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3
2021_Algebra_I.tex
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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}
\lecturer{Prof.~Dr.~Jens Franke}

4
algebra.sty
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@ -1,7 +1,7 @@
\ProvidesPackage{algebra}[2022/02/10 - Style file for notes of Algebra I]
\RequirePackage{mkessler-math}
\RequirePackage[english]{mkessler-math}
\RequirePackage{mkessler-refproof}
\RequirePackage[number in = section]{fancythm}
@ -11,7 +11,7 @@
\RequirePackage[utf8x]{inputenc}
\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{pdflscape}

70
inputs/finiteness_conditions.tex
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@ -122,11 +122,11 @@
\begin{definition}[Generated (sub)algebra, algebra of finite type]
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] \\
P = \sum_{\beta \in \N^m} p_\beta X^{\beta} & \longmapsto \sum_{\beta \in \N^m} \alpha(p_\beta)
X^{\beta}
\end{align}
\end{align*}
is a ring homomorphism.
We will sometimes write $P(a_1,\ldots,a_m)$ instead of
$(\alpha(P))(a_1,\ldots,a_m)$.
@ -285,10 +285,10 @@ commutative rings with $1$.
and finite over $R$.
Let $(b_i)_{i=1}^{n}$ generate $B$ as an
$R$-module.
\begin{align}
\begin{align*}
q: R^n & \longrightarrow B \\
(r_1,\ldots,r_n) & \longmapsto \sum_{i=1}^{n} r_i b_i
\end{align}
\end{align*}
is surjective.
Thus there are $\rho_{i} = \left( r_{i,j} \right)_{j=1}^n \in R^n$
such that
@ -478,10 +478,10 @@ commutative rings with $1$.
Then there are $a = (a_i)_{i=1}^{n} \in A$ which
are algebraically independent
over $K$, i.e. the ring homomorphism
\begin{align}
\begin{align*}
\ev_a: K[X_1,\ldots,X_n] &
\longrightarrow A \\ P & \longmapsto P(a_1,\ldots,a_n)
\end{align}
\end{align*}
is
injective.
$n$ 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 $a_i$.
If suffices to show that the $a_i$ are algebraically independent.
Since $A$ is of finite type over $K$ and thus over $\tilde{A}$,
by fact
\ref{ftaiimplf} (integral and finite type $\implies$ finite) $A$ is finite over
$\tilde{A}$.
Thus we only need to show that the $a_i$ are algebraically independent over
$K$.
Assume there is $P \in K[X_1,\ldots,X_n] \setminus \{0\} $ such
that
$P(a_1,\ldots,a_n) = 0$.
by fact \ref{ftaiimplf} (integral and finite type $\implies$ finite),
$A$ is finite over $\tilde{A}$.
Thus we only need to show that the $a_i$ are algebraically independent
over $K$.
Assume there is $P \in K[X_1,\ldots,X_n] \setminus \{0\}$
such that $P(a_1,\ldots,a_n) = 0$.
Let $P = \sum_{\alpha \in \N^n} p_\alpha X^{\alpha}$ and
$S = \{ \alpha \in
\N^n | p_\alpha \neq 0\}$.
For $\vec{k} = (k_i)_{i=1}^{n}
\in \N^n$ and $\alpha \in \N^n$ we define
$w_{\vec{k}}(\alpha) \coloneqq \sum_{i=1}^{n}
k_i\alpha_i$.
$S = \{ \alpha \in \N^n | p_\alpha \neq 0\}$.
For $\vec{k} = (k_i)_{i=1}^{n} \in \N^n$ and $\alpha \in \N^n$ we define
$w_{\vec{k}}(\alpha) \coloneqq \sum_{i=1}^{n} k_i\alpha_i$.
By
\ref{nntechlemma} it is possible to choose $\vec{k}
\in \N^n$ such that $k_1
= 1$ and for $\alpha \neq \beta \in S$ we have
$w_{\vec{k}}(\alpha) \neq
w_{\vec{k}}(\beta)$.
