diff --git a/2021_Algebra_I.tex b/2021_Algebra_I.tex index 37f28ef..026d9c5 100644 --- a/2021_Algebra_I.tex +++ b/2021_Algebra_I.tex @@ -1,4 +1,4 @@ -\documentclass[english, fancyfoot% , git +\documentclass[english, fancyfoot %, git ]{mkessler-script} \course{Algebra I} @@ -33,10 +33,8 @@ \section{Varieties} \input{inputs/varieties} -\iffalse - \section{Übersicht} - \input{inputs/uebersicht} -\fi +% \section{Übersicht} +% \input{inputs/uebersicht} \cleardoublepage \printvocabindex diff --git a/inputs/finiteness_conditions.tex b/inputs/finiteness_conditions.tex index e8d257f..6274d79 100644 --- a/inputs/finiteness_conditions.tex +++ b/inputs/finiteness_conditions.tex @@ -5,9 +5,9 @@ Then the following sets coincide \begin{enumerate} \item - $\left\{ \sum_{s \in S'} r_{s} \cdot s ~ |~ S \subseteq S' \text{finite}, r_s \in R, \right\}$, + \[\left\{ \sum_{s \in S'} r_{s} \cdot s ~ |~ S \subseteq S' \text{ finite}, r_s \in R\right\},\] \item - $\bigcap_{\substack{S \subseteq N \subseteq M\\N \text{submodule}}} N$, + \[\bigcap_{\substack{S \subseteq N \subseteq M\\N \text{submodule}}} N,\] \item The $\subseteq$-smallest submodule of $M$ containing $S$. \end{enumerate} diff --git a/inputs/nullstellensatz_and_zariski_topology.tex b/inputs/nullstellensatz_and_zariski_topology.tex index 59d57ff..ce9c932 100644 --- a/inputs/nullstellensatz_and_zariski_topology.tex +++ b/inputs/nullstellensatz_and_zariski_topology.tex @@ -366,7 +366,7 @@ For $f \in R$ let $V(f) = V(fR)$. \begin{proof} Suppose $f$ vanishes on all zeros of $I$. Let $R' \coloneqq \mathfrak{k}[X_1,\ldots,X_n,T]$, - $g(X_1,\ldots,X_n,T) \coloneqq 1 - T \cdot f(X_1,\ldots,X_n)$ + \[g(X_1,\ldots,X_n,T) \coloneqq 1 - T \cdot f(X_1,\ldots,X_n)\] and $J \subseteq R'$ the ideal generated by $g$ and the elements of $I$ (viewed as elements of $R'$ which are constant in the $T$-direction). @@ -600,7 +600,7 @@ Let $X$ be a topological space. irreducible. The remaining assertion follows from the fact, - that the bijection is $\subseteq$-antimonotonic + that the bijection is $\subseteq$-anti\-monotonic and thus maximal ideals correspond to minimal irreducible closed subsets, which are the one-point subsets as $\mathfrak{k}^n$ is T${}_1$. \end{proof} @@ -609,7 +609,7 @@ Let $X$ be a topological space. Let $Z$ be an irreducible subset of the topological space $X$. Let $\codim(Z,X)$ be the maximum of the length $n$ of strictly increasing chains - $Z \subseteq Z_0 \subsetneq Z_1 \subsetneq \ldots \subsetneq Z_n$ + \[Z \subseteq Z_0 \subsetneq Z_1 \subsetneq \ldots \subsetneq Z_n\] of irreducible closed subsets of $X$ containing $Z$ or $\infty$ if such chains can be found for arbitrary $n$. Let @@ -617,7 +617,7 @@ Let $X$ be a topological space. \dim X \coloneqq \begin{cases} - \infty & \text{if } X = \emptyset,\\ - \sup_{\substack{Z \subseteq X \\ Z \text{ irreducible}}} \codim(Z,X) & \text{otherwise}. + \sup\limits_{\substack{Z \subseteq X \\ Z \text{ irreducible}}} \codim(Z,X) & \text{otherwise}. \end{cases} \] \end{definition} @@ -730,7 +730,7 @@ In general, these inequalities may be strict. it is clear that $\codim(\{0\}, \mathfrak{k}^n) \ge n$. Translation by $x \in \mathfrak{k}^n$ gives us - $\codim(\{x\}, \mathfrak{k}^n) \ge n$. + \[\codim(\{x\}, \mathfrak{k}^n) \ge n.