From f4bed5efb3276a5e6e525f9eb8e28a8a5d911262 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Sat, 22 Jul 2023 21:52:48 +0200 Subject: [PATCH] some small changes --- 2021_Algebra_I.tex | 3 +- algebra.sty | 4 +- inputs/finiteness_conditions.tex | 72 ++-- .../nullstellensatz_and_zariski_topology.tex | 307 ++++++++---------- inputs/preamble.tex | 8 +- inputs/projective_spaces.tex | 123 ++++--- inputs/uebersicht.tex | 2 +- inputs/varieties.tex | 90 +++-- 8 files changed, 270 insertions(+), 339 deletions(-) diff --git a/2021_Algebra_I.tex b/2021_Algebra_I.tex index acf0b7e..37f28ef 100644 --- a/2021_Algebra_I.tex +++ b/2021_Algebra_I.tex @@ -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} diff --git a/algebra.sty b/algebra.sty index a655fc0..af02eb0 100644 --- a/algebra.sty +++ b/algebra.sty @@ -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} diff --git a/inputs/finiteness_conditions.tex b/inputs/finiteness_conditions.tex index 15c0ca0..de9d9d4 100644 --- a/inputs/finiteness_conditions.tex +++ b/inputs/finiteness_conditions.tex @@ -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 $\beta_{\alpha, l} \in B$. @@ -551,10 +539,9 @@ commutative rings with $1$. Q(T) = p_{\alpha} T^{w_{\vec k}(\alpha)} + \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 - $p_\alpha \neq 0$. + with $q_j \in B$ and + $\alpha$ 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} - diff --git a/inputs/nullstellensatz_and_zariski_topology.tex b/inputs/nullstellensatz_and_zariski_topology.tex index 55299a1..2a67953 100644 --- a/inputs/nullstellensatz_and_zariski_topology.tex +++ b/inputs/nullstellensatz_and_zariski_topology.tex @@ -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] - $I_\lambda \subseteq \sum_{\lambda - \in \Lambda} I_\lambda \implies - \Va(\sum_{\lambda \in \Lambda} I_\lambda) - \subseteq \Va(I_\lambda)$. - 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 $\bigcap_{\lambda \in \Lambda} \Va(I_\lambda) \subseteq - \Va(\sum_{\lambda - \in \Lambda} I_\lambda)$. + $I_\lambda \subseteq \sum_{\lambda \in \Lambda} I_\lambda + \implies \Va(\sum_{\lambda \in \Lambda} I_\lambda) + \subseteq \Va(I_\lambda)$. + 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 + $\bigcap_{\lambda \in \Lambda} \Va(I_\lambda) + \subseteq \Va(\sum_{\lambda \in \Lambda} I_\lambda)$. \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 diff --git a/inputs/preamble.tex b/inputs/preamble.tex index 3904d25..c9c2654 100644 --- a/inputs/preamble.tex +++ b/inputs/preamble.tex @@ -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 diff --git a/inputs/projective_spaces.tex b/inputs/projective_spaces.tex index b63accd..901476c 100644 --- a/inputs/projective_spaces.tex +++ b/inputs/projective_spaces.tex @@ -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! - diff --git a/inputs/uebersicht.tex b/inputs/uebersicht.tex index 3058692..0fa7ff1 100644 --- a/inputs/uebersicht.tex +++ b/inputs/uebersicht.tex @@ -44,7 +44,7 @@ % Proofs -Def of integrality (<=>) +Def of integrality (<=>) Fact about integrality and field: diff --git a/inputs/varieties.tex b/inputs/varieties.tex index 1d1d05a..6dccdb7 100644 --- a/inputs/varieties.tex +++ b/inputs/varieties.tex @@ -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 \\ - \gamma & \longmapsto \gamma_x \coloneqq (U, \gamma) / \sim - \end{align} + \begin{align*} + \cdot_x : \mathcal{G}(U) & \longrightarrow \mathcal{G}_x\\ + \gamma & \longmapsto \gamma_x \coloneqq (U, \gamma) / \sim + \end{align*} \end{definition} \begin{fact}