From dffe97ef2a21a6e9dc80998675e6aef693aec2a6 Mon Sep 17 00:00:00 2001 From: Matthis Date: Sat, 2 Dec 2023 21:28:01 +0100 Subject: [PATCH] added lec06 --- lec06.tex | 162 ++++++++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 162 insertions(+) create mode 100644 lec06.tex diff --git a/lec06.tex b/lec06.tex new file mode 100644 index 0000000..3f273ea --- /dev/null +++ b/lec06.tex @@ -0,0 +1,162 @@ +\documentclass{lecture} + +\begin{document} + +\begin{lemma} + Let $X/k$ be of locally finite type. + Then $X_\text{sm}$ is constructible. +\end{lemma} + +\begin{proof} + WLOG $X$ is affine. Then the assertion follows from B53, B72c and OCG13Z. +\end{proof} + +\begin{satz} + If $f:X\to Y$ is a morphism of schemes and $|Y|$ is discrete, then $f$ is universally open. (cf. Corollary \ref{cor:lfp+discrete-target->univ-open}) +\end{satz} + +\begin{proof} + Universal openness is local on the target, therefore wlog $\#Y=1$. + Since, in addition, universal openness is a topological condition, we can assume $Y$ to be reduced. Therefore let $Y=\Spec k$ for $k$ a field. + + Let $Y'\to Y$ be arbitrary. Since openness is local on the domain, assume $X=\Spec A$; $Y'=\Spec B$ and therefore $X\times_Y Y'=\Spec A\otimes_kB$. Write $A=\operatorname{colim}_\alpha A_\alpha$ as colimit over the finitely generated subalgebras $A_\alpha\subseteq A$. Then + \[ A\otimes_kB = \operatorname{colim}_\alpha(A_\alpha\otimes_kB)\,. \] + + Let $t\in B$ and denote by $f'$ the base change of $f$. We show that $U=f'(D(t))$ is open in $\Spec B$. Let $t\in A_\alpha\otimes_k B$ for suitable $\alpha$. Applying Corollary \ref{cor:flat+lfp->univ-open} shows that $f'':\Spec A_\alpha\otimes_kB\to\Spec B$ is open. Therefore $U'=f''(D(t))\subseteq\Spec B$ is open, so it suffices to check $U=U'$. + + We have $U\subseteq U''$ by assumption.Let $y\in U'$. It suffices to show + \[ (f')^{-1}(y)\cap D(t)\neq\emptyset\,. \] + But $(f')^{-1}=g^{-1}(W)$ with $g:(f')^{-1}(y)=\Spec(B'\otimes_B\kappa(y)), (f'')^{-1}(y)=\Spec (A_\alpha\otimes_k\kappa(y))$ and $W=(f'')^{-1}(y)\cap D(t)$. Since $\kappa(y)/k$ is flat, we have an injection $A_\alpha\otimes_k\kappa(y)\hookrightarrow A_\alpha\otimes_k\kappa(y)$ and $g$ is dominant. This implies $g^{-1}(W)\neq\emptyset$, since W is open and non empty. +\end{proof} + +\section{Differentials and Smoothness} + +\begin{definition} + $A\to B$, $M\in B \operatorname{-Mod}$. Define + \[ \operatorname{Der}_A(B,M)=\{D\in\\operatorname{Hom}_A(B,M)\mid D(Fg)=fD(g)+gD(f) \quad\forall f,g\in B\} \,.\] + The module of Kähler differentials of $B/A$ is a pair $(\Omega^1_{B/A},d_{B/A})$ with $\Omega^1_{B/A}\in B\operatorname{-Mod},d_{B/A}\in\operatorname{Der}_A(B,\Omega^1_{B/A})$ such that $d_{B/A,\ast}:\operatorname{Hom}_B(\Omega^1_{B/A},M)\xrightarrow{\cong}\operatorname{Der}_A(B,M)$ is an isomorphism. +\end{definition} + + +\begin{lemma} + For $A\to B$, we have + \[\Omega^1_{B/A}\cong \bigoplus_{b\in B} db B / \Big\langle\begin{aligned} + d(bb')=dbdb'+b'db \\ d(b+b')=db+db'\,, \quad da=0 + \end{aligned} \mid b,b'\in B, a\in A\Big\rangle\,.\] + The universal differential $d_{B/A}$ is given by $b\mapsto[db]$. For $I=\operatorname{ker}(B\otimes_AB\to B)$, we have an isomorphism + \[\Omega^1_{B/A}\to I/I^2\,, \quad [db]\mapsto \overline{1\otimes b-b\otimes 1} \,.