1$ we have $a_i = b_{i-1} + a_1^{k_i}$.
Thus it suffices to show this for $a_1$. Thus it suffices to show this for $a_1$.
Define $Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots, Define
b_{n-1} + T^{k_n}) \in $Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots,
B[T]$. b_{n-1} + T^{k_n}) \in B[T]$.
We have $0 = P(a_1,\ldots, a_n) = Q(a_1)$. We have $0 = P(a_1,\ldots, a_n) = Q(a_1)$.
Hence it suffices to show that the leading coefficient of $Q$ is a unit. Hence it suffices to show that the leading coefficient of $Q$ is a unit.
We have We have
\[ \[
T^{\alpha_1} \prod_{i=1}^{n-1} (b_i + T^{k_i + 1})^{\alpha_{i+1}} T^{\alpha_1} \prod_{i=1}^{n-1} (b_i + T^{k_i + 1})^{\alpha_{i+1}} =
=
T^{w_{\vec k}(\alpha)} + T^{w_{\vec k}(\alpha)} +
\sum_{l = 0}^{w_{\vec k}(\alpha) - 1} \sum_{l = 0}^{w_{\vec k}(\alpha) - 1} \beta_{\alpha, l} T^l
\beta_{\alpha,
l} T^l
\] \]
with suitable $\beta_{\alpha, l} \in B$. with suitable $\beta_{\alpha, l} \in B$.
@ -551,10 +539,9 @@ commutative rings with $1$.
Q(T) = p_{\alpha} T^{w_{\vec k}(\alpha)} Q(T) = p_{\alpha} T^{w_{\vec k}(\alpha)}
+ \sum_{j=0}^{w_{\vec k}(\alpha) - 1} q_j T^j + \sum_{j=0}^{w_{\vec k}(\alpha) - 1} q_j T^j
\] \]
with $q_j \in B$ and with $q_j \in B$ and
$\alpha$ such that $w_{\vec k }(\alpha)$ is maximal subject $\alpha$ such that $w_{\vec k}(\alpha)$ is maximal subject
to the condition to the condition $p_\alpha \neq 0$.
$p_\alpha \neq 0$.
Thus the leading coefficient of $Q$ is a unit. Thus the leading coefficient of $Q$ is a unit.
\end{subproof} \end{subproof}
@ -562,4 +549,3 @@ commutative rings with $1$.
elements $b_i$. elements $b_i$.
\end{proof} \end{proof}

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@ -6,21 +6,19 @@ Let $\mathfrak{k}$ be a field, $R \coloneqq
\begin{definition}[zero] \begin{definition}[zero]
$x \in \mathfrak{k}^n$ is \vocab[Ideal!zero]{a zero of $I$} $x \in \mathfrak{k}^n$ is \vocab[Ideal!zero]{a zero of $I$}
if $\forall x \in I: P(x) = 0$. if $\forall x \in I: P(x) = 0$.
Let $\Va(I)$ denote the set of zeros if $I$ in Let $\Va(I)$ denote the set of zeros if $I$ in $\mathfrak{k}^n$.
$\mathfrak{k}^n$.
The \vocab[Ideal! The \vocab[Ideal!zero]{zero in a field extension %
zero]{zero in a field extension $\mathfrak{i}$ of $\mathfrak{k}$} is defined similarly. $\mathfrak{i}$ of $\mathfrak{k}$} is defined similarly.
\end{definition} \end{definition}
\begin{remark}[Set of zeros and generators] \begin{remark}[Set of zeros and generators]
Let $I$ be generated by $S$. Let $I$ be generated by $S$.
Then $\{x \in R | \forall s \in S: s(x) = 0\} = \Va(I)$. Then $\{x \in R | \forall s \in S: s(x) = 0\} = \Va(I)$.
Thus zero sets of ideals correspond to solutions sets to systems of polynomial Thus zero sets of ideals correspond to solutions sets to systems of
equations. polynomial equations.
If $S, \tilde{S}$ generate the same ideal $I$ they have the same If $S, \tilde{S}$ generate the same ideal $I$ they have the same
set of set of solutions.
solutions.
Therefore we only consider zero sets of ideals. Therefore we only consider zero sets of ideals.
\end{remark} \end{remark}
@ -32,16 +30,15 @@ Let $\mathfrak{k}$ be a field, $R \coloneqq
\end{theorem} \end{theorem}
\begin{remark} \begin{remark}
Will be shown later (see proof of Will be shown later (see proof of \ref{hns1b}).
\ref{hns1b}). It is trivial if $n = 1$:
Trivial if $n = 1$: $R$ is a PID, thus $I = pR$ for some $p \in R$. $R$ is a PID, thus $I = pR$ for some $p \in R$.
Since $I \neq R$ $p = 0$ or $P$ is non-constant. Since $I \neq R$ $p = 0$ or $P$ is non-constant.
$\mathfrak{k}$ algebraically closed $\leadsto$ there exists a zero of $\mathfrak{k}$ algebraically closed
$p$.\\ $\leadsto$ there exists a zero of $p$.\\
If $\mathfrak{k}$ is not algebraically closed and $n > 0$, the If $\mathfrak{k}$ is not algebraically closed and $n > 0$,
theorem fails the theorem fails (consider $I = p(X_1) R$).
(consider $I = p(X_1) R$).
\end{remark} \end{remark}
Equivalent\footnote{used in a vague sense here} formulation: Equivalent\footnote{used in a vague sense here} formulation:
@ -55,82 +52,68 @@ Equivalent\footnote{used in a vague sense here} formulation:
\begin{proof} \begin{proof}
\begin{itemize} \begin{itemize}
\item[$\implies$] \item[$\implies$]
If $(l_i)_{i=1}^{m}$ is a base of $L$ as a If $(l_i)_{i=1}^{m}$ is a base of $L$ as a $K$-vector space,
$K$-vector space, then $L$ is generated by the $l_i$ as a $K$-algebra. then $L$ is generated by the $l_i$ as a $K$-algebra.
\item[$\impliedby$ ] \item[$\impliedby$ ]
Apply the Noether normalization theorem ( Apply the Noether normalization theorem (\ref{noenort}) to $A = L$.
\ref{noenort}) to $A = L$.
This yields an injective ring homomorphism $\ev_a: This yields an injective ring homomorphism $\ev_a:
K[X_1,\ldots,X_n] \to A$ K[X_1,\ldots,X_n] \to A$
such that $A$ is finite over the image of $\ev_a$. such that $A$ is finite over the image of $\ev_a$.
By the fact about integrality and fields ( By the fact about integrality and fields (\ref{fintaf}),
\ref{fintaf}), the the isomorphic image
isomorphic image
of $\ev_a$ is a field. of $\ev_a$ is a field.
Thus $K[X_1,\ldots, X_n]$ is a field $\implies n = 0$. Thus $K[X_1,\ldots, X_n]$ is a field $\implies n = 0$.
Thus $L / K$ is a finite ring extension, hence a finite field extension. Thus $L / K$ is a finite ring extension,
hence a finite field extension.
\end{itemize} \end{itemize}
\end{proof} \end{proof}
\begin{remark} \begin{remark}
We will see several additional proofs of this theorem. We will see several additional proofs of this theorem.
See See \ref{hns2unc} and \ref{rfuncnft}.
\ref{hns2unc} and
\ref{rfuncnft}.
All will be accepted in the exam. All will be accepted in the exam.
\ref{hns3} and \ref{hns3} and \ref{hnsp} are closely related.
\ref{hnsp} are closely related.
