fix tizkcd figure
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@ -851,14 +851,12 @@ The following will lead to another proof of the Nullstellensatz, which uses the
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\label{artintate}
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\label{artintate}
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Let $A$ be a subalgebra of the $R$-algebra $B$, where $R$ is Noetherian. If $ B / R$ is of finite type and $B / A$ is finite, then $A / R$ is also of finite type.
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Let $A$ be a subalgebra of the $R$-algebra $B$, where $R$ is Noetherian. If $ B / R$ is of finite type and $B / A$ is finite, then $A / R$ is also of finite type.
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\begin{figure}[H]
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\[
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\centering
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\begin{tikzcd}
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\begin{tikzcd}
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A \arrow[hookrightarrow]{rr}{\subseteq}& & B \\
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A \arrow[hookrightarrow]{rr}{\subseteq}& & B \\
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&R \arrow{ul}{\alpha} \arrow{ur}{\alpha} \text{~(Noeth.)}
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&R \arrow{ul}{\alpha} \arrow{ur}{\alpha} \text{~(Noeth.)}
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\end{tikzcd}
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\end{tikzcd}
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\end{figure}
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\]
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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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Let $(b_i)_{i=1}^{m}$ generate $B$ as an $A$-module and $(\beta_j)_{j=1}^m$ as an $R$-algebra.
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Let $(b_i)_{i=1}^{m}$ generate $B$ as an $A$-module and $(\beta_j)_{j=1}^m$ as an $R$-algebra.
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