\] The opposite inequality follows from \ref{upperbounddim} ($Z = \mathfrak{k}^n$, @@ -805,13 +805,13 @@ In general, these inequalities may be strict. \begin{enumerate} \item[M] If $m,n \in X$ and $A \subseteq X$ then - $m \in \mathcal{H}( \{n\} \cup A) \setminus \mathcal{H}(A) % - \iff n \in \mathcal{H}(\{m\} \cup A) \setminus \mathcal{H}(A)$, + \[m \in \mathcal{H}( \{n\} \cup A) \setminus \mathcal{H}(A) % + \iff n \in \mathcal{H}(\{m\} \cup A) \setminus \mathcal{H}(A),\] \item[F] $\mathcal{H}(A) = % \bigcup_{F \subseteq A \text{ finite}} \mathcal{H}(F)$. \end{enumerate} - In this case, $S \subseteq X$ is called \vocab{Independent subset}, + In this case, $S \subseteq X$ is called \vocab{independent subset}, if $s \not\in \mathcal{H}(S \setminus \{s\})$ for all $s \in S$ and \vocab[Generating subset]{generating} if $X = \mathcal{H}(S)$. $S$ is called a \vocab{base}, if it is both generating and independent. @@ -1944,9 +1944,9 @@ extensions: let $L$ be a field extension of $Q(A)$. By \ref{intclosure} the set of elements of $L$ integral over $A$ is a subring of $L$, the \vocab{integral closure} of $A$ in $L$. - $A$ is \vocab{Domain!integrally closed} in $L$ + $A$ is \vocab[Domain!integrally closed]{integrally closed} in $L$ if the integral closure of $A$ in $L$ equals $A$. - $A$ is \vocab{Domain!normal} if it is integrally closed in $Q(A)$. + $A$ is \vocab[Domain!normal]{normal} if it is integrally closed in $Q(A)$. \end{definition} \begin{proposition} @@ -2029,7 +2029,8 @@ extensions: (\ref{characfixnormalfe}), there is a positive integer $k$ with $y^k \in K$. As $A$ is normal, we have $y^k \in K \cap B = A$. - Thus $y^k \in (A \cap \fq) \setminus (A \cap \fr) = \fp \setminus \fp = \emptyset \lightning$. + Thus + \[y^k \in (A \cap \fq) \setminus (A \cap \fr) = \fp \setminus \fp = \emptyset \lightning.\] If $L / K$ is not finite, one applies Zorn's lemma to the poset of pairs $(M, \sigma)$ where $M$ is an intermediate field @@ -2116,7 +2117,7 @@ extensions: Japanese. \end{remark} -\begin{align*}+[Counterexample to going down] +\begin{example}+[Counterexample to going down] Let $R = \mathfrak{k}[X,Y]$ and $A = \mathfrak{k}[X,Y, \frac{X}{Y}]$. Then going down does not hold for $A / R$: @@ -2127,7 +2128,7 @@ extensions: and thus proper, we have $(X,Y)_R = \fq \cap R$. The prime ideal $(\frac{X}{Y},Y)_A = (\frac{X}{Y}, X,Y)_A$ is lying over $(X,Y)_R$, so going down is violated. -\end{align*} +\end{example} \subsubsection{Proof of \texorpdfstring{$\codim(\{y\},Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$}{codim(\{y\},Y) = trdeg(K(Y) /k)}} @@ -2280,7 +2281,7 @@ We will use the following $(a_i)_{i = 1}^m$ for $K / \mathfrak{l}$ with $a_i \in A$ for $1 \le i \le m$. \end{lemma} -\begin{align*} +\begin{proof} The proof is similar to the proof of \ref{ltrdegresfieldtrbase}. There are a natural number $m \ge n$ and elements $(a_i)_{i = n+1}^m \in A^{m-n}$ which generate $K$ @@ -2296,7 +2297,7 @@ We will use the following But then $K$ is algebraic over its subfield generated by $\mathfrak{l}$ and the $(a_i)_{i=1}^{m-1}$, contradicting the minimality of $m$. -\end{align*} +\end{proof} \begin{theorem} \label{htandtrdeg} @@ -2724,13 +2725,15 @@ 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*} - \codim(C, \mathfrak{k}^n) = n - \dim(C) = n - \dim(C') - = n - \dim(A \times B) + \codim(C', A \times B) \\ - \overset{\text{\ref{corpithm}}}{\le }2n - \dim(A \times B) - \overset{\text{\ref{dimprod}}}{=} 2n - \dim(A) - \dim(B) - = \codim(A,\mathfrak{k}^n) + \codim(B, \mathfrak{k}^n) - \end{align*} + \begin{IEEEeqnarray*}{rCl} + \codim(C, \mathfrak{k}^n) + &=& n - \dim(C)\\ + &=& n - \dim(C')\\ + &=& n - \dim(A \times B) + \codim(C', A \times B) \\ + &\overset{\text{\ref{corpithm}}}{\le }&2n - \dim(A \times B)\\ + &\overset{\text{\ref{dimprod}}}{=}& 2n - \dim(A) - \dim(B)\\ + &=& \codim(A,\mathfrak{k}^n) + \codim(B, \mathfrak{k}^n) + \end{IEEEeqnarray*} by the general properties of dimension and codimension, \ref{corpithm} applied to $(X_i - Y_i)_{i=1}^n$, @@ -2778,7 +2781,7 @@ this more general theorem. \subsection{The Jacobson radical} \begin{proposition} For a ring $A$, - $\bigcap_{\mathfrak{m} \in \MaxSpec A} \mathfrak{m} = \{a \in A | \forall x \in A ~ 1 - ax \in A^{\times }\} \text{\reflectbox{$\coloneqq$}} \rad(A)$, + \[\bigcap_{\mathfrak{m} \in \MaxSpec A} \mathfrak{m} = \{a \in A | \forall x \in A ~ 1 - ax \in A^{\times }\} \text{\reflectbox{$\coloneqq$}} \rad(A).\] the \vocab{Jacobson radical} of $A$. \end{proposition} \begin{proof} diff --git a/inputs/projective_spaces.tex b/inputs/projective_spaces.tex index d513500..0627499 100644 --- a/inputs/projective_spaces.tex +++ b/inputs/projective_spaces.tex @@ -65,12 +65,12 @@ Let $\mathfrak{l}$ be any field. We call the $r_d$ the \vocab{homogeneous components} of $r$. An ideal $I \subseteq A$ is called \vocab{homogeneous} if - $r \in I \implies \forall d \in \mathbb{I} ~ r_d \in I_d$ + \[r \in I \implies \forall d \in \mathbb{I} ~ r_d \in I_d\] where $I_d \coloneqq I \cap A_d$. By a \vocab{graded ring} we understand an $\N$-graded ring. - Tin this case, - $A_{+} \coloneqq \bigoplus_{d=1}^{\infty} A_d = \{r \in A | r_0 = 0\} $ + In this case, + \[A_{+} \coloneqq \bigoplus_{d=1}^{\infty} A_d = \{r \in A | r_0 = 0\}\] is called the \vocab{augmentation ideal} of $A$. \end{definition} \begin{remark}[Decomposition of $1$] @@ -133,16 +133,18 @@ Let $\mathfrak{l}$ be any field. \subsubsection{The Zariski topology on $\mathbb{P}^n$} \begin{notation} Recall that for - $\alpha \in \N^{n+1}$ $|\alpha| = \sum_{i=0}^{n} \alpha_i$ - and $x^\alpha = x_0^{\alpha_0} \cdot \ldots \cdot x_n^{\alpha_n}$. + $\alpha \in \N^{n+1}$ + \[|\alpha| = \sum_{i=0}^{n} \alpha_i \text{ and } + x^\alpha = x_0^{\alpha_0} \cdot \ldots \cdot x_n^{\alpha_n}.\] \end{notation} \begin{definition}[Homogeneous polynomials] Let $R$ be any ring and - $f = \sum_{\alpha \in \N^{n+1}} f_\alpha X^{\alpha}\in R[X_0,\ldots,X_n]$. + \[f = \sum_{\alpha \in \N^{n+1}} f_\alpha X^{\alpha}\in R[X_0,\ldots,X_n].\] We say that $f$ is \vocab{homogeneous of degree $d$} - if $|\alpha| \neq d \implies f_\alpha = 0$ . + if + \[|\alpha| \neq d \implies f_\alpha = 0.\] We denote the subset of homogeneous polynomials of degree $d$ by - $R[X_0,\ldots,X_n]_d \subseteq R[X_0,\ldots,X_n]$. + \[R[X_0,\ldots,X_n]_d \subseteq R[X_0,\ldots,X_n].