\end{remark} \end{remark}
\begin{theorem}[Hilbert's Nullstellensatz (1b)] \begin{theorem}[Hilbert's Nullstellensatz (1b)]
\label{hns1b} \label{hns1b}
Let $\mathfrak{l}$ be a field and $I \subset R = Let $\mathfrak{l}$ be a field and
\mathfrak{l}[X_1,\ldots,X_m]$ $I \subset R = \mathfrak{l}[X_1,\ldots,X_m]$ a proper ideal.
a proper ideal. Then there are a finite field extension $\mathfrak{i}$ of $\mathfrak{l}$
Then there are a finite field extension $\mathfrak{i}$ of and a zero of $I$ in $\mathfrak{i}^m$.
$\mathfrak{l}$ and a
zero of $I$ in $\mathfrak{i}^m$.
\end{theorem} \end{theorem}
\begin{proof} \begin{proof}
(HNS2 ( (HNS2 (\ref{hns2}) $\implies$ HNS1b (\ref{hns1b}))
\ref{hns2}) $\implies$ HNS1b ( $I \subseteq \mathfrak{m}$ for some maximal ideal.
\ref{hns1b})) $R /\mathfrak{m}$ is a field,
$I \subseteq \mathfrak{m}$ for some maximal ideal. $R / since $\mathfrak{m}$ is maximal.
\mathfrak{m}$ is a field, since $\mathfrak{m}$ is maximal. $R / \mathfrak{m}$ is of finite type,
$R / \mathfrak{m}$ is of finite type, since the images of the $X_i$ since the images of the $X_i$ generate it as a $\mathfrak{l}$-algebra.
generate it as a $\mathfrak{l}$-algebra. There are thus a field extension $\mathfrak{i} / \mathfrak{l}$
There are thus a field extension $\mathfrak{i} / and an isomorphism $R / \mathfrak{m} \xrightarrow{\iota} \mathfrak{i}$ of
\mathfrak{l}$ and an
isomorphism $R / \mathfrak{m} \xrightarrow{\iota}
\mathfrak{i}$ of
$\mathfrak{l}$-algebras. $\mathfrak{l}$-algebras.
By HNS2 ( By HNS2 (\ref{hns2}),
\ref{hns2}), $\mathfrak{i} / $\mathfrak{i} / \mathfrak{l}$ is a finite field extension.
\mathfrak{l}$ is a finite field
extension.
Let $x_i \coloneqq \iota (X_i \mod \mathfrak{m})$. Let $x_i \coloneqq \iota (X_i \mod \mathfrak{m})$.
\[ \[
P(x_1,\ldots,x_m) = \iota(P \mod \mathfrak{m}) P(x_1,\ldots,x_m) = \iota(P \mod \mathfrak{m})
\] \]
Both sides are morphisms $R \to \mathfrak{i}$ of Both sides are morphisms $R \to \mathfrak{i}$ of $\mathfrak{l}$-algebras.
$\mathfrak{l}$-algebras.
For for $P = X_i$ the equality is trivial. For for $P = X_i$ the equality is trivial.
It follows in general, since the $X_i$ generate $R$ as a It follows in general, since the $X_i$ generate $R$ as a
$\mathfrak{l}$-algebra. $\mathfrak{l}$-algebra.
Thus $(x_1,\ldots,x_m)$ is a zero of $I$ (since $P \mod \mathfrak{m} Thus $(x_1,\ldots,x_m)$ is a zero of $I$
= 0$ for (since $P \mod \mathfrak{m} = 0$ for $P \in I \subseteq \mathfrak{m}$).
$P \in I \subseteq \mathfrak{m}$). HNS1 (\ref{hns1}) can easily be derived from HNS1b.
HNS1 (
\ref{hns1}) can easily be derived from HNS1b.
\end{proof} \end{proof}
\subsubsection{Nullstellensatz for uncountable fields} % from lecture 5 Yet another proof of the Nullstellensatz \subsubsection{Nullstellensatz for uncountable fields}
% from lecture 5 Yet another proof of the Nullstellensatz
The following proof of the Nullstellensatz only works for uncountable fields, The following proof of the Nullstellensatz only works for uncountable fields,
but will be accepted in the exam. but will be accepted in the exam.
\begin{lemma} \begin{lemma}
\label{dimrfunc} \label{dimrfunc}
If $K$ is an uncountable field, then $\dim_K K(T)$ is uncountable. If $K$ is an uncountable field, then $\dim_K K(T)$ is uncountable.
@ -140,23 +123,20 @@ but will be accepted in the exam.
K\right\} $ is $K$-linearly independent. K\right\} $ is $K$-linearly independent.
It follows that $\dim_K K(T) \ge \#S > \aleph_0$. It follows that $\dim_K K(T) \ge \#S > \aleph_0$.
Suppose Suppose $(x_{\kappa})_{\kappa \in K}$ is a selection
$(x_{\kappa})_{\kappa \in K}$ is a of coefficients from $K$
selection of coefficients from $K$ such that $I \coloneqq \{\kappa \in K | x_{\kappa} \neq 0\}$
such that $I \coloneqq \{\kappa \in K | x_{\kappa} \neq 0\} is finite and
$ is finite and
\[ \[
g \coloneqq \sum_{\kappa \in K} \frac{x_\kappa}{T-\kappa} = 0 g \coloneqq \sum_{\kappa \in K} \frac{x_\kappa}{T-\kappa} = 0
\] \]
Let $d Let $d \coloneqq \prod_{\kappa \in I} (T - \kappa) $.
\coloneqq \prod_{\kappa \in I} (T - \kappa) $.
Then for $\lambda \in I$ we have Then for $\lambda \in I$ we have
\[ \[
0 = (dg)(\lambda) = x_\lambda \prod_{\kappa 0 = (dg)(\lambda) =
\in I \setminus \{\lambda\} } (\lambda - \kappa) x_\lambda \prod_{\kappa \in I \setminus \{\lambda\}}(\lambda - \kappa).
\] \]
This is a contradiction as This is a contradiction as $x_\lambda \neq 0$.
$x_\lambda \neq 0$.
\end{proof} \end{proof}
\begin{theorem}[Hilbert's Nullstellensatz for uncountable fields] \begin{theorem}[Hilbert's Nullstellensatz for uncountable fields]
@ -165,78 +145,64 @@ but will be accepted in the exam.
type as a $K$-algebra, then this field extension is finite. type as a $K$-algebra, then this field extension is finite.
\end{theorem} \end{theorem}
\begin{proof} \begin{proof}
If $(x_i)_{i=1}^{n}$ generate $L$ as an If $(x_i)_{i=1}^{n}$ generate $L$ as an $K$-algebra,
$K$-algebra, then the countably many then the countably many monomials
monomials $x^{\alpha} = \prod_{i = 1}^{n} $x^{\alpha} = \prod_{i = 1}^{n} x_i^{\alpha_i}$
x_i^{\alpha_i} $ in the $x_i$ with in the $x_i$ with $\alpha \in \N^n$
$\alpha \in \N^n$ generate $L$ as a $K$-vector space. generate $L$ as a $K$-vector space.
Thus $\dim_K L \le \aleph_0$ and the same holds for any intermediate field $K Thus $\dim_K L \le \aleph_0$
\subseteq M \subseteq L$ . and the same holds for any intermediate field $K \subseteq M \subseteq L$.
If $l \in L$ is transcendent over $K$ and $M = K(l)$, then $M \cong K(T)$ has If $l \in L$ is transcendent over $K$ and $M = K(l)$,
uncountable dimension by then $M \cong K(T)$ has uncountable dimension by \ref{dimrfunc}.
\ref{dimrfunc}.
Thus $L / K$ is algebraic, hence integral, hence finite Thus $L / K$ is algebraic, hence integral, hence finite
( (\ref{ftaiimplf}).
\ref{ftaiimplf}).
