From 4b55cffc5926d30e5460c59291349f05aa61d7e0 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Maximilian=20Ke=C3=9Fler?= Date: Wed, 16 Feb 2022 01:33:04 +0100 Subject: [PATCH] fix tikzcd --- 2021_Algebra_I.tex | 5 ++--- 1 file changed, 2 insertions(+), 3 deletions(-) diff --git a/2021_Algebra_I.tex b/2021_Algebra_I.tex index f53507e..034a9b4 100644 --- a/2021_Algebra_I.tex +++ b/2021_Algebra_I.tex @@ -863,13 +863,12 @@ The following will lead to another proof of the Nullstellensatz, which uses the 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 $B$. Hence $B / \tilde{A}$ is finite. Since $A \subseteq B, A / \tilde{A}$ is finite (\ref{noethersubalg}). Hence $A / \tilde{A}$ is of finite type. By the transitivity of ``of finite type'', it follows that $A / R$ is of finite type. - \begin{figure}[H] - \centering + \[ \begin{tikzcd} \tilde A \arrow[hookrightarrow]{r}{\subseteq}& A \arrow[hookrightarrow]{r}{\subseteq} & B \\ &R \arrow[bend left, swap]{ul}{\alpha} \arrow{u}{\alpha} \arrow[bend right]{ur}{\alpha} \end{tikzcd} - \end{figure} + \] \end{proof} \subsubsection{Artin-Tate proof of the Nullstellensatz}