\] +\end{lemma} + +\begin{proof} + This formal calculation. +\end{proof} + +\begin{bsp} + \begin{enumerate} + \item Let $B=A[T_1,\dots,T_n]$. Then $\Omega^1_{B/A}=\bigoplus_{i=1}^n dT_i B$. + \item Let $L/K$ be an separable extension. Then $\Omega^1_{L/K}=0$. + \end{enumerate} +\end{bsp} + +\begin{lemma} + Let $A', B$ be $A$-algebras and $S\subseteq A$ multiplicatively closed. Then we have $S^{-1}\Omega^1_{B/A}=\Omega^1{S^{-1}B/A}$ and $\Omega^1{B/A}\otimes_B B' = \Omega^1{B'/A'}$. +\end{lemma} + +\begin{lemma} + $f:A\to B, g:B\to C$. Then we have an exact sequence + \[ \Omega^1_{B/A}\otimes_BC\to \Omega^1_{C/A}\to\Omega^1_{C/B}\to 0 \] + of $C$-modules. If $g$ is surjective with kernel $I$, the sequence + \[ I/I^2\to\Omega^1_{B/A}\otimes_BC\to \Omega^1_{C/A}\to 0 \] + is exact. +\end{lemma} + +\begin{bsp} + Let $A$ be a ring and $B=A[T_1,\dots,T_n]/(f_1,\dots,f_n)$. Then $$\Omega^1_{B/A}\cong \bigoplus_{i=1}^ndT_iB/\langle df_i\mid i=1,\dots,n\rangle\,,$$ where $df_i=\sum_k\frac{\partial f_i}{\partial T_k}T_k$. +\end{bsp} + +\begin{definition} + Let $i:Y\hookrightarrow X$ be an immersion $Y\xrightarrow{\text{closed}}U\xrightarrow{\text{open}}X$ and $I$ the associated ideal sheaf. Then define $\omega_{Y/X}=I/I^2$ as an $\mathcal{O}_Y$-module. + + For $f:X\to S$ the module of Kähler differentials is given by + \[\Omega^1_{X/S}:=\omega_{X}/X\times_SX\] + with $\Delta:X\to X\times_SX$. +\end{definition} + +\begin{bem} + $X\to S$ mono implies $\Omega^1_{X/S}=0$. +\end{bem} + +\begin{satz} + \begin{enumerate} + \item For $X\xrightarrow{f}Y\to S$ we have an exact sequence + \[f^\ast\Omega^1_{Y/S}\to\Omega^1_{X/S}\to\Omega^1_{X/Y}\to 0 \,.\] + + \item Given $X\to S \leftarrow Y$, we have an isomorphism + \[ p_1^\ast\Omega^1_{X/S}\oplus p_2^\ast\Omega^1_{Y/S}\cong\Omega^1_{X\times_SY} \,. \] + + \item Given $Z\xrightarrow{\text{closed}}X\to S$, we have + \[\omega_{Z/Y}\to i^\ast \Omega^1_{X/S}\to\Omega^1_{Z/S}\to 0\,.\] + + \item If $X\to S$ is of locally finite type, then $\Omega^1_{X/S}$ is of finite presentation. + \end{enumerate} +\end{satz} + +\begin{satz}[\cite{gw}, 18.64] + Let $k$ be a field, $X/k$ of locally finite type and $x\in X$. + + Then $X$ is smooth in $X$ iff $\Omega^1_{X/S}$ is a free $\mathcal{O}_{X,x}$-module of dimension $\operatorname{dim} X$. +\end{satz} + +\begin{satz} + Let $X'=X\times_S\times S'$ be cartesian. Then there exists a canonical isomorphism + \[ h^\ast \Omega^1_{X/S}\xrightarrow{\cong}\Omega^1_{X/S} \,. \] +\end{satz} + +\begin{proof} + exercise sheet numero sei +\end{proof} + +\begin{satz} + Let $\pi:G\to S$ be a group scheme with unit $e\in G(S)$. Then there are the following isomorphisms of $\mathcal{O}_G$-modules + \[ \Omega^1_{G/S}\cong \pi^\ast e^\ast\Omega^1_{G/S}\cong \pi^\ast \underbrace{\omega_{S/G}}_\text{via $e$} \,. \] +\end{satz} + +\begin{proof} + First, consider the cartesian diagram + \[ \begin{tikzcd} + G\times_S G \ar[r,"m"] \ar[d,"p_i"] & G \ar[d,"\pi"] \\ G \ar[r,"\pi"] & S + \end{tikzcd}\] + for $i=1,2$. It yields + \[ m^\ast \Omega^1_{G/S}\cong\Omega^1_{G\times_SG/G}\cong p_i^\ast \Omega^1_{G/S} \,. \] + Consider also $i=(e\pi,\operatorname{id}_G):G\to G\times_SG$. Then + \[ \Omega^1_{G/S}\cong\operatorname{id}_G^\ast \Omega^1_{G/S}=i^\ast p_2^\ast \Omega^1_{G/S} = i^\ast m^\ast \Omega^1_{G/S} = i^\ast p_1^\ast\Omega^1_{G/S} =(e\pi)^\ast\Omega^1_{G/S}=\pi^\ast e^\ast \Omega^1_{G/S}\,.\] + + Secondly, consider the diagram of sections + \[\begin{tikzcd} + S \ar[d, "e"] \ar[r, "e"] &G \ar[d,"\Delta"] \\ + G \ar[u,"\pi", bend left] \ar[r,"i"] & G\times_SG \,, \ar[u,"p_1",bend left] + \end{tikzcd}\] + where $i=(e\pi,\operatorname{id}_G)$. We deduce $e^\ast\Omega^1_{G/S}\cong\omega_e$ and $\pi^\ast$ yields $\pi^\ast e^\ast \Omega^1_{G/S}\cong\pi^\ast \omega_e$. +\end{proof} + + + + + + + + + + + + + + + + + + + + + +\end{document} \ No newline at end of file