\end{proof} \end{proof}
\subsection{The Zariski topology} \subsection{The Zariski topology}
\subsubsection{Operations on ideals and \texorpdfstring{$\Va\left( I \right) $}{V(I)}} \subsubsection{Operations on ideals and \texorpdfstring{$\Va\left( I \right) $}{V(I)}}
Let $R$ be a ring and $I,J, I_\lambda \subseteq R$ ideals, $\lambda \in Let $R$ be a ring and $I,J, I_\lambda \subseteq R$ ideals, $\lambda \in \Lambda$.
\Lambda$.
\begin{definition}[Radical, product and sum of ideals] \begin{definition}[Radical, product and sum of ideals]
\[ \[
\sqrt{I} \coloneqq \bigcap_{n=0} ^{\infty} \{ f \in R | f^n \in \sqrt{I} \coloneqq \bigcap_{n=0}^{\infty} \{f \in R | f^n \in I\}
I\}
\] \]
\[ \[
I \cdot J \coloneqq \langle\{ i \cdot j | i \in I , j \in J\}\rangle_R I \cdot J \coloneqq \langle\{ i \cdot j | i \in I , j \in J\}\rangle_R
\] \]
\[ \[
\sum_{\lambda \in \Lambda} \sum_{\lambda \in \Lambda} I_\lambda
I_\lambda \coloneqq \left\{\sum_{\lambda \in \Lambda'} i_\lambda | \Lambda' \coloneqq \left\{\sum_{\lambda \in \Lambda'} i_\lambda |
\subseteq \Lambda \text{ finite}\right\} \Lambda' \subseteq \Lambda \text{ finite}\right\}
\] \]
\end{definition} \end{definition}
\begin{fact} \begin{fact}
The The radical is an ideal in $R$ and $\sqrt{\sqrt{I} } = \sqrt{I}$.\\
radical is an ideal in $R$ and $\sqrt{\sqrt{I} } =
\sqrt{I}$.
\\
$I \cdot J$ is an ideal.\\ $I \cdot J$ is an ideal.\\
$\sum_{\lambda \in \Lambda} $\sum_{\lambda \in \Lambda} I_\lambda$ coincides with
I_\lambda$ coincides with the ideal generated by $\bigcap_{\lambda \in the ideal generated by $\bigcap_{\lambda \in \Lambda} I_\lambda$ in $R$.\\
\Lambda} $\bigcap_{\lambda \in \Lambda} I_\lambda$ is an ideal.
I_\lambda$ in $R$.
\\
$\bigcap_{\lambda \in \Lambda}
I_\lambda$ is an ideal.
\end{fact} \end{fact}
Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ where $\mathfrak{k}$ is an Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ where $\mathfrak{k}$ is an
algebraically algebraically closed field.
closed field.
\begin{fact} \begin{fact}
\label{fvop} \label{fvop}
Let $I, J, Let $I, J, (I_{\lambda})_{\lambda \in \Lambda}$ be
(I_{\lambda})_{\lambda \in \Lambda}$ be
ideals in $R$. ideals in $R$.
$\Lambda$ may be infinite. $\Lambda$ may be infinite.
\begin{enumerate}[A] \begin{enumerate}[A]
\item \item
$\Va(I) = \Va(\sqrt{I})$ $\Va(I) = \Va(\sqrt{I})$,
\item \item
$\sqrt{J} \subseteq \sqrt{I} \implies $\sqrt{J} \subseteq \sqrt{I} \implies \Va(I) \subseteq \Va(J)$,
\Va(I) \subseteq \Va(J)$
\item \item
$\Va(R) = \emptyset, \Va(\{0\} =\mathfrak{k}^n$ $\Va(R) = \emptyset, \Va(\{0\} = \mathfrak{k}^n$,
\item \item
$\Va(I \cap J) = \Va(I\cdot J) = \Va(I) \cup \Va(J)$ $\Va(I \cap J) = \Va(I\cdot J) = \Va(I) \cup \Va(J)$,
\item \item
$\Va(\sum_{\lambda \in \Lambda} $\Va(\sum_{\lambda \in \Lambda} I_\lambda) =
I_\lambda) = \bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$ \bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$.
\end{enumerate} \end{enumerate}
\end{fact} \end{fact}
\begin{proof} \begin{proof}
@ -244,46 +210,44 @@ closed field.
\item[A-C] \item[A-C]
trivial trivial
\item[D] \item[D]
$I \cdot $I \cdot J \subseteq I \cap J \subseteq I$.
J \subseteq I \cap J \subseteq I$. Thus $\Va(I) \subseteq \Va(I \cap J) \subseteq \Va(I \cdot J)$.
Thus $\Va(I) \subseteq \Va(I \cap J) \subseteq By symmetry we have
\Va(I \cdot J)$. $\Va(I) \cup \Va(J) \subseteq \Va(I \cap J)
By symmetry we have $\Va(I) \cup \Va(J) \subseteq \Va(I \cdot J)$.
\subseteq \Va(I \cap J) \subseteq \Va(I
\cdot J)$.
Let $x \not\in \Va(I) \cup \Va(J)$. Let $x \not\in \Va(I) \cup \Va(J)$.
Then there are $f \in I, g \in J$ such that $f(x) \neq 0, g(x) \neq 0$ thus Then there are $f \in I, g \in J$
$(f \cdot g)(x) \neq 0 \implies x \not\in \Va(I\cdot J)$. such that $f(x) \neq 0, g(x) \neq 0$
thus $(f \cdot g)(x) \neq 0 \implies x \not\in \Va(I\cdot J)$.
Therefore Therefore
\[ \[
\Va(I) \cup \Va(J) \subseteq \Va(I) \cup \Va(J) \subseteq \Va(I \cap J)
\Va(I \cap J) \subseteq \Va(I \cdot \subseteq \Va(I \cdot J)
J) \subseteq \subseteq \Va(I) \cup \Va(J).
\Va(I) \cup \Va(J)
\] \]
\item[E] \item[E]
$I_\lambda \subseteq \sum_{\lambda $I_\lambda \subseteq \sum_{\lambda \in \Lambda} I_\lambda
\in \Lambda} I_\lambda \implies \implies \Va(\sum_{\lambda \in \Lambda} I_\lambda)
\Va(\sum_{\lambda \in \Lambda} I_\lambda) \subseteq \Va(I_\lambda)$.
\subseteq \Va(I_\lambda)$. Thus
Thus $\Va(\sum_{\lambda \in \Lambda} I_\lambda) \subseteq \bigcap_{\lambda \in $\Va(\sum_{\lambda \in \Lambda} I_\lambda)
\Lambda} \subseteq \bigcap_{\lambda \in \Lambda} \Va(I_\lambda)$.
\Va(I_\lambda)$. On the other hand if $f \in \sum_{\lambda \in \Lambda} I_\lambda$
On the other hand if $f \in \sum_{\lambda \in \Lambda} I_\lambda$ we have $f = we have $f = \sum_{\lambda \in \Lambda} f_\lambda$.
\sum_{\lambda \in \Lambda} f_\lambda$. Thus $f$ vanishes on
Thus $f$ vanishes on $\bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$ and we $\bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$
have $\bigcap_{\lambda \in \Lambda} \Va(I_\lambda) \subseteq and we have
\Va(\sum_{\lambda $\bigcap_{\lambda \in \Lambda} \Va(I_\lambda)
\in \Lambda} I_\lambda)$. \subseteq \Va(\sum_{\lambda \in \Lambda} I_\lambda)$.
\end{enumerate} \end{enumerate}
\end{proof} \end{proof}
\begin{remark} \begin{remark}
There is no similar way to describe $\Va(\bigcap_{\lambda \in \Lambda} There is no similar way to describe
I_\lambda)$ in terms of the $\Va(\bigcap_{\lambda \in \Lambda} I_\lambda)$
$\Va(I_{\lambda})$ when $\Lambda$ is infinite. in terms of the $\Va(I_{\lambda})$ when $\Lambda$ is infinite.