By \ref{nntechlemma} it is possible to choose $\vec{k} \in \N^n$
such that $k_1 = 1$ and for $\alpha \neq \beta \in S$ we have
$w_{\vec{k}}(\alpha) \neq w_{\vec{k}}(\beta)$.
Define $b_i \coloneqq a_{i+1} -
a^{k_{i+1}}_1$ for $1 \le i < n$.
Define $b_i \coloneqq a_{i+1} - a^{k_{i+1}}_1$ for $1 \le i < n$.
\begin{claim}
$A$ is integral over the subalgebra $B$ generated by the $b_i$.
\end{claim}
\begin{subproof}
By the transitivity of integrality, it is sufficient to show that the $a_i$ are
integral over $B$.
By the transitivity of integrality,
it is sufficient to show that the $a_i$ are integral over $B$.
For $i > 1$ we have $a_i = b_{i-1} + a_1^{k_i}$.
Thus it suffices to show this for $a_1$.
Define $Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots,
b_{n-1} + T^{k_n}) \in
B[T]$.
Define
$Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots,
b_{n-1} + T^{k_n}) \in B[T]$.
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.
We have
\[
T^{\alpha_1} \prod_{i=1}^{n-1} (b_i + T^{k_i + 1})^{\alpha_{i+1}}
=
T^{\alpha_1} \prod_{i=1}^{n-1} (b_i + T^{k_i + 1})^{\alpha_{i+1}} =
T^{w_{\vec k}(\alpha)} +
\sum_{l = 0}^{w_{\vec k}(\alpha) - 1}
\beta_{\alpha,
l} T^l
\sum_{l = 0}^{w_{\vec k}(\alpha) - 1} \beta_{\alpha, l} T^l
\]
with suitable $\beta_{\alpha, l} \in B$.
@ -552,9 +540,8 @@ commutative rings with $1$.
+ \sum_{j=0}^{w_{\vec k}(\alpha) - 1} q_j T^j
\]
with $q_j \in B$ and
$\alpha$ such that $w_{\vec k }(\alpha)$ is maximal subject
to the condition
$p_\alpha \neq 0$.
$\alpha$ such that $w_{\vec k}(\alpha)$ is maximal subject
to the condition $p_\alpha \neq 0$.
Thus the leading coefficient of $Q$ is a unit.
\end{subproof}
@ -562,4 +549,3 @@ commutative rings with $1$.
elements $b_i$.
\end{proof}

305
inputs/nullstellensatz_and_zariski_topology.tex
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@ -6,21 +6,19 @@ Let $\mathfrak{k}$ be a field, $R \coloneqq
\begin{definition}[zero]
$x \in \mathfrak{k}^n$ is \vocab[Ideal!zero]{a zero of $I$}
if $\forall x \in I: P(x) = 0$.
Let $\Va(I)$ denote the set of zeros if $I$ in
$\mathfrak{k}^n$.
Let $\Va(I)$ denote the set of zeros if $I$ in $\mathfrak{k}^n$.
The \vocab[Ideal!
zero]{zero in a field extension $\mathfrak{i}$ of $\mathfrak{k}$} is defined similarly.
The \vocab[Ideal!zero]{zero in a field extension %
$\mathfrak{i}$ of $\mathfrak{k}$} is defined similarly.
\end{definition}
\begin{remark}[Set of zeros and generators]
Let $I$ be generated by $S$.
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
equations.
Thus zero sets of ideals correspond to solutions sets to systems of
polynomial equations.
If $S, \tilde{S}$ generate the same ideal $I$ they have the same
set of
solutions.
set of solutions.
Therefore we only consider zero sets of ideals.
\end{remark}
@ -32,16 +30,15 @@ Let $\mathfrak{k}$ be a field, $R \coloneqq
\end{theorem}
\begin{remark}
Will be shown later (see proof of
\ref{hns1b}).
Trivial if $n = 1$: $R$ is a PID, thus $I = pR$ for some $p \in R$.
Will be shown later (see proof of \ref{hns1b}).
