fix \eps
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@ -894,7 +894,7 @@ Let $K$ be a field and $R = K[X_1,\ldots,X_n]$.
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\[
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b^Ng \in R \tag{+} \label{bNginR}
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\]
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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 $P_i$,
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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 $P_i$,
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then \eqref{bNginR} fails for any $N \in \N$.
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\end{proof}
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@ -1955,10 +1955,10 @@ Let $\mathfrak{l}$ be any field.
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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 $A$.
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\end{definition}
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\begin{remark}[Decomposition of $1$]
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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 $\eps_a\eps_b \in A_{a+b}$.
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By the uniqueness of the decomposition into homogeneous components, $\eps_a \eps_0 = \eps_a$ and $b \neq 0 \implies \eps_a \eps_b = 0$.
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Applying the last equation with $a = 0$ gives $b\neq 0 \implies \eps_b = \eps_0 \eps _b = 0$.
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Thus $1 = \eps_0 \in A_0$.
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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 $\varepsilon_a\varepsilon_b \in A_{a+b}$.
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By the uniqueness of the decomposition into homogeneous components, $\varepsilon_a \varepsilon_0 = \varepsilon_a$ and $b \neq 0 \implies \varepsilon_a \varepsilon_b = 0$.
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Applying the last equation with $a = 0$ gives $b\neq 0 \implies \varepsilon_b = \varepsilon_0 \varepsilon _b = 0$.
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Thus $1 = \varepsilon_0 \in A_0$.
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\end{remark}
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\begin{remark}
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The augmentation ideal of a graded ring is a homogeneous ideal.
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