For instance if $n = 1, I_k \coloneqq X_1^k R$ then For instance if $n = 1, I_k \coloneqq X_1^k R$ then
$\bigcap_{k=0}^\infty I_k = $\bigcap_{k=0}^\infty I_k = \{0\}$
\{0\} $ but $\bigcup_{k=0}^{\infty} \Va(I_k) = \{0\}$. but $\bigcup_{k=0}^{\infty} \Va(I_k) = \{0\}$.
\end{remark} \end{remark}
\subsubsection{Definition of the Zariski topology} \subsubsection{Definition of the Zariski topology}
Let $\mathfrak{k}$ be algebraically closed, $R = Let $\mathfrak{k}$ be algebraically closed, $R =
@ -444,11 +408,11 @@ For $f \in R$ let $V(f) = V(fR)$.
\begin{corollary} \begin{corollary}
\label{antimonbij} \label{antimonbij}
\begin{align} \begin{align*}
f: \{I \subseteq R | I \text{ ideal}, I = \sqrt{I} \} & \longrightarrow \{A \subseteq \mathfrak{k}^n | A \text{ Zariski-closed}\} \\ f: \{I \subseteq R | I \text{ ideal}, I = \sqrt{I} \} & \longrightarrow \{A \subseteq \mathfrak{k}^n | A \text{ Zariski-closed}\} \\
I & \longmapsto V(I) \\ I & \longmapsto V(I) \\
\{f \in R | A \subseteq V(f)\} & \longmapsfrom A \{f \in R | A \subseteq V(f)\} & \longmapsfrom A
\end{align} \end{align*}
is a $\subseteq$-antimonotonic bijection. is a $\subseteq$-antimonotonic bijection.
\end{corollary} \end{corollary}
\begin{corollary} \begin{corollary}
@ -618,12 +582,12 @@ Let $X$ be a topological space.
\label{bijiredprim} \label{bijiredprim}
By By
\ref{antimonbij} there exists a bijection \ref{antimonbij} there exists a bijection
\begin{align} \begin{align*}
f: \{I \subseteq R | f: \{I \subseteq R |
I \text{ ideal}, I = \sqrt{I} \} & \longrightarrow \{A \subseteq \mathfrak{k}^n I \text{ ideal}, I = \sqrt{I} \} & \longrightarrow \{A \subseteq \mathfrak{k}^n
| A \text{ Zariski-closed}\} \\ I & \longmapsto V(I)\\ \{f \in R | A | A \text{ Zariski-closed}\} \\ I & \longmapsto V(I)\\ \{f \in R | A
\subseteq V(f)\} & \longmapsfrom A \subseteq V(f)\} & \longmapsfrom A
\end{align} \end{align*}
Under this correspondence $A \subseteq \mathfrak{k}^n$ is Under this correspondence $A \subseteq \mathfrak{k}^n$ is
irreducible iff $I irreducible iff $I
@ -708,12 +672,17 @@ Let $X$ be a topological space.
Let $Y \subseteq X$ be irreducible and $U \subseteq X$ an open subset such that Let $Y \subseteq X$ be irreducible and $U \subseteq X$ an open subset such that
$U \cap Y \neq \emptyset$. $U \cap Y \neq \emptyset$.
Then we have a bijection Then we have a bijection
\begin{align} \begin{IEEEeqnarray*}{rl}
f: \{A \subseteq X | A \text{ f: &\{A \subseteq X |
irreducible, closed and } Y \subseteq A\} & \longrightarrow \{B \subseteq U | A \text{ irreducible, closed and } Y \subseteq A\}\\
B \text{ irreducible, closed and } Y \cap U \subseteq B\} \\ A & \longmapsto A & \longrightarrow \{B \subseteq U |
\cap U \\ \overline{B} & \longmapsfrom B B \text{ irreducible, closed and } Y \cap U \subseteq B\}\\
\end{align} \end{IEEEeqnarray*}
given by
\begin{align*}
A & \longmapsto A \cap U\\
\overline{B} & \longmapsfrom B
\end{align*}
where $\overline{B}$ denotes where $\overline{B}$ denotes
the closure in $X$. the closure in $X$.
\end{fact} \end{fact}
@ -931,9 +900,8 @@ inequalities may be strict.
\[ \[
Q(X,Y) \coloneqq \sum_{i=0}^{\infty} X^i Q_i(Y) = \sum_{i,j=0}^{\infty} Q(X,Y) \coloneqq \sum_{i=0}^{\infty} X^i Q_i(Y) = \sum_{i,j=0}^{\infty}
q_{i,j}X^i Y^j = \sum_{j=0}^{\infty} Y^j q_{i,j}X^i Y^j = \sum_{j=0}^{\infty} Y^j
\hat{Q_j}(X) \in K[X,Y] \hat{Q_j}(X) \in K[X,Y].
\] \]
.
Then $Q(x,y) = 0$. Then $Q(x,y) = 0$.
Let $\hat{p_j} \coloneqq \hat{Q_j}(x)$. Let $\hat{p_j} \coloneqq \hat{Q_j}(x)$.
Then $\hat{P}(y) = 0$. Then $\hat{P}(y) = 0$.
@ -974,8 +942,7 @@ transcendence degree.
If the ring extension $B / A$ is finite (i.e. $B$ finitely generated as an If the ring extension $B / A$ is finite (i.e. $B$ finitely generated as an
$A$-module) then $A$ is Noetherian. $A$-module) then $A$ is Noetherian.
\end{theorem} \end{theorem}
\begin{fact} \begin{fact}+
+
\label{noethersubalg} \label{noethersubalg}
Let $R$ be Noetherian and let $B$ be a finite $R$-algebra. Let $R$ be Noetherian and let $B$ be a finite $R$-algebra.
Then every $R$-subalgebra $A \subseteq B$ is finite over $R$. Then every $R$-subalgebra $A \subseteq B$ is finite over $R$.
@ -1308,11 +1275,11 @@ $S$-saturated ideal in $R$ when $J$ is an ideal in $R_S$.
\begin{proposition} \begin{proposition}
\label{idealslocbij} \label{idealslocbij}
\begin{align} \begin{align*}
f: \{I \subseteq R | I \text{ $S$-saturated ideal}\} & \longrightarrow \left\{J \subseteq R_S | J \text{ ideal}\right\} \\ f: \{I \subseteq R | I \text{ $S$-saturated ideal}\} & \longrightarrow \left\{J \subseteq R_S | J \text{ ideal}\right\} \\
I & \longmapsto I_S \coloneqq \left\{\frac{i}{s} | i \in I, s \in S\right\} \\ I & \longmapsto I_S \coloneqq \left\{\frac{i}{s} | i \in I, s \in S\right\} \\
J \sqcap R & \longmapsfrom J \\ J \sqcap R & \longmapsfrom J \\
\end{align} \end{align*}
is a bijection. is a bijection.
Under this bijection $I$ is a prime ideal iff $f(I)$ is. Under this bijection $I$ is a prime ideal iff $f(I)$ is.
\end{proposition} \end{proposition}
@ -1637,11 +1604,11 @@ Localization at a prime ideal is a technique to reduce a problem to this case.
S\} $. S\} $.