It is trivial if $n = 1$:
$R$ is a PID, thus $I = pR$ for some $p \in R$.
Since $I \neq R$ $p = 0$ or $P$ is non-constant.
$\mathfrak{k}$ algebraically closed $\leadsto$ there exists a zero of
$p$.\\
$\mathfrak{k}$ algebraically closed
$\leadsto$ there exists a zero of $p$.\\
If $\mathfrak{k}$ is not algebraically closed and $n > 0$, the
theorem fails
(consider $I = p(X_1) R$).
If $\mathfrak{k}$ is not algebraically closed and $n > 0$,
the theorem fails (consider $I = p(X_1) R$).
\end{remark}
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{itemize}
\item[$\implies$]
If $(l_i)_{i=1}^{m}$ is a base of $L$ as a
$K$-vector space, then $L$ is generated by the $l_i$ as a $K$-algebra.
If $(l_i)_{i=1}^{m}$ is a base of $L$ as a $K$-vector space,
then $L$ is generated by the $l_i$ as a $K$-algebra.
\item[$\impliedby$ ]
Apply the Noether normalization theorem (
\ref{noenort}) to $A = L$.
Apply the Noether normalization theorem (\ref{noenort}) to $A = L$.
This yields an injective ring homomorphism $\ev_a:
K[X_1,\ldots,X_n] \to A$
such that $A$ is finite over the image of $\ev_a$.
By the fact about integrality and fields (
\ref{fintaf}), the
isomorphic image
By the fact about integrality and fields (\ref{fintaf}),
the isomorphic image
of $\ev_a$ is a field.
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{proof}
\begin{remark}
We will see several additional proofs of this theorem.
See
\ref{hns2unc} and
\ref{rfuncnft}.
See \ref{hns2unc} and \ref{rfuncnft}.
All will be accepted in the exam.
\ref{hns3} and
\ref{hnsp} are closely related.
\ref{hns3} and \ref{hnsp} are closely related.
\end{remark}
\begin{theorem}[Hilbert's Nullstellensatz (1b)]
\label{hns1b}
Let $\mathfrak{l}$ be a field and $I \subset R =
\mathfrak{l}[X_1,\ldots,X_m]$
a proper ideal.
Then there are a finite field extension $\mathfrak{i}$ of
$\mathfrak{l}$ and a
zero of $I$ in $\mathfrak{i}^m$.
Let $\mathfrak{l}$ be a field and
$I \subset R = \mathfrak{l}[X_1,\ldots,X_m]$ a proper ideal.
Then there are a finite field extension $\mathfrak{i}$ of $\mathfrak{l}$
and a zero of $I$ in $\mathfrak{i}^m$.
\end{theorem}
\begin{proof}
(HNS2 (
\ref{hns2}) $\implies$ HNS1b (
\ref{hns1b}))
$I \subseteq \mathfrak{m}$ for some maximal ideal. $R /
\mathfrak{m}$ is a field, since $\mathfrak{m}$ is maximal.
$R / \mathfrak{m}$ is of finite type, since the images of the $X_i$
generate it as a $\mathfrak{l}$-algebra.
There are thus a field extension $\mathfrak{i} /
\mathfrak{l}$ and an
isomorphism $R / \mathfrak{m} \xrightarrow{\iota}
\mathfrak{i}$ of
(HNS2 (\ref{hns2}) $\implies$ HNS1b (\ref{hns1b}))
$I \subseteq \mathfrak{m}$ for some maximal ideal.
$R /\mathfrak{m}$ is a field,
since $\mathfrak{m}$ is maximal.
$R / \mathfrak{m}$ is of finite type,
since the images of the $X_i$ generate it as a $\mathfrak{l}$-algebra.
There are thus a field extension $\mathfrak{i} / \mathfrak{l}$
and an isomorphism $R / \mathfrak{m} \xrightarrow{\iota} \mathfrak{i}$ of
$\mathfrak{l}$-algebras.
