From 3d4fa3154e1bfe48b88105e7630c8533bd6714e5 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Maximilian=20Ke=C3=9Fler?= <git@maximilian-kessler.de>
Date: Wed, 16 Feb 2022 01:34:08 +0100
Subject: [PATCH] fix \eps

---
 2021_Algebra_I.tex | 10 +++++-----
 1 file changed, 5 insertions(+), 5 deletions(-)

diff --git a/2021_Algebra_I.tex b/2021_Algebra_I.tex
index 034a9b4..bfab4aa 100644
--- a/2021_Algebra_I.tex
+++ b/2021_Algebra_I.tex
@@ -894,7 +894,7 @@ Let $K$ be a field and $R = K[X_1,\ldots,X_n]$.
     \[
         b^Ng \in R \tag{+} \label{bNginR}
     \]
-    However, if $b = \eps \prod_{i=1}^{l} P_i$ is a decomposition of $b$ into prime factors $P_i$ and a unit $\eps$ in $R$ and $g = \frac{1}{P}$, wehere $P \in R$ is a prime element not multiplicatively equvalent to any $P_i$, 
+    However, if $b = \varepsilon \prod_{i=1}^{l} P_i$ is a decomposition of $b$ into prime factors $P_i$ and a unit $\varepsilon$ in $R$ and $g = \frac{1}{P}$, wehere $P \in R$ is a prime element not multiplicatively equvalent to any $P_i$, 
     then \eqref{bNginR} fails for any $N \in \N$.
 \end{proof}
 
@@ -1955,10 +1955,10 @@ Let $\mathfrak{l}$ be any field.
     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 $1 = \sum_{d \in \bI}  \eps_d$ is the decomposition into homogeneous components, then $\eps_a = 1 \cdot \eps_a = \sum_{b \in \bI}  \eps_a\eps_b$ with $\eps_a\eps_b \in A_{a+b}$.
-    By the uniqueness of the decomposition into homogeneous components, $\eps_a \eps_0 = \eps_a$ and $b \neq 0 \implies \eps_a \eps_b = 0$.
-    Applying the last equation with $a = 0$ gives $b\neq 0 \implies \eps_b = \eps_0 \eps _b = 0$.
-    Thus $1 = \eps_0 \in A_0$.
+    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 $\varepsilon_a\varepsilon_b \in A_{a+b}$.
+    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 $1 = \varepsilon_0 \in A_0$.
 \end{remark}
 \begin{remark}
     The augmentation ideal of a graded ring is a homogeneous ideal.