fix tikzcd
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@ -863,13 +863,12 @@ The following will lead to another proof of the Nullstellensatz, which uses the
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There are $a_{ijk} \in A$ such that $b_i b_j = \sum_{k=1}^{m} a_{ijk}b_k$. And $\alpha_{ij} \in A$ such that $\beta_i = \sum_{j=1}^{m} \alpha_{ij}b_j$. Let $\tilde{A}$ be the sub- $R$-algebra of $A$ generated by the $a_{ijk}$ and $\alpha_{ij}$. $\tilde{A}$ is of finite type over $ R$, hence Noetherian. The $\tilde{A}$-submodule generated by $1$ and the $b_i$ is a sub-$R$-algebra containing the $\beta_i$ and thus coincides with $B$.
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Hence $B / \tilde{A}$ is finite. Since $A \subseteq B, A / \tilde{A}$ is finite (\ref{noethersubalg}).
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Hence $A / \tilde{A}$ is of finite type. By the transitivity of ``of finite type'', it follows that $A / R$ is of finite type.
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\begin{figure}[H]
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\centering
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\[
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\begin{tikzcd}
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\tilde A \arrow[hookrightarrow]{r}{\subseteq}& A \arrow[hookrightarrow]{r}{\subseteq} & B \\
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&R \arrow[bend left, swap]{ul}{\alpha} \arrow{u}{\alpha} \arrow[bend right]{ur}{\alpha}
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\end{tikzcd}
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\end{figure}
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\]
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\end{proof}
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\subsubsection{Artin-Tate proof of the Nullstellensatz}
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