By HNS2 (
\ref{hns2}), $\mathfrak{i} /
\mathfrak{l}$ is a finite field
extension.
By HNS2 (\ref{hns2}),
$\mathfrak{i} / \mathfrak{l}$ is a finite field extension.
Let $x_i \coloneqq \iota (X_i \mod \mathfrak{m})$.
\[
P(x_1,\ldots,x_m) = \iota(P \mod \mathfrak{m})
\]
Both sides are morphisms $R \to \mathfrak{i}$ of
$\mathfrak{l}$-algebras.
Both sides are morphisms $R \to \mathfrak{i}$ of $\mathfrak{l}$-algebras.
For for $P = X_i$ the equality is trivial.
It follows in general, since the $X_i$ generate $R$ as a
$\mathfrak{l}$-algebra.
Thus $(x_1,\ldots,x_m)$ is a zero of $I$ (since $P \mod \mathfrak{m}
= 0$ for
$P \in I \subseteq \mathfrak{m}$).
HNS1 (
\ref{hns1}) can easily be derived from HNS1b.
Thus $(x_1,\ldots,x_m)$ is a zero of $I$
(since $P \mod \mathfrak{m} = 0$ for $P \in I \subseteq \mathfrak{m}$).
HNS1 (\ref{hns1}) can easily be derived from HNS1b.
\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,
but will be accepted in the exam.
\begin{lemma}
\label{dimrfunc}
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.
It follows that $\dim_K K(T) \ge \#S > \aleph_0$.
Suppose
$(x_{\kappa})_{\kappa \in K}$ is a
selection of coefficients from $K$
such that $I \coloneqq \{\kappa \in K | x_{\kappa} \neq 0\}
$ is finite and
Suppose $(x_{\kappa})_{\kappa \in K}$ is a selection
of coefficients from $K$
such that $I \coloneqq \{\kappa \in K | x_{\kappa} \neq 0\}$
is finite and
\[
g \coloneqq \sum_{\kappa \in K} \frac{x_\kappa}{T-\kappa} = 0
\]
Let $d
\coloneqq \prod_{\kappa \in I} (T - \kappa) $.
Let $d \coloneqq \prod_{\kappa \in I} (T - \kappa) $.
Then for $\lambda \in I$ we have
\[
0 = (dg)(\lambda) = x_\lambda \prod_{\kappa
\in I \setminus \{\lambda\} } (\lambda - \kappa)
0 = (dg)(\lambda) =
x_\lambda \prod_{\kappa \in I \setminus \{\lambda\}}(\lambda - \kappa).
\]
This is a contradiction as
$x_\lambda \neq 0$.
This is a contradiction as $x_\lambda \neq 0$.
\end{proof}
\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.
\end{theorem}
\begin{proof}
If $(x_i)_{i=1}^{n}$ generate $L$ as an
$K$-algebra, then the countably many
monomials $x^{\alpha} = \prod_{i = 1}^{n}
x_i^{\alpha_i} $ in the $x_i$ with
$\alpha \in \N^n$ generate $L$ as a $K$-vector space.
Thus $\dim_K L \le \aleph_0$ 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
uncountable dimension by
\ref{dimrfunc}.
If $(x_i)_{i=1}^{n}$ generate $L$ as an $K$-algebra,
then the countably many monomials
$x^{\alpha} = \prod_{i = 1}^{n} x_i^{\alpha_i}$
in the $x_i$ with $\alpha \in \N^n$
generate $L$ as a $K$-vector space.
Thus $\dim_K L \le \aleph_0$
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 uncountable dimension by \ref{dimrfunc}.
Thus $L / K$ is algebraic, hence integral, hence finite
(
\ref{ftaiimplf}).
(\ref{ftaiimplf}).
\end{proof}
\subsection{The Zariski topology}
\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, $\lambda \in
\Lambda$.