We have a bijection We have a bijection
\begin{align} \begin{align*}
f: \Spec A_S & \longrightarrow \{\fq \in f: \Spec A_S & \longrightarrow \{\fq \in
\Spec A | \fq \subseteq \fp\} \\ \fr & \longmapsto \fr \sqcap A\\ \fq_S \Spec A | \fq \subseteq \fp\} \\ \fr & \longmapsto \fr \sqcap A\\ \fq_S
\coloneqq \left\{\frac{q}{s} | q \in \fq, s \in S\right\} & \longmapsfrom \fq \coloneqq \left\{\frac{q}{s} | q \in \fq, s \in S\right\} & \longmapsfrom \fq
\end{align} \end{align*}
\end{proposition} \end{proposition}
\begin{proof} \begin{proof}
It is clear that $S$ is a It is clear that $S$ is a
@ -1911,10 +1878,10 @@ This is part of the proof of
\mathfrak{k}$. \mathfrak{k}$.
\begin{align} \begin{align*}
\mathfrak{k}[X_1,\ldots,X_d] & \longrightarrow R \\ \mathfrak{k}[X_1,\ldots,X_d] & \longrightarrow R \\
P & \longmapsto P(f_1,\ldots,f_d) P & \longmapsto P(f_1,\ldots,f_d)
\end{align} \end{align*}
is an isomorphism and in $\mathfrak{k}[X_1,\ldots,X_d]$ there is a strictly is an isomorphism and in $\mathfrak{k}[X_1,\ldots,X_d]$ there is a strictly
ascending chain of prime ideals corresponding to $\mathfrak{k}^d ascending chain of prime ideals corresponding to $\mathfrak{k}^d
\supsetneq \{0\} \times \mathfrak{k}^{d-1} \supsetneq \ldots \supsetneq \supsetneq \{0\} \times \mathfrak{k}^{d-1} \supsetneq \ldots \supsetneq
@ -2445,12 +2412,12 @@ following:
\subseteq \fp$ of $A$ contained in $\fp$ and the prime ideals and the prime \subseteq \fp$ of $A$ contained in $\fp$ and the prime ideals and the prime
ideals $\tilde \fr \in \Spec B$ with $\fq \subseteq \tilde \fr \subseteq \tilde ideals $\tilde \fr \in \Spec B$ with $\fq \subseteq \tilde \fr \subseteq \tilde
\fp$: \fp$:
\begin{align} \begin{align*}
f: \{\fr \in \Spec A | \fr \subseteq \fp \} f: \{\fr \in \Spec A | \fr \subseteq \fp \}
& \longrightarrow \{\tilde \fr \in \Spec B | \fq \subseteq \tilde \fr \subseteq & \longrightarrow \{\tilde \fr \in \Spec B | \fq \subseteq \tilde \fr \subseteq
\tilde \fp\} \\ \fr & \longmapsto \pi_{B, \fq}^{-1}(\fr)\\ \tilde \fr / \fq \tilde \fp\} \\ \fr & \longmapsto \pi_{B, \fq}^{-1}(\fr)\\ \tilde \fr / \fq
& \longmapsfrom \tilde \fr & \longmapsfrom \tilde \fr
\end{align} \end{align*}
By By
\ref{bijiredprim}, the $\tilde \fr$ \ref{bijiredprim}, the $\tilde \fr$
are in canonical bijection with the irreducible subsets $Z$ of $Y$ containing are in canonical bijection with the irreducible subsets $Z$ of $Y$ containing
@ -2467,11 +2434,11 @@ following:
Let $A$ be an arbitrary ring. Let $A$ be an arbitrary ring.
One can show that there is a bijection between $\Spec A$ and the set of One can show that there is a bijection between $\Spec A$ and the set of
irreducible subsets $Y \subseteq \Spec A$: irreducible subsets $Y \subseteq \Spec A$:
\begin{align} \begin{align*}
f: \Spec A f: \Spec A
& \longrightarrow \{Y \subseteq \Spec A | Y\text{irreducible}\} \\ \fp & \longrightarrow \{Y \subseteq \Spec A | Y\text{irreducible}\} \\ \fp
& \longmapsto \Vs(\fp) \\ \bigcup_{\fp \in Y} \fp & \longmapsfrom Y & \longmapsto \Vs(\fp) \\ \bigcup_{\fp \in Y} \fp & \longmapsfrom Y
\end{align} \end{align*}
Thus, the chains $\fp = \fp_0 \supsetneq \ldots \supsetneq \fp_k$ are in Thus, the chains $\fp = \fp_0 \supsetneq \ldots \supsetneq \fp_k$ are in
canonical bijection with the chains $V(\fp) = X_0 \subsetneq X_1 \subsetneq canonical bijection with the chains $V(\fp) = X_0 \subsetneq X_1 \subsetneq
\ldots \subsetneq X_k \subseteq \Spec A$ of irreducible subsets, and \ldots \subsetneq X_k \subseteq \Spec A$ of irreducible subsets, and
@ -2800,11 +2767,11 @@ We will use the following
\begin{proposition} \begin{proposition}
\label{bijspecideal} \label{bijspecideal}
There is a bijection There is a bijection
\begin{align} \begin{align*}
f: \{A \subseteq \Spec R | A\text{ closed}\} f: \{A \subseteq \Spec R | A\text{ closed}\}
& \longrightarrow \{I \subseteq R | I \text{ ideal and } I = \sqrt{I} \} \\ A & \longrightarrow \{I \subseteq R | I \text{ ideal and } I = \sqrt{I} \} \\ A
& \longmapsto \bigcap_{\fp \in A} \fp \\ \Vspec(I) & \longmapsfrom I & \longmapsto \bigcap_{\fp \in A} \fp \\ \Vspec(I) & \longmapsfrom I
\end{align} \end{align*}
Under this bijection, the irreducible subsets correspond to the prime ideals Under this bijection, the irreducible subsets correspond to the prime ideals
and the closed points $\{\mathfrak{m}\}, \mathfrak{m} and the closed points $\{\mathfrak{m}\}, \mathfrak{m}
\in \Spec A$ to the \in \Spec A$ to the
@ -3012,7 +2979,7 @@ this more general theorem.
defines a homeomorphism between $(A \times B) \cap \Delta$ and $A \cap B$. defines a homeomorphism between $(A \times B) \cap \Delta$ and $A \cap B$.
Thus, $C$ is homeomorphic to an irreducible component $C'$ of $(A \times B) Thus, $C$ is homeomorphic to an irreducible component $C'$ of $(A \times B)
\cap \Delta$ and \cap \Delta$ and
\begin{align} \begin{align*}
\codim(C, \mathfrak{k}^n) = n - \dim(C) = n - \codim(C, \mathfrak{k}^n) = n - \dim(C) = n -
\dim(C') = n - \dim(A \times B) + \codim(C', A \times B) \\ \dim(C') = n - \dim(A \times B) + \codim(C', A \times B) \\
\overset{\text{ \overset{\text{
@ -3020,7 +2987,7 @@ this more general theorem.
\overset{\text{ \overset{\text{
\ref{dimprod}}}{=} 2n - \dim(A) - \dim(B) = \ref{dimprod}}}{=} 2n - \dim(A) - \dim(B) =
\codim(A,\mathfrak{k}^n) + \codim(B, \mathfrak{k}^n) \codim(A,\mathfrak{k}^n) + \codim(B, \mathfrak{k}^n)
\end{align} \end{align*}
by the general by the general
properties of dimension and codimension, properties of dimension and codimension,
\ref{corpithm} applied to \ref{corpithm} applied to

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@ -1,11 +1,7 @@
\begin{warning} \begin{warning}
This is not an official script! This is not an official script.
This document was written in preparation for the oral exam. There is no guarantee for completeness or correctness.
It mostly follows the way \textsc{Prof.
Franke} presented the material in his
lecture rather closely.
There are probably errors.
