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replace \fM

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Maximilian Keßler 2022-02-16 01:23:12 +01:00
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2021_Algebra_I.tex
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@ -53,7 +53,7 @@ Fields which are not assumed to be algebraically closed have been renamed (usual
\begin{enumerate} \begin{enumerate}
\item Every submodule $N \subseteq M$ is finitely generated. \item Every submodule $N \subseteq M$ is finitely generated.
\item Every sequence $N_0 \subset N_1 \subset \ldots$ of submodules terminates \item Every sequence $N_0 \subset N_1 \subset \ldots$ of submodules terminates
\item Every set $\fM \neq \emptyset$ of submodules of $M$ has a $\subseteq$-largest element. \item Every set $\mathfrak{M} \neq \emptyset$ of submodules of $M$ has a $\subseteq$-largest element.
\end{enumerate} \end{enumerate}
\end{definition} \end{definition}
\begin{proposition}[Hilbert's Basissatz]\label{basissatz} \begin{proposition}[Hilbert's Basissatz]\label{basissatz}
@ -640,8 +640,8 @@ Let $X$ be a topological space.
\begin{proof} \begin{proof}
% i = ic % i = ic
Let $\fM$ be the set of closed subsets of $X$ which cannot be decomposed as a union of finitely many irreducible subsets. Let $\mathfrak{M}$ be the set of closed subsets of $X$ which cannot be decomposed as a union of finitely many irreducible subsets.
Suppose $\fM \neq \emptyset$. Then there exists a $\subseteq$-minimal $Y \in \fM$. $Y$ cannot be empty or irreducible. Hence $Y = A \cup B$ where $A,B$ are proper closed subsets of $ Y$. By the minimality of $Y$, $A$ and $B$ can be written as a union of proper closed subsets $\lightning$. Suppose $\mathfrak{M} \neq \emptyset$. Then there exists a $\subseteq$-minimal $Y \in \mathfrak{M}$. $Y$ cannot be empty or irreducible. Hence $Y = A \cup B$ where $A,B$ are proper closed subsets of $ Y$. By the minimality of $Y$, $A$ and $B$ can be written as a union of proper closed subsets $\lightning$.
Let $X = \bigcup_{i = 1}^n Z_i$, where there are no inclusions between the $Z_i$. If $Y$ is an irreducible subsets of $X$, $Y = \bigcup_{i = 1}^n (Y \cap Z_i)$ and there exists $1 \le i \le n$ such that $Y = Y \cap Z_i$. Let $X = \bigcup_{i = 1}^n Z_i$, where there are no inclusions between the $Z_i$. If $Y$ is an irreducible subsets of $X$, $Y = \bigcup_{i = 1}^n (Y \cap Z_i)$ and there exists $1 \le i \le n$ such that $Y = Y \cap Z_i$.
Hence $Y \subseteq Z_i$. Thus the $Z_i$ are irreducible components. Conversely, if $Y$ is an irreducible component of $X$, $Y \subseteq Z_i$ for some $i$ and $Y = Z_i$ by the definition of irreducible component. Hence $Y \subseteq Z_i$. Thus the $Z_i$ are irreducible components. Conversely, if $Y$ is an irreducible component of $X$, $Y \subseteq Z_i$ for some $i$ and $Y = Z_i$ by the definition of irreducible component.
@ -2072,7 +2072,7 @@ Let $\mathfrak{l}$ be any field.
\item $R$ is Noetherian. \item $R$ is Noetherian.
\item Every homogeneous ideal of $R_{\bullet}$ is finitely generated. \item Every homogeneous ideal of $R_{\bullet}$ is finitely generated.
\item Every chain $I_0\subseteq I_1 \subseteq \ldots$ of homogeneous ideals terminates. \item Every chain $I_0\subseteq I_1 \subseteq \ldots$ of homogeneous ideals terminates.
\item Every set $\fM \neq \emptyset$ of homogeneous ideals has a $\subseteq$-maximal element. \item Every set $\mathfrak{M} \neq \emptyset$ of homogeneous ideals has a $\subseteq$-maximal element.
\item $R_0$ is Noetherian and the ideal $R_+$ is finitely generated. \item $R_0$ is Noetherian and the ideal $R_+$ is finitely generated.
\item $R_0$ is Noetherian and $R / R_0$ is of finite type. \item $R_0$ is Noetherian and $R / R_0$ is of finite type.
\end{enumerate} \end{enumerate}