Let $R$ be a ring and $I,J, I_\lambda \subseteq R$ ideals, $\lambda \in \Lambda$.
\begin{definition}[Radical, product and sum of ideals]
\[
\sqrt{I} \coloneqq \bigcap_{n=0} ^{\infty} \{ f \in R | f^n \in
I\}
\sqrt{I} \coloneqq \bigcap_{n=0}^{\infty} \{f \in R | f^n \in I\}
\]
\[
I \cdot J \coloneqq \langle\{ i \cdot j | i \in I , j \in J\}\rangle_R
\]
\[
\sum_{\lambda \in \Lambda}
I_\lambda \coloneqq \left\{\sum_{\lambda \in \Lambda'} i_\lambda | \Lambda'
\subseteq \Lambda \text{ finite}\right\}
\sum_{\lambda \in \Lambda} I_\lambda
\coloneqq \left\{\sum_{\lambda \in \Lambda'} i_\lambda |
\Lambda' \subseteq \Lambda \text{ finite}\right\}
\]
\end{definition}
\begin{fact}
The
radical is an ideal in $R$ and $\sqrt{\sqrt{I} } =
\sqrt{I}$.
\\
The radical is an ideal in $R$ and $\sqrt{\sqrt{I} } = \sqrt{I}$.\\
$I \cdot J$ is an ideal.\\
$\sum_{\lambda \in \Lambda}
I_\lambda$ coincides with the ideal generated by $\bigcap_{\lambda \in
\Lambda}
I_\lambda$ in $R$.
\\
$\bigcap_{\lambda \in \Lambda}
I_\lambda$ is an ideal.
$\sum_{\lambda \in \Lambda} I_\lambda$ coincides with
the ideal generated by $\bigcap_{\lambda \in \Lambda} I_\lambda$ in $R$.\\
$\bigcap_{\lambda \in \Lambda} I_\lambda$ is an ideal.
\end{fact}
Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ where $\mathfrak{k}$ is an
algebraically
closed field.
algebraically closed field.
\begin{fact}
\label{fvop}
Let $I, J,
(I_{\lambda})_{\lambda \in \Lambda}$ be
Let $I, J, (I_{\lambda})_{\lambda \in \Lambda}$ be
ideals in $R$.
$\Lambda$ may be infinite.
\begin{enumerate}[A]
\item
$\Va(I) = \Va(\sqrt{I})$
$\Va(I) = \Va(\sqrt{I})$,
\item
$\sqrt{J} \subseteq \sqrt{I} \implies
\Va(I) \subseteq \Va(J)$
$\sqrt{J} \subseteq \sqrt{I} \implies \Va(I) \subseteq \Va(J)$,
\item
$\Va(R) = \emptyset, \Va(\{0\} =\mathfrak{k}^n$
$\Va(R) = \emptyset, \Va(\{0\} = \mathfrak{k}^n$,
\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
$\Va(\sum_{\lambda \in \Lambda}
I_\lambda) = \bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$
$\Va(\sum_{\lambda \in \Lambda} I_\lambda) =
\bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$.
\end{enumerate}
\end{fact}
\begin{proof}
@ -244,46 +210,44 @@ closed field.
\item[A-C]
trivial
\item[D]
$I \cdot
J \subseteq I \cap J \subseteq I$.
Thus $\Va(I) \subseteq \Va(I \cap J) \subseteq
\Va(I \cdot J)$.
By symmetry we have $\Va(I) \cup \Va(J)
\subseteq \Va(I \cap J) \subseteq \Va(I
\cdot J)$.
$I \cdot J \subseteq I \cap J \subseteq I$.
Thus $\Va(I) \subseteq \Va(I \cap J) \subseteq \Va(I \cdot J)$.
By symmetry we have
$\Va(I) \cup \Va(J) \subseteq \Va(I \cap J)
\subseteq \Va(I \cdot J)$.
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
$(f \cdot g)(x) \neq 0 \implies x \not\in \Va(I\cdot J)$.
Then there are $f \in I, g \in J$
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
\[
\Va(I) \cup \Va(J) \subseteq
\Va(I \cap J) \subseteq \Va(I \cdot
J) \subseteq
\Va(I) \cup \Va(J)
\Va(I) \cup \Va(J) \subseteq \Va(I \cap J)
\subseteq \Va(I \cdot J)
\subseteq \Va(I) \cup \Va(J).