\end{warning} \end{warning}
\noindent The \LaTeX template by \textsc{Maximilian Kessler} is published under the \noindent The \LaTeX template by \textsc{Maximilian Kessler} is published under the

View file

@ -384,12 +384,12 @@ Let $\mathfrak{l}$ be any field.
\begin{proposition} \begin{proposition}
\label{bijproj} \label{bijproj}
There is a bijection There is a bijection
\begin{align} \begin{align*}
f: \{I \subseteq A_+ | I \text{ homogeneous f: \{I \subseteq A_+ | I \text{ homogeneous
ideal}, I = \sqrt{I}\} & \longrightarrow \{X \subseteq \mathbb{P}^n | X \text{ 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 closed}\} \\ I & \longmapsto \Vp(I)\\ \langle \{f \in A_d | d > 0, X
\subseteq \Vp(f)\} \rangle & \longmapsfrom X \subseteq \Vp(f)\} \rangle & \longmapsfrom X
\end{align} \end{align*}
Under this bijection, Under this bijection,
the irreducible subsets correspond to the elements of the irreducible subsets correspond to the elements of
$\Proj(A_\bullet)$. $\Proj(A_\bullet)$.
@ -528,11 +528,11 @@ Let $\mathfrak{l}$ be any field.
\codim(Y \codim(Y
\cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$. \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$.
Thus Thus
\begin{align} \begin{align*}
\codim(X,Y) + \codim(Y,Z) & = \codim(X \cap \mathbb{A}^n, Y \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) \\ & = \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) \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n) \\ & = \codim(X, Z)
\end{align} \end{align*}
because $\mathfrak{k}^n$ is catenary and the first point follows. because $\mathfrak{k}^n$ is catenary and the first point follows.
The remaining assertions can easily be derived from the first two. The remaining assertions can easily be derived from the first two.
\end{proof} \end{proof}
@ -566,8 +566,9 @@ Let $\mathfrak{l}$ be any field.
\begin{proof} \begin{proof}
The first assertion follows from The first assertion follows from
\ref{bijproj} and \ref{bijproj} and
\ref{bijiredprim} (bijection \ref{bijiredprim}
of irreducible subsets and prime ideals in the projective and affine case). (bijection of irreducible subsets and prime ideals in the projective
and affine case).
Let $d = \dim(X)$ and Let $d = \dim(X)$ and
\[ \[
@ -575,101 +576,92 @@ Let $\mathfrak{l}$ be any field.
X_{d+1} \subsetneq \ldots \subsetneq X_n = X_{d+1} \subsetneq \ldots \subsetneq X_n =
\mathbb{P}^n \mathbb{P}^n
\] \]
be a chain of be a chain of irreducible subsets of $\mathbb{P}^n$.
irreducible subsets of $\mathbb{P}^n$.
Then Then
\[ \[
\{0\} \subsetneq C(X_0) \subsetneq \ldots \subsetneq C(X_d) = C(X) \{0\} \subsetneq C(X_0) \subsetneq \ldots \subsetneq C(X_d) = C(X)
\subsetneq \ldots \subsetneq C(X_n) = \mathfrak{k}^{n+1} \subsetneq \ldots \subsetneq C(X_n) = \mathfrak{k}^{n+1}
\] \]
is a chain of is a chain of irreducible subsets of $\mathfrak{k}^{n+1}$.
irreducible subsets of $\mathfrak{k}^{n+1}$. Hence $\dim(C(X)) \ge 1 + d$ and
Hence $\dim(C(X)) \ge 1 + d$ and $\codim(C(X), \mathfrak{k}^{n+1}) \ge $\codim(C(X), \mathfrak{k}^{n+1}) \ge n-d$.
n-d$. Since
Since $\dim(C(X)) + \codim(C(X), \mathfrak{k}^{n+1}) = $\dim(C(X)) + \codim(C(X), \mathfrak{k}^{n+1}) = \dim(\mathfrak{k}^{n+1})
\dim(\mathfrak{k}^{n+1}) = n+1$,
= n+1$, the two inequalities must be equalities. the two inequalities must be equalities.
\end{proof} \end{proof}
\subsubsection{Application to hypersurfaces in $\mathbb{P}^n$} \subsubsection{Application to hypersurfaces in $\mathbb{P}^n$}
\begin{definition}[Hypersurface] \begin{definition}[Hypersurface]
Let $n > 0$. Let $n > 0$.
By a \vocab{hypersurface} in $\mathbb{P}^n$ or By a \vocab{hypersurface} in $\mathbb{P}^n$ or
$\mathbb{A}^n$ we understand an $\mathbb{A}^n$ we understand an irreducible closed subset
irreducible closed subset of codimension $1$. of codimension $1$.
\end{definition} \end{definition}
\begin{corollary} \begin{corollary}
If $P \in A_d$ is a prime element, then $H = \Vp(P)$ is a If $P \in A_d$ is a prime element,
hypersurface in then $H = \Vp(P)$ is a hypersurface in $\mathbb{P}^n$
$\mathbb{P}^n$ and every hypersurface $H$ in and every hypersurface $H$ in $\mathbb{P}^n$ can be obtained in this way.
$\mathbb{P}^n$ can be obtained in
this way.
\end{corollary} \end{corollary}
\begin{proof} \begin{proof}
If $H = \Vp(P)$ then $C(H) = \Va(P)$ is a If $H = \Vp(P)$ then $C(H) = \Va(P)$
hypersurface in $\mathfrak{k}^{n+1}$ is a hypersurface in $\mathfrak{k}^{n+1}$
by by \ref{irredcodimone}.
\ref{irredcodimone}. By \ref{conedim}, $H$ is irreducible and of codimension $1$.
By
\ref{conedim}, $H$ is irreducible and of codimension $1$.
Conversely, let $H$ be a hypersurface in $\mathbb{P}^n$. Conversely, let $H$ be a hypersurface in $\mathbb{P}^n$.
By By \ref{conedim}, $C(H)$ is a hypersurface in
\ref{conedim}, $C(H)$ is a hypersurface in $\mathfrak{k}^{n+1}$,
$\mathfrak{k}^{n+1}$, hence $C(H) hence $C(H) = \Vp(P)$ for some prime element $P \in A$
= \Vp(P)$ for some prime element $P \in A$ (again by (again by \ref{irredcodimone}).
\ref{irredcodimone}). We have $H = \Vp(\fp)$ for some $\fp \in \Proj(A)$ and $C(H) = \Va(\fp)$.
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 By the bijection between closed subsets of $\mathfrak{k}^{n+1}$ and ideals
$I = $I = \sqrt{I} \subseteq A$ (\ref{antimonbij}),
\sqrt{I} \subseteq A$ ( $\fp = P \cdot A$.
\ref{antimonbij}), $\fp = P \cdot
A$.
Let $P = \sum_{k=0}^{d}P_k$ with $P_d \neq 0$ be the decomposition Let $P = \sum_{k=0}^{d}P_k$ with $P_d \neq 0$ be the decomposition
into into homogeneous components.
homogeneous components. If $P_e $ with $e < d$ was $\neq 0$,
If $P_e $ with $e < d$ was $\neq 0$, it could not be a multiple of $P$ it could not be a multiple of $P$ contradicting the homogeneity of
contradicting the homogeneity of $\fp = P \cdot A$. $\fp = P \cdot A$.
Thus, $P$ is homogeneous of degree $d$. Thus, $P$ is homogeneous of degree $d$.
\end{proof} \end{proof}
\begin{definition} \begin{definition}
A hypersurface $H \subseteq \mathbb{P}^n$ has A hypersurface $H \subseteq \mathbb{P}^n$ has \vocab{degree $d$}
\vocab{degree $d$} if $H = if $H = \Vp(P)$,
\Vp(P)$ where $P \in A_d$ is an irreducible polynomial. where $P \in A_d$ is an irreducible polynomial.