\] \end{definition} \begin{remark} This definition gives $R$ the structure of a graded ring. @@ -151,8 +153,8 @@ Let $\mathfrak{l}$ be any field. \label{ztoppn} Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.% \footnote{As always, $\mathfrak{k}$ is algebraically closed} - For $f \in A_d = \mathfrak{k}[X_0,\ldots, - _n]_d$, the validity of the equation $f(x_0,\ldots,x_{n}) = 0$ + For $f \in A_d = \mathfrak{k}[X_0,\ldots,X_n]_d$, + the validity of the equation $f(x_0,\ldots,x_{n}) = 0$ does not depend on the choice of homogeneous coordinates, as \[ f(\lambda x_0,\ldots, \lambda x_n) 0 \lambda^d f(x_0,\ldots,x_n). @@ -320,8 +322,7 @@ Let $\mathfrak{l}$ be any field. \begin{proof} $\impliedby$ is clear. Let $\Vp(I) \subseteq \Vp(f)$. - If $x = (x_0, - ldots,x_n) \in \Va(I)$, then either $x = 0$ in + If $x = (x_0,\ldots,x_n) \in \Va(I)$, then either $x = 0$ in which case $f(x) = 0$ since $d > 0$ or the point $[x_0,\ldots,x_n] \in \mathbb{P}^n$ is well-defined and belongs to $\Vp(I) \subseteq \Vp(f)$, @@ -330,8 +331,8 @@ Let $\mathfrak{l}$ be any field. (\ref{hns3}). \end{proof} -\begin{definition} - \footnote{This definition is not too important, the characterization in the following remark suffices.}. +\begin{definition}% + \footnote{This definition is not too important, the characterization in the following remark suffices.} For a graded ring $R_\bullet$, let $\Proj(R_\bullet)$ be the set of $\fp \in \Spec R$ such that $\fp$ is a homogeneous ideal and $\fp \not\supseteq R_+$. @@ -432,17 +433,18 @@ Let $\mathfrak{l}$ be any field. \end{proof} \begin{remark} - If $R_\bullet$ is $\N$-graded and $\fp \in \Spec R_0$, - then $\fp \oplus R_+ = \{r \in R | r_0 \in \fp\} $ + If $R_\bullet$ is $\N$-graded and $\fp \in \Spec R_0$, then + \[\fp \oplus R_+ = \{r \in R | r_0 \in \fp\}\] is a homogeneous prime ideal of $R$. - \[ - \{\fp \in \Spec R | \fp \text{ is a homogeneous ideal of } R_\bullet\} - = \Proj(R_\bullet) \sqcup \{\fp \oplus R_+ | \fp \in \Spec R_0\}. - \] + \begin{IEEEeqnarray*}{rCl} + &&\{\fp \in \Spec R | \fp \text{ is a homogeneous ideal of } R_\bullet\}\\ + &=&\Proj(R_\bullet) \sqcup \{\fp \oplus R_+ | \fp \in \Spec R_0\}. + \end{IEEEeqnarray*} \end{remark} +\pagebreak \subsection{Dimension of $\mathbb{P}^n$} -\begin{proposition} +\begin{proposition}\, \begin{itemize} \item $\mathbb{P}^n$ is catenary. @@ -450,11 +452,11 @@ Let $\mathfrak{l}$ be any field. $\dim(\mathbb{P}^n) = n$. Moreover, $\codim(\{x\} ,\mathbb{P}^n) = n$ for every $x \in \mathbb{P}^n$. \item - If $X \subseteq \mathbb{P}^n$ is irreducible and $x \in X$, - then $\codim(\{x\}, X) = \dim(X) = n - \codim(X, \mathbb{P}^n)$. + If $X \subseteq \mathbb{P}^n$ is irreducible and $x \in X$, then + \[\codim(\{x\}, X) = \dim(X) = n - \codim(X, \mathbb{P}^n).\] \item If $X \subseteq Y \subseteq \mathbb{P}^n$ are irreducible subsets, - then $\codim(X,Y) = \dim(Y) - \dim(X)$. + then \[\codim(X,Y) = \dim(Y) - \dim(X).