From 6423c3c08e68c1b2e138e22ab5a50b51b7faa3b4 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Maximilian=20Ke=C3=9Fler?= Date: Wed, 16 Feb 2022 02:06:52 +0100 Subject: [PATCH] replace \bI --- 2021_Algebra_I.tex | 18 +++++++++--------- 1 file changed, 9 insertions(+), 9 deletions(-) diff --git a/2021_Algebra_I.tex b/2021_Algebra_I.tex index 2022e60..e427871 100644 --- a/2021_Algebra_I.tex +++ b/2021_Algebra_I.tex @@ -1940,19 +1940,19 @@ Let $\mathfrak{l}$ be any field. \subsubsection{Graded rings and homogeneous ideals} \begin{notation} - Let $\bI = \N$ or $\bI = \Z$. + Let $\mathbb{I} = \N$ or $\mathbb{I} = \Z$. \end{notation} \begin{definition} - By an \vocab[Graded ring]{$\bI$-graded ring} $A_\bullet$ we understand a ring $A$ with a collection $(A_d)_{d \in \bI}$ of subgroups of the additive group $(A, +)$ such that $A_a \cdot A_b \subseteq A_{a + b}$ for $a,b \in \bI$ and such that $A = \bigoplus_{d \in \bI} A_d$ in the sense that every $r \in A$ has a unique decomposition $r = \sum_{d \in \bI} r_d$ with $r_d \in A_d$ and but finitely many $r_d \neq 0$. + By an \vocab[Graded ring]{$\mathbb{I}$-graded ring} $A_\bullet$ we understand a ring $A$ with a collection $(A_d)_{d \in \mathbb{I}}$ of subgroups of the additive group $(A, +)$ such that $A_a \cdot A_b \subseteq A_{a + b}$ for $a,b \in \mathbb{I}$ and such that $A = \bigoplus_{d \in \mathbb{I}} A_d$ in the sense that every $r \in A$ has a unique decomposition $r = \sum_{d \in \mathbb{I}} r_d$ with $r_d \in A_d$ and but finitely many $r_d \neq 0$. We call the $r_d$ the \vocab{homogeneous components} of $r$. - An ideal $I \subseteq A$ is called \vocab{homogeneous} if $r \in I \implies \forall d \in \bI ~ r_d \in I_d$ where $I_d \coloneqq I \cap A_d$. + An ideal $I \subseteq A$ is called \vocab{homogeneous} if $r \in I \implies \forall d \in \mathbb{I} ~ r_d \in I_d$ 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\} $ is called the \vocab{augmentation ideal} of $A$. \end{definition} \begin{remark}[Decomposition of $1$] - If $1 = \sum_{d \in \bI} \varepsilon_d$ is the decomposition into homogeneous components, then $\varepsilon_a = 1 \cdot \varepsilon_a = \sum_{b \in \bI} \varepsilon_a\varepsilon_b$ with $\varepsilon_a\varepsilon_b \in A_{a+b}$. + If $1 = \sum_{d \in \mathbb{I}} \varepsilon_d$ is the decomposition into homogeneous components, then $\varepsilon_a = 1 \cdot \varepsilon_a = \sum_{b \in \mathbb{I}} \varepsilon_a\varepsilon_b$ with $\varepsilon_a\varepsilon_b \in A_{a+b}$. By the uniqueness of the decomposition into homogeneous components, $\varepsilon_a \varepsilon_0 = \varepsilon_a$ and $b \neq 0 \implies \varepsilon_a \varepsilon_b = 0$. Applying the last equation with $a = 0$ gives $b\neq 0 \implies \varepsilon_b = \varepsilon_0 \varepsilon _b = 0$. Thus $1 = \varepsilon_0 \in A_0$. @@ -1972,9 +1972,9 @@ Let $\mathfrak{l}$ be any field. \end{proposition} \begin{proof} Most assertions are trivial. We only show that $J$ homogeneous $\implies \sqrt{J} $ homogeneous. - Let $A$ be $\bI$-graded, $f \in \sqrt{J} $ and $f = \sum_{d \in \bI} f_d$ the decomposition. - To show that all $f_d \in \sqrt{J} $, we use induction on $N_f \coloneqq \# \{d \in \bI | f_d \neq 0\}$. - $N_f = 0$ is trivial. Suppose $N_f > 0$ and $e \in \bI$ is maximal with $f_e \neq 0$. + Let $A$ be $\mathbb{I}$-graded, $f \in \sqrt{J} $ and $f = \sum_{d \in \mathbb{I}} f_d$ the decomposition. + To show that all $f_d \in \sqrt{J} $, we use induction on $N_f \coloneqq \# \{d \in \mathbb{I} | f_d \neq 0\}$. + $N_f = 0$ is trivial. Suppose $N_f > 0$ and $e \in \mathbb{I}$ is maximal with $f_e \neq 0$. For $l \in \N$, the $le$-th homogeneous component of $f^l$ is $f_e^l$. Choosing $l$ large enough such that $f^l \in J$ and using the homogeneity of $J$, we find $f_e \in \sqrt{J}$. As $\sqrt{J} $ is an ideal, $\tilde f \coloneqq f - f_e \in \sqrt{J} $. As $N_{\tilde f} = N_f -1$, the induction assumption may be applied to $\tilde f$ and shows $f_d \in \sqrt{J} $ for $d \neq e$. \end{proof} @@ -2153,13 +2153,13 @@ Let $A = \mathfrak{k}[X_0,\ldots,X_n]$. \subsection{Some remarks on homogeneous prime ideals} \begin{lemma}\label{homprime} - Let $R_\bullet$ be an $\bI$ graded ring ($\bI = \N$ or $\bI = \Z$). + Let $R_\bullet$ be an $\mathbb{I}$ graded ring ($\mathbb{I} = \N$ or $\mathbb{I} = \Z$). A homogeneous ideal $I \subseteq R$ is a prime ideal iff $1 \not\in I$ and for homogeneous elements $f, g \in R , fg \in I \implies f \in I \lor g \in I$. \end{lemma} \begin{proof} $\implies$ is trivial. It suffices to show that for arbitrary $f,g \in R fg \in I \implies f \in I \lor g \in I$. - Let $f = \sum_{d \in \bI} f_d, g = \sum_{d \in \bI} g_d $ be the decompositions into homogeneous components. + Let $f = \sum_{d \in \mathbb{I}} f_d, g = \sum_{d \in \mathbb{I}} g_d $ be the decompositions into homogeneous components. If $f \not\in I$ and $g \not\in I$ there are $d,e \in I$ with $f_d \in I, g_e \in I$, and they may assumed to be maximal with this property. As $I$ is homogeneous and $fg \in I$, we have $(fg)_{d+e} \in I$ but \[