\end{definition} \end{definition}
\subsubsection{Application to intersections in $\mathbb{P}^n$ and Bezout's \subsubsection{Application to intersections in $\mathbb{P}^n$
theorem} and Bezout's theorem}
\begin{corollary} \begin{corollary}
Let $A \subseteq \mathbb{P}^n$ and $B \subseteq Let $A \subseteq \mathbb{P}^n$ and $B \subseteq \mathbb{P}^n$ be
\mathbb{P}^n$ be irreducible irreducible subsets of dimensions $a$ and $b$.
subsets of dimensions $a$ and $b$. If $a+ b \ge n$,
If $a+ b \ge n$, then $A \cap B \neq \emptyset$ and every irreducible component then $A \cap B \neq \emptyset$
of $A \cap B$ as dimension $\ge a + b - n$. and every irreducible component of $A \cap B$
has dimension $\ge a + b - n$.
\end{corollary} \end{corollary}
\begin{remark} \begin{remark}
This shows that $\mathbb{P}^n$ indeed fulfilled the goal of This shows that $\mathbb{P}^n$ indeed fulfilled the goal of allowing for
allowing for nicer nicer results of algebraic geometry because ``solutions at infinity''
results of algebraic geometry because ``solutions at infinity'' to systems of to systems of algebraic equations are present in $\mathbb{P}^n$
algebraic equations are present in $\mathbb{P}^n$ (see (see \ref{affineproblem}).
\ref{affineproblem}).
\end{remark} \end{remark}
\begin{proof} \begin{proof}
The lower bound on the dimension of irreducible components of $A \cap B$ is 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 easily derived from the similar affine result
theorem, (corollary of the principal ideal theorem, \ref{codimintersection}).
\ref{codimintersection}).
From the definition of the affine cone it follows that $C(A \cap B) = C(A) \cap From the definition of the affine cone it follows that
C(B)$. $C(A \cap B) = C(A) \cap C(B)$.
We have $\dim(C(A)) = a+1$ and $\dim(C(B)) = b + 1$ by We have $\dim(C(A)) = a+1$ and $\dim(C(B)) = b + 1$ by
\ref{conedim}. \ref{conedim}.
If $A \cap B = \emptyset$, then $C(A) \cap C(B) = \{0\}$ with $\{0\} $ as an 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 irreducible component, contradicting the lower bound $a + b + 1 - n > 0$ for
the dimension of irreducible components of $C(A) \cap C(B)$ (again the dimension of irreducible components of $C(A) \cap C(B)$
\ref{codimintersection}). (again \ref{codimintersection}).
\end{proof} \end{proof}
\begin{remark}[Bezout's theorem] \begin{remark}[Bezout's theorem]
If $A \neq B$ are hypersurfaces of degree $a$ and $b$ If $A \neq B$ are hypersurfaces of degree $a$ and $b$
@ -682,4 +674,3 @@ Let $\mathfrak{l}$ be any field.
% SLIDE APPLICATION TO HYPERSURFACES IN $\P^n$ % SLIDE APPLICATION TO HYPERSURFACES IN $\P^n$
%ERROR: C(H) = V_A(P) %ERROR: C(H) = V_A(P)
%If n = 0, P = 0, V_P(P) = \emptyset is a problem! %If n = 0, P = 0, V_P(P) = \emptyset is a problem!

View file

@ -44,7 +44,7 @@
% Proofs % Proofs
Def of integrality (<=>) Def of integrality (<=>)
Fact about integrality and field: Fact about integrality and field:

View file

@ -29,12 +29,12 @@
r_{U_j, U_i \cap U_j}(f_j)$ for all $i,j \in I$. r_{U_j, U_i \cap U_j}(f_j)$ for all $i,j \in I$.
Consider the map Consider the map
\begin{align} \begin{align*}
\phi_{U, (U_i)_{i \in I}}: \mathcal{G}(U) \phi_{U, (U_i)_{i \in I}}: \mathcal{G}(U)
& \longrightarrow \{(f_i)_{i \in I} \in \prod_{i \in I} \mathcal{G}(U_i) | & \longrightarrow \{(f_i)_{i \in I} \in \prod_{i \in I} \mathcal{G}(U_i) |
r_{U_i, U_i \cap U_j}(f_i) = r_{U_j, U_i \cap U_j}(f_j) \text{ for } i,j \in I r_{U_i, U_i \cap U_j}(f_i) = r_{U_j, U_i \cap U_j}(f_j) \text{ for } i,j \in I
\} \\ f & \longmapsto (r_{U, U_i}( f))_{i \in I} \} \\ f & \longmapsto (r_{U, U_i}( f))_{i \in I}
\end{align} \end{align*}
A presheaf is called \vocab[Presheaf! A presheaf is called \vocab[Presheaf!
separated]{separated} if $\phi_{U, (U_i)_{i \in I}}$ is separated]{separated} if $\phi_{U, (U_i)_{i \in I}}$ is
@ -163,10 +163,10 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$.
Let $X = V(I)$ where $I = \sqrt{I} \subseteq R$ is an ideal. Let $X = V(I)$ where $I = \sqrt{I} \subseteq R$ is an ideal.
Let $A = R / I$. Let $A = R / I$.
Then Then
\begin{align} \begin{align*}
\phi: A & \longrightarrow \mathcal{O}_X(X) \\ f \mod I \phi: A & \longrightarrow \mathcal{O}_X(X) \\ f \mod I
& \longmapsto f\defon{X} & \longmapsto f\defon{X}
\end{align} \end{align*}
is an isomorphism. is an isomorphism.
\end{proposition} \end{proposition}
@ -176,8 +176,7 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$.
ring homomorphism. ring homomorphism.
Its injectivity follows from the Nullstellensatz and $I = Its injectivity follows from the Nullstellensatz and $I =
\sqrt{I}$ \sqrt{I}$
( (\ref{hns3}).
\ref{hns3}).
Let $\phi \in \mathcal{O}_X(X)$. Let $\phi \in \mathcal{O}_X(X)$.
@ -211,8 +210,7 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$.
As the $U_i = X \setminus V(g_i)$ cover $X$, $V(I) \cap As the $U_i = X \setminus V(g_i)$ cover $X$, $V(I) \cap
\bigcap_{i=1}^m V(g_i) \bigcap_{i=1}^m V(g_i)
= X \cap \bigcap_{i=1}^m V(g_i) = \emptyset$. = X \cap \bigcap_{i=1}^m V(g_i) = \emptyset$.
By the Nullstellensatz ( By the Nullstellensatz (\ref{hns1}) the ideal of $R$ generated by
\ref{hns1}) the ideal of $R$ generated by
$I$ and the $I$ and the
$a_i$ equals $R$. $a_i$ equals $R$.
There are thus $n \ge m \in \N$ and elements There are thus $n \ge m \in \N$ and elements
@ -366,8 +364,7 @@ The following is somewhat harder than in the affine case:
The category of topological spaces The category of topological spaces
\item \item
The category $\Var_\mathfrak{k}$ of varieties over The category $\Var_\mathfrak{k}$ of varieties over
$\mathfrak{k}$ (see $\mathfrak{k}$ (see \ref{defvariety})
\ref{defvariety})
\item \item
If $\mathcal{A}$ is a category, then the \vocab{opposite category} If $\mathcal{A}$ is a category, then the \vocab{opposite category}
or \vocab{dual category} is defined by $\Ob(\mathcal{A}\op) = or \vocab{dual category} is defined by $\Ob(\mathcal{A}\op) =
@ -469,24 +466,21 @@ The following is somewhat harder than in the affine case:
\begin{definition}[Algebraic variety] \begin{definition}[Algebraic variety]
\label{defvariety} \label{defvariety}
An \vocab{algebraic variety} or \vocab{prevariety} over An \vocab{algebraic variety} or \vocab{prevariety} over
$\mathfrak{k}$ is a $\mathfrak{k}$ is a pair $(X, \mathcal{O}_X)$,
pair $(X, \mathcal{O}_X)$, where $X$ is a topological space and where $X$ is a topological space and $\mathcal{O}_X$
$\mathcal{O}_X$
a subsheaf of the sheaf of $\mathfrak{k}$-valued functions on $X$ a subsheaf of the sheaf of $\mathfrak{k}$-valued functions on $X$
such that such that for every $x \in X$,
for every $x \in X$, there are a neighbourhood $U_x$ of $x$ in $X$, an open there are a neighbourhood $U_x$ of $x$ in $X$,
subset $V_x$ of a closed subset $Y_x$ of an open subset $V_x$ of a closed subset $Y_x$ of $\mathfrak{k}^{n_x}$%
$\mathfrak{k}^{n_x}$\footnote{By the \footnote{By the result of \ref{affopensubtopbase},
result of it can be assumed that $V_x = Y_x$ without altering the definition.}
\ref{affopensubtopbase} it can be assumed that $V_x = Y_x$ without and a homeomorphism $V_x
altering the definition.
