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get rid of legacy macros

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Maximilian Keßler 2022-02-16 02:52:05 +01:00
parent 55477f0984
commit 8263aa6d10
2 changed files with 6 additions and 8 deletions
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10
algebra.sty
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@ -37,14 +37,9 @@
\newcommand{\Vp}{\ensuremath V_{\mathbb{P}}}%\Spec}} \newcommand{\Vp}{\ensuremath V_{\mathbb{P}}}%\Spec}}
\newcommand{\Pn}{\bP^n}%\Spec}} \newcommand{\Pn}{\bP^n}%\Spec}}
\newcommand{\npr}{\footnote{Not relevant for the exam.}}
\newcommand{\limrel}{\footnote{Limited relevance for the exam.}} % may appear in 3x questions
%\DeclareMathOperator{\ev}{ev}
\DeclareMathOperator{\Ker}{Ker} \DeclareMathOperator{\Ker}{Ker}
\DeclareMathOperator{\nil}{\mathfrak{nil}} \DeclareMathOperator{\nil}{\mathfrak{nil}}
%\DeclareMathOperator{\rad}{\mathfrak{rad}}
\RequirePackage{stackengine}
\stackMath
\usetikzlibrary{arrows.meta, \usetikzlibrary{arrows.meta,
quotes, babel} quotes, babel}
@ -53,3 +48,6 @@
\def\existsone{\exists!} \def\existsone{\exists!}
\def\defon#1{\upharpoonright_{#1}} \def\defon#1{\upharpoonright_{#1}}
\DeclareMathOperator{\Id}{Id} \DeclareMathOperator{\Id}{Id}
% \RequirePackage{stackengine}
% \stackMath

4
inputs/nullstellensatz_and_zariski_topology.tex
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@ -1549,7 +1549,7 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ and $I \subseteq R$ an ideal.
\end{remark} \end{remark}
\subsubsection{Application to the property of being a UFD} \subsubsection{Application to the property of being a UFD}
\begin{proposition}\limrel \begin{proposition}
Let $R$ be a Noetherian domain. Then $R$ is a UFD iff every $\fp \in \Spec R$ with $\hght(\fp)= 1$\footnote{In other words, every $\subseteq$-minimal element of the set of non-zero prime ideals of $R$ } is a principal ideal. Let $R$ be a Noetherian domain. Then $R$ is a UFD iff every $\fp \in \Spec R$ with $\hght(\fp)= 1$\footnote{In other words, every $\subseteq$-minimal element of the set of non-zero prime ideals of $R$ } is a principal ideal.
\end{proposition} \end{proposition}
\begin{proof} \begin{proof}
@ -1565,7 +1565,7 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ and $I \subseteq R$ an ideal.
Thus $\fp = pR$ for some prime element $p$. We have $p | f$ since $f \in \fp$. As $f$ is irreducible, $p$ and $f$ are multiplicatively equivalent. Thus $f$ is a prime element. Thus $\fp = pR$ for some prime element $p$. We have $p | f$ since $f \in \fp$. As $f$ is irreducible, $p$ and $f$ are multiplicatively equivalent. Thus $f$ is a prime element.
\end{proof} \end{proof}
\subsection{The Jacobson radical}\limrel \subsection{The Jacobson radical}
\begin{proposition} \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$}} \rad(A)$, the \vocab{Jacobson radical} of $A$. 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$}} \rad(A)$, the \vocab{Jacobson radical} of $A$.
\end{proposition} \end{proposition}