get rid of legacy macros
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2 changed files with 6 additions and 8 deletions
10
algebra.sty
10
algebra.sty
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@ -37,14 +37,9 @@
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\newcommand{\Vp}{\ensuremath V_{\mathbb{P}}}%\Spec}}
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\newcommand{\Vp}{\ensuremath V_{\mathbb{P}}}%\Spec}}
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\newcommand{\Pn}{\bP^n}%\Spec}}
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\newcommand{\Pn}{\bP^n}%\Spec}}
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\newcommand{\npr}{\footnote{Not relevant for the exam.}}
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\newcommand{\limrel}{\footnote{Limited relevance for the exam.}} % may appear in 3x questions
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%\DeclareMathOperator{\ev}{ev}
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\DeclareMathOperator{\Ker}{Ker}
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\DeclareMathOperator{\Ker}{Ker}
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\DeclareMathOperator{\nil}{\mathfrak{nil}}
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\DeclareMathOperator{\nil}{\mathfrak{nil}}
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%\DeclareMathOperator{\rad}{\mathfrak{rad}}
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\RequirePackage{stackengine}
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\stackMath
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\usetikzlibrary{arrows.meta,
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\usetikzlibrary{arrows.meta,
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quotes, babel}
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quotes, babel}
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@ -53,3 +48,6 @@
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\def\existsone{\exists!}
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\def\existsone{\exists!}
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\def\defon#1{\upharpoonright_{#1}}
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\def\defon#1{\upharpoonright_{#1}}
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\DeclareMathOperator{\Id}{Id}
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\DeclareMathOperator{\Id}{Id}
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% \RequirePackage{stackengine}
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% \stackMath
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@ -1549,7 +1549,7 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ and $I \subseteq R$ an ideal.
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\end{remark}
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\end{remark}
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\subsubsection{Application to the property of being a UFD}
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\subsubsection{Application to the property of being a UFD}
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\begin{proposition}\limrel
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\begin{proposition}
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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.
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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.
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\end{proposition}
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\end{proposition}
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\begin{proof}
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\begin{proof}
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@ -1565,7 +1565,7 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ and $I \subseteq R$ an ideal.
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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.
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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.
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\end{proof}
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\end{proof}
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\subsection{The Jacobson radical}\limrel
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\subsection{The Jacobson radical}
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\begin{proposition}
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\begin{proposition}
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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 $A$.
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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 $A$.
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\end{proposition}
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\end{proposition}
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