\]
\item[E]
$I_\lambda \subseteq \sum_{\lambda
\in \Lambda} I_\lambda \implies
\Va(\sum_{\lambda \in \Lambda} I_\lambda)
$I_\lambda \subseteq \sum_{\lambda \in \Lambda} I_\lambda
\implies \Va(\sum_{\lambda \in \Lambda} I_\lambda)
\subseteq \Va(I_\lambda)$.
Thus $\Va(\sum_{\lambda \in \Lambda} I_\lambda) \subseteq \bigcap_{\lambda \in
\Lambda}
\Va(I_\lambda)$.
On the other hand if $f \in \sum_{\lambda \in \Lambda} I_\lambda$ we have $f =
\sum_{\lambda \in \Lambda} f_\lambda$.
Thus $f$ vanishes on $\bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$ and we
have $\bigcap_{\lambda \in \Lambda} \Va(I_\lambda) \subseteq
\Va(\sum_{\lambda
\in \Lambda} I_\lambda)$.
Thus
$\Va(\sum_{\lambda \in \Lambda} I_\lambda)
\subseteq \bigcap_{\lambda \in \Lambda} \Va(I_\lambda)$.
On the other hand if $f \in \sum_{\lambda \in \Lambda} I_\lambda$
we have $f = \sum_{\lambda \in \Lambda} f_\lambda$.
Thus $f$ vanishes on
$\bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$
and we have
$\bigcap_{\lambda \in \Lambda} \Va(I_\lambda)
\subseteq \Va(\sum_{\lambda \in \Lambda} I_\lambda)$.
\end{enumerate}
\end{proof}
\begin{remark}
There is no similar way to describe $\Va(\bigcap_{\lambda \in \Lambda}
I_\lambda)$ in terms of the
$\Va(I_{\lambda})$ when $\Lambda$ is infinite.
There is no similar way to describe
$\Va(\bigcap_{\lambda \in \Lambda} I_\lambda)$
in terms of the $\Va(I_{\lambda})$ when $\Lambda$ is infinite.
For instance if $n = 1, I_k \coloneqq X_1^k R$ then
$\bigcap_{k=0}^\infty I_k =
\{0\} $ but $\bigcup_{k=0}^{\infty} \Va(I_k) = \{0\}$.
$\bigcap_{k=0}^\infty I_k = \{0\}$
but $\bigcup_{k=0}^{\infty} \Va(I_k) = \{0\}$.
\end{remark}
\subsubsection{Definition of the Zariski topology}
Let $\mathfrak{k}$ be algebraically closed, $R =
@ -444,11 +408,11 @@ For $f \in R$ let $V(f) = V(fR)$.
\begin{corollary}
\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}\} \\
I & \longmapsto V(I) \\
\{f \in R | A \subseteq V(f)\} & \longmapsfrom A
\end{align}
\end{align*}
is a $\subseteq$-antimonotonic bijection.
\end{corollary}
\begin{corollary}
@ -618,12 +582,12 @@ Let $X$ be a topological space.
\label{bijiredprim}
By
\ref{antimonbij} there exists a bijection
\begin{align}
\begin{align*}
f: \{I \subseteq R |
I \text{ ideal}, I = \sqrt{I} \} & \longrightarrow \{A \subseteq \mathfrak{k}^n
| A \text{ Zariski-closed}\} \\ I & \longmapsto V(I)\\ \{f \in R | A
\subseteq V(f)\} & \longmapsfrom A
\end{align}
\end{align*}
Under this correspondence $A \subseteq \mathfrak{k}^n$ is
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
$U \cap Y \neq \emptyset$.