\] \end{itemize} \end{proposition} \begin{proof} @@ -472,12 +474,13 @@ Let $\mathfrak{l}$ be any field. $\codim(X,Z) = \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$ and $\codim(Y,Z) = \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$. Thus - \begin{align*} + \begin{IEEEeqnarray*}{rCl} \codim(X,Y) + \codim(Y,Z) - & = \codim(X \cap \mathbb{A}^n, Y \cap \mathbb{A}^n) + \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n) \\ - & = \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)\\ - & = \codim(X, Z) - \end{align*} + & = &\codim(X \cap \mathbb{A}^n, Y \cap \mathbb{A}^n)\\ + & & + \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n) \\ + & = &\codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)\\ + & = &\codim(X, Z) + \end{IEEEeqnarray*} because $\mathfrak{k}^n$ is catenary and the first point follows. The remaining assertions can easily be derived from the first two. \end{proof} @@ -492,17 +495,15 @@ Let $\mathfrak{l}$ be any field. If $X = \Vp(I)$ where $I \subseteq A_+ = \mathfrak{k}[X_0,\ldots,X_n]_+$ is homogeneous, then $C(X) = \Va(I)$. \end{definition} -\begin{proposition} +\begin{proposition}\, \label{conedim} \begin{itemize} \item $C(X)$ is irreducible iff $X$ is irreducible or $X = \emptyset$. \item If $X$ is irreducible, then - - $\dim(C(X)) = \dim(X) + 1$ and - - $\codim(C(X), \mathfrak{k}^{n+1}) = \codim(X, \mathbb{P}^n)$ + \[\dim(C(X)) = \dim(X) + 1\] and + \[\codim(C(X), \mathfrak{k}^{n+1}) = \codim(X, \mathbb{P}^n).\] \end{itemize} \end{proposition} \begin{proof} @@ -525,7 +526,7 @@ Let $\mathfrak{l}$ be any field. Hence $\dim(C(X)) \ge 1 + d$ and $\codim(C(X), \mathfrak{k}^{n+1}) \ge n-d$. Since - $\dim(C(X)) + \codim(C(X), \mathfrak{k}^{n+1}) = \dim(\mathfrak{k}^{n+1}) = n+1$, + \[\dim(C(X)) + \codim(C(X), \mathfrak{k}^{n+1}) = \dim(\mathfrak{k}^{n+1}) = n+1,\] the two inequalities must be equalities. \end{proof} \subsubsection{Application to hypersurfaces in $\mathbb{P}^n$} diff --git a/inputs/varieties.tex b/inputs/varieties.tex index fa8471e..2470f5c 100644 --- a/inputs/varieties.tex +++ b/inputs/varieties.tex @@ -47,7 +47,7 @@ \vocab{sheaf axiom}. \end{definition} \begin{trivial}+ - A presheaf is a contravariant functor $\mathcal{G} :\mathcal{O}(X) \to C$ + A presheaf is a contravariant functor $\mathcal{G}:\mathcal{O}(X) \to C$ where $\mathcal{O}(X)$ denotes the category of open subsets of $X$ with inclusions as morphisms and $C$ is the category of sets, rings or (abelian) groups. @@ -315,7 +315,7 @@ The following is somewhat harder than in the affine case: \end{itemize} \end{remark} \subsubsection{Examples of categories} -\begin{example} +\begin{example}\, \begin{itemize} \item The category of sets. @@ -357,7 +357,7 @@ The following is somewhat harder than in the affine case: addition $\Hom_\mathcal{B}(X,Y) = \Hom_\mathcal{A}(X,Y)$ for arbitrary $X,Y \in \Ob(\mathcal{B})$. \end{definition} -\begin{example} +\begin{example}\, \begin{itemize} \item The category of abelian groups is a full subcategory of the @@ -407,7 +407,7 @@ The following is somewhat harder than in the affine case: It is called an \vocab{equivalence of categories} if it is full, faithful and essentially surjective. \end{definition} -\begin{example} +\begin{example}\, \begin{itemize} \item There are \vocab[Functor!forgetful]{forgetful functors} @@ -552,7 +552,7 @@ The following is somewhat harder than in the affine case: \end{proof} -\begin{proposition}[About affine varieties] +\begin{proposition}[About affine varieties]\, \label{propaffvar} \begin{itemize} \item