rav: fix errors

ws2022/rav/lecture/rav11.tex

@@ -40,23 +40,21 @@ at $x$: For all $h \in k^{n}$:
 \end{bem}
 
 \begin{definition}
-    We set
-    \[
-    \mathcal{I}(X)_x^{*} \coloneqq \{ P_x^{*} \colon P \in \mathcal{I}(X) \} 
-    .\] 
+    We set $\mathcal{I}(X)_x^{*}$ to be the ideal generated
+    by $P_x^{*}$ for all $P \in \mathcal{I}(X)$.
 \end{definition}
 
-\begin{satz}
-    The set $\mathcal{I}(X)_x^{*}$ is an ideal of $k[T_1, \ldots, T_n]$.
-\end{satz}
-
-\begin{proof}
-    By definition, $0 \in \mathcal{I}(X)_x^{*}$. Let $P_x^{*}, Q_x^{*}$ be elements
-    of $\mathcal{I}(X)_x^{*}$ coming from $P, Q \in \mathcal{I}(X)$. Then
-    $P_x^{*} - Q_x^{*}$ is of the form $R_x^{*}$ for some $R \in \mathcal{I}(X)$, where
-    $R = 0$, $R = P$, $R = Q$ or $R = P-Q$. Moreover, for $Q \in k[T_1, \ldots, T_n]$,
-    we have $P_x^{*} Q = (PQ)_x^{*} \in \mathcal{I}(X)_x^{*}$.
-\end{proof}
+%\begin{satz}
+%    The set $\mathcal{I}(X)_x^{*}$ is an ideal of $k[T_1, \ldots, T_n]$.
+%\end{satz}
+%
+%\begin{proof}
+%    By definition, $0 \in \mathcal{I}(X)_x^{*}$. Let $P_x^{*}, Q_x^{*}$ be elements
+%    of $\mathcal{I}(X)_x^{*}$ coming from $P, Q \in \mathcal{I}(X)$. Then
+%    $P_x^{*} - Q_x^{*}$ is of the form $R_x^{*}$ for some $R \in \mathcal{I}(X)$, where
+%    $R = 0$, $R = P$, $R = Q$ or $R = P-Q$. Moreover, for $Q \in k[T_1, \ldots, T_n]$,
+%    we have $P_x^{*} Q = (PQ)_x^{*} \in \mathcal{I}(X)_x^{*}$.
+%\end{proof}
 
 \begin{bem}[]
     The ideal $\mathcal{I}(X)^{*}$ is finitely generated. However,
@@ -324,29 +322,29 @@ consider the Zariski tangent space to $X$ at a point $x \in X$.
     may vary with $x$.
 \end{bem}
 
-\begin{satz}[a Jacobian criterion]
-    If $(P_1, \ldots, P_m)$ are polynomials such that
-    $\mathcal{I}(X) = (P_1, \ldots, P_m)$ and $\operatorname{rk } P'(x) = m$, where
-    $P = (P_1, \ldots, P_m)$, then $x$ is a non-singular point of $X$.
-\end{satz}
-
-\begin{proof}
-    By \ref{kor:cone-in-tangent-space} and \ref{kor:tangent-kernel-jacobian} it suffices to show that
-    \[
-    \mathcal{C}_x(X) \supseteq x + \bigcap_{i=1} ^{m} \text{ker } P_i'(x)
-    .\] By definition
-    \[
-    \mathcal{C}_x(X) = x + \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x^{*})
-    \] and $\mathcal{I}(X)_x^{*} = \{ Q_x^{*} : Q \in \mathcal{I}(X)\} $. If $Q \in \mathcal{I}(X)$,
-    there exist polynomials $Q_1, \ldots, Q_m$ such that
-    $Q = \sum_{i=1}^{m} Q_i P_i$, so $Q_x^{*}$ is a linear combination of the $(P_i)_x^{*}$.
-    Since $\text{rk }(P_1'(x), \ldots, P_m'(x)) = m$, we have
-    $P_i'(x) \neq 0$ for all $i$. So $(P_i)_x^{*} = P_i'(x)$ in the Taylor expansion
-    of $P_i$ at $x$. So $Q_x^{*}$ is a linear combination
-    of $(P_1'(x), \ldots, P_m'(x))$,
-    which proves that if $h \in \bigcap_{i=1}^{m} \text{ker } P_i'(x)$, then
-    $Q_x^{*}(h) = 0$ for all $Q \in \mathcal{I}(X)$, hence
-    $x + h \in \mathcal{C}_x(X)$.
-\end{proof}
+%\begin{satz}[a Jacobian criterion]
+%    If $(P_1, \ldots, P_m)$ are polynomials such that
+%    $\mathcal{I}(X) = (P_1, \ldots, P_m)$ and $\operatorname{rk } P'(x) = m$, where
+%    $P = (P_1, \ldots, P_m)$, then $x$ is a non-singular point of $X$.
+%\end{satz}
+%
+%\begin{proof}
+%    By \ref{kor:cone-in-tangent-space} and \ref{kor:tangent-kernel-jacobian} it suffices to show that
+%    \[
+%    \mathcal{C}_x(X) \supseteq x + \bigcap_{i=1} ^{m} \text{ker } P_i'(x)
+%    .\] By definition
+%    \[
+%    \mathcal{C}_x(X) = x + \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x^{*})
+%    \] and $\mathcal{I}(X)_x^{*} = \{ Q_x^{*} : Q \in \mathcal{I}(X)\} $. If $Q \in \mathcal{I}(X)$,
+%    there exist polynomials $Q_1, \ldots, Q_m$ such that
+%    $Q = \sum_{i=1}^{m} Q_i P_i$, so $Q_x^{*}$ is a linear combination of the $(P_i)_x^{*}$.
+%    Since $\text{rk }(P_1'(x), \ldots, P_m'(x)) = m$, we have
+%    $P_i'(x) \neq 0$ for all $i$. So $(P_i)_x^{*} = P_i'(x)$ in the Taylor expansion
+%    of $P_i$ at $x$. So $Q_x^{*}$ is a linear combination
+%    of $(P_1'(x), \ldots, P_m'(x))$,
+%    which proves that if $h \in \bigcap_{i=1}^{m} \text{ker } P_i'(x)$, then
+%    $Q_x^{*}(h) = 0$ for all $Q \in \mathcal{I}(X)$, hence
+%    $x + h \in \mathcal{C}_x(X)$.
+%\end{proof}
 
 \end{document}