Then we have a bijection
\begin{align}
f: \{A \subseteq X | A \text{
irreducible, closed and } Y \subseteq A\} & \longrightarrow \{B \subseteq U |
B \text{ irreducible, closed and } Y \cap U \subseteq B\} \\ A & \longmapsto A
\cap U \\ \overline{B} & \longmapsfrom B
\end{align}
\begin{IEEEeqnarray*}{rl}
f: &\{A \subseteq X |
A \text{ irreducible, closed and } Y \subseteq A\}\\
& \longrightarrow \{B \subseteq U |
B \text{ irreducible, closed and } Y \cap U \subseteq B\}\\
\end{IEEEeqnarray*}
given by
\begin{align*}
A & \longmapsto A \cap U\\
\overline{B} & \longmapsfrom B
\end{align*}
where $\overline{B}$ denotes
the closure in $X$.
\end{fact}
@ -931,9 +900,8 @@ inequalities may be strict.
\[
Q(X,Y) \coloneqq \sum_{i=0}^{\infty} X^i Q_i(Y) = \sum_{i,j=0}^{\infty}
q_{i,j}X^i Y^j = \sum_{j=0}^{\infty} Y^j
\hat{Q_j}(X) \in K[X,Y]
\hat{Q_j}(X) \in K[X,Y].
\]
.
Then $Q(x,y) = 0$.
Let $\hat{p_j} \coloneqq \hat{Q_j}(x)$.
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
$A$-module) then $A$ is Noetherian.
\end{theorem}
\begin{fact}
+
\begin{fact}+
\label{noethersubalg}
Let $R$ be Noetherian and let $B$ be a finite $R$-algebra.
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}
\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\} \\
I & \longmapsto I_S \coloneqq \left\{\frac{i}{s} | i \in I, s \in S\right\} \\
J \sqcap R & \longmapsfrom J \\
\end{align}
\end{align*}
is a bijection.
Under this bijection $I$ is a prime ideal iff $f(I)$ is.
\end{proposition}
@ -1637,11 +1604,11 @@ Localization at a prime ideal is a technique to reduce a problem to this case.
S\} $.
We have a bijection
\begin{align}
\begin{align*}
f: \Spec A_S & \longrightarrow \{\fq \in
\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
\end{align}
\end{align*}
\end{proposition}
\begin{proof}
It is clear that $S$ is a
@ -1911,10 +1878,10 @@ This is part of the proof of
\mathfrak{k}$.
\begin{align}
\begin{align*}
\mathfrak{k}[X_1,\ldots,X_d] & \longrightarrow R \\
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
ascending chain of prime ideals corresponding to $\mathfrak{k}^d
\supsetneq \{0\} \times \mathfrak{k}^{d-1} \supsetneq \ldots \supsetneq
@ -2445,12 +2412,12 @@ following:
\subseteq \fp$ of $A$ contained in $\fp$ and the prime ideals and the prime
ideals $\tilde \fr \in \Spec B$ with $\fq \subseteq \tilde \fr \subseteq \tilde
\fp$:
\begin{align}
\begin{align*}
f: \{\fr \in \Spec A | \fr \subseteq \fp \}
& \longrightarrow \{\tilde \fr \in \Spec B | \fq \subseteq \tilde \fr \subseteq
\tilde \fp\} \\ \fr & \longmapsto \pi_{B, \fq}^{-1}(\fr)\\ \tilde \fr / \fq
& \longmapsfrom \tilde \fr
\end{align}
\end{align*}
By
\ref{bijiredprim}, the $\tilde \fr$
are in canonical bijection with the irreducible subsets $Z$ of $Y$ containing
@ -2467,11 +2434,11 @@ following:
Let $A$ be an arbitrary ring.
