replace \bI

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
     \[