\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}