One can show that there is a bijection between $\Spec A$ and the set of
irreducible subsets $Y \subseteq \Spec A$:
\begin{align}
\begin{align*}
f: \Spec A
& \longrightarrow \{Y \subseteq \Spec A | Y\text{irreducible}\} \\ \fp
& \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
canonical bijection with the chains $V(\fp) = X_0 \subsetneq X_1 \subsetneq
\ldots \subsetneq X_k \subseteq \Spec A$ of irreducible subsets, and
@ -2800,11 +2767,11 @@ We will use the following
\begin{proposition}
\label{bijspecideal}
There is a bijection
\begin{align}
\begin{align*}
f: \{A \subseteq \Spec R | A\text{ closed}\}
& \longrightarrow \{I \subseteq R | I \text{ ideal and } I = \sqrt{I} \} \\ A
& \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
and the closed points $\{\mathfrak{m}\}, \mathfrak{m}
\in \Spec A$ to the
@ -3012,7 +2979,7 @@ this more general theorem.
defines a homeomorphism between $(A \times B) \cap \Delta$ and $A \cap B$.
Thus, $C$ is homeomorphic to an irreducible component $C'$ of $(A \times B)
\cap \Delta$ and
\begin{align}
\begin{align*}
\codim(C, \mathfrak{k}^n) = n - \dim(C) = n -
\dim(C') = n - \dim(A \times B) + \codim(C', A \times B) \\
\overset{\text{
@ -3020,7 +2987,7 @@ this more general theorem.
\overset{\text{
\ref{dimprod}}}{=} 2n - \dim(A) - \dim(B) =
\codim(A,\mathfrak{k}^n) + \codim(B, \mathfrak{k}^n)
\end{align}
\end{align*}
by the general
properties of dimension and codimension,
\ref{corpithm} applied to

8
inputs/preamble.tex
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@ -1,11 +1,7 @@
\begin{warning}
This is not an official script!
This document was written in preparation for the oral exam.
It mostly follows the way \textsc{Prof.
Franke} presented the material in his
lecture rather closely.
There are probably errors.
This is not an official script.
There is no guarantee for completeness or correctness.
\end{warning}
\noindent The \LaTeX template by \textsc{Maximilian Kessler} is published under the

123
inputs/projective_spaces.tex
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@ -384,12 +384,12 @@ Let $\mathfrak{l}$ be any field.
\begin{proposition}
\label{bijproj}
There is a bijection
\begin{align}
\begin{align*}
f: \{I \subseteq A_+ | I \text{ homogeneous
ideal}, I = \sqrt{I}\} & \longrightarrow \{X \subseteq \mathbb{P}^n | X \text{
closed}\} \\ I & \longmapsto \Vp(I)\\ \langle \{f \in A_d | d > 0, X
\subseteq \Vp(f)\} \rangle & \longmapsfrom X
\end{align}
\end{align*}
Under this bijection,
the irreducible subsets correspond to the elements of
$\Proj(A_\bullet)$.
@ -528,11 +528,11 @@ Let $\mathfrak{l}$ be any field.
\codim(Y
\cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$.
Thus
\begin{align}
\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}
\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}
@ -566,8 +566,9 @@ Let $\mathfrak{l}$ be any field.
\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).
\ref{bijiredprim}
(bijection of irreducible subsets and prime ideals in the projective
and affine case).
Let $d = \dim(X)$ and
\[
@ -575,101 +576,92 @@ Let $\mathfrak{l}$ be any field.
X_{d+1} \subsetneq \ldots \subsetneq X_n =
\mathbb{P}^n
\]
be a chain of
irreducible subsets of $\mathbb{P}^n$.
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.
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 $1$.
$\mathbb{A}^n$ 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.
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 $1$.
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 \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}), $\fp = P \cdot
A$.
$I = \sqrt{I} \subseteq A$ (\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 $\fp = P \cdot A$.
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.
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}
\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 $A \cap B$ as dimension $\ge a + b - n$.
Let $A \subseteq \mathbb{P}^n$ and $B \subseteq \mathbb{P}^n$ be
irreducible subsets of dimensions $a$ and $b$.
If $a+ b \ge n$,
then $A \cap B \neq \emptyset$
and every irreducible component of $A \cap B$
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}).
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 $C(A \cap B) = C(A) \cap
C(B)$.
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}).
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 $b$
@ -682,4 +674,3 @@ Let $\mathfrak{l}$ be any field.
% 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!

88
inputs/varieties.tex
View file

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