josia-notes/s21-algebra-1
1
0
Fork 0
Code Issues Pull requests Projects Releases Packages Wiki Activity Actions

fix tikzcd

Browse source
This commit is contained in:
Maximilian Keßler 2022-02-16 01:33:04 +01:00
parent 2204004e63
commit 4b55cffc59
1 changed files with 2 additions and 3 deletions
Show stats Download patch file Download diff file Expand all files Collapse all files

5
2021_Algebra_I.tex
View file

@ -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 $B$. 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 $B / \tilde{A}$ 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. 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] \[
\centering
\begin{tikzcd} \begin{tikzcd}
\tilde A \arrow[hookrightarrow]{r}{\subseteq}& A \arrow[hookrightarrow]{r}{\subseteq} & B \\ \tilde A \arrow[hookrightarrow]{r}{\subseteq}& A \arrow[hookrightarrow]{r}{\subseteq} & B \\
&R \arrow[bend left, swap]{ul}{\alpha} \arrow{u}{\alpha} \arrow[bend right]{ur}{\alpha} &R \arrow[bend left, swap]{ul}{\alpha} \arrow{u}{\alpha} \arrow[bend right]{ur}{\alpha}
\end{tikzcd} \end{tikzcd}
\end{figure} \]
\end{proof} \end{proof}
\subsubsection{Artin-Tate proof of the Nullstellensatz} \subsubsection{Artin-Tate proof of the Nullstellensatz}