} and a homeomorphism $V_x
\xrightarrow{\iota_x} \xrightarrow{\iota_x}
U_x$ such that for every open subset $V \subseteq U_x$ and every function U_x$ such that for every open subset $V \subseteq U_x$ and every function
$V\xrightarrow{f} \mathfrak{k}$, we have $f \in $V\xrightarrow{f} \mathfrak{k}$, we have $f \in
\mathcal{O}_X(V) \iff \mathcal{O}_X(V) \iff
\iota^{\ast}_x(f) \in \iota^{\ast}_x(f) \in
\mathcal{O}_{Y_x}(\iota_x^{-1}(V))$, \mathcal{O}_{Y_x}(\iota_x^{-1}(V))$.
In this, the \vocab{pull-back} $\iota_x^{\ast}(f)$ of $f$ is In this, the \vocab{pull-back} $\iota_x^{\ast}(f)$ of $f$ is
defined by defined by
@ -514,29 +508,26 @@ The following is somewhat harder than in the affine case:
If $X$ is a closed subset of $\mathfrak{k}^n$ or If $X$ is a closed subset of $\mathfrak{k}^n$ or
$\mathbb{P}^n$, then $(X, \mathcal{O}_X)$ is a $\mathbb{P}^n$, then $(X, \mathcal{O}_X)$ is a
variety, where $\mathcal{O}_X$ is the structure sheaf on $X$ variety, where $\mathcal{O}_X$ is the structure sheaf on $X$
( (\ref{structuresheafkn}, reps. \ref{structuresheafpn}).
\ref{structuresheafkn}, reps. A variety is called \vocab[Variety!affine]{affine}
\ref{structuresheafpn}). (resp. \vocab[Variety!projective]{projective})
A variety is called \vocab[Variety! if it is isomorphic to a variety of this form,
affine]{affine} (resp. \vocab[Variety!projective]{projective}) if it is isomorphic to a variety of with $X $ closed in $\mathfrak{k}^n$ (resp. $\mathbb{P}^n$).
this form, with $X $ closed in $\mathfrak{k}^n$ (resp. A variety which is isomorphic to and open subvariety of $X$
$\mathbb{P}^n$). is called \vocab[Variety!quasi-affine]{quasi-affine}
A variety which is isomorphic to and open subvariety of $X$ is called (resp. \vocab[Variety!quasi-projective]{quasi-projective}).
\vocab[Variety!
quasi-affine]{quasi-affine} (resp. \vocab[Variety!quasi-projective]{quasi-projective}).
\item \item
If $X = V(X^2 - Y^3) \subseteq \mathfrak{k}^2$ then If $X = V(X^2 - Y^3) \subseteq \mathfrak{k}^2$
$\mathfrak{k} \xrightarrow{t \mapsto (t^3,t^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 X$ is a morphism which is a homeomorphism of topological spaces
isomorphism of varieties. but not an isomorphism of varieties.
% TODO % TODO
\item \item
The composition of two morphisms $X \to Y \to Z$ of varieties is a morphism of The composition of two morphisms $X \to Y \to Z$ of varieties
varieties. is a morphism of varieties.
\item \item
$X\xrightarrow{\Id_X} $X\xrightarrow{\Id_X} X$ is a morphism of varieties.
X$ is a morphism of varieties.
\end{itemize} \end{itemize}
\end{example} \end{example}
@ -563,8 +554,7 @@ The following is somewhat harder than in the affine case:
\item \item
The property is local on $U$, hence it is sufficient to show it in the The property is local on $U$, hence it is sufficient to show it in the
quasi-affine case. quasi-affine case.
This was done in This was done in \ref{structuresheafcontinuous}.
\ref{structuresheafcontinuous}.
\item \item
For the second part, let $\lambda_x \coloneqq \lambda \defon{V_x} For the second part, let $\lambda_x \coloneqq \lambda \defon{V_x}
$. $.
@ -603,21 +593,21 @@ The following is somewhat harder than in the affine case:
\item \item
Let $X,Y$ be varieties over $\mathfrak{k}$. Let $X,Y$ be varieties over $\mathfrak{k}$.
Then the map Then the map
\begin{align} \begin{align*}
\phi: \Hom_{\Var_\mathfrak{k}}(X,Y) & \longrightarrow \phi: \Hom_{\Var_\mathfrak{k}}(X,Y) & \longrightarrow
\Hom_{\Alg_\mathfrak{k}}(\mathcal{O}_Y(Y), \mathcal{O}_X(X)) \\ (X \Hom_{\Alg_\mathfrak{k}}(\mathcal{O}_Y(Y), \mathcal{O}_X(X)) \\ (X
\xrightarrow{f} Y) & \longmapsto (\mathcal{O}_Y(Y) \xrightarrow{f^{\ast}} \xrightarrow{f} Y) & \longmapsto (\mathcal{O}_Y(Y) \xrightarrow{f^{\ast}}
\mathcal{O}_X(X)) \mathcal{O}_X(X))
\end{align} \end{align*}
is injective when $Y$ is quasi-affine and is injective when $Y$ is quasi-affine and
bijective when $Y$ is affine. bijective when $Y$ is affine.
\item \item
The contravariant functor The contravariant functor
\begin{align} \begin{align*}
F: \Var_\mathfrak{k} & \longrightarrow \Alg_\mathfrak{k} \\ X & \longmapsto F: \Var_\mathfrak{k} & \longrightarrow \Alg_\mathfrak{k} \\ X & \longmapsto
\mathcal{O}_X(X) \\ (X\xrightarrow{f} Y) & \longmapsto (\mathcal{O}_X(X) \mathcal{O}_X(X) \\ (X\xrightarrow{f} Y) & \longmapsto (\mathcal{O}_X(X)
\xrightarrow{f^{\ast}} \mathcal{O}_Y(Y)) \xrightarrow{f^{\ast}} \mathcal{O}_Y(Y))
\end{align} \end{align*}
restricts to an restricts to an
equivalence of categories between the category of affine varieties over equivalence of categories between the category of affine varieties over
$\mathfrak{k}$ and the full subcategory $\mathcal{A}$ of $\mathfrak{k}$ and the full subcategory $\mathcal{A}$ of
@ -899,10 +889,10 @@ The following is somewhat harder than in the affine case:
If $U$ is an open neighbourhood of $x \in X$, then we have a map (resp. If $U$ is an open neighbourhood of $x \in X$, then we have a map (resp.
homomorphism) homomorphism)
\begin{align} \begin{align*}
\cdot_x : \mathcal{G}(U) & \longrightarrow \mathcal{G}_x \\ \cdot_x : \mathcal{G}(U) & \longrightarrow \mathcal{G}_x\\
\gamma & \longmapsto \gamma_x \coloneqq (U, \gamma) / \sim \gamma & \longmapsto \gamma_x \coloneqq (U, \gamma) / \sim
\end{align} \end{align*}
\end{definition} \end{definition}
\begin{fact} \begin{fact}