Lecture21.tex

 \stepcounter{lecture}
 \setcounter{lecture}{21}
 \sektion{Lecture 21}

Recall the previous definition.

\begin{proposition} If $f_1,\dots, f_r$ is a regular sequence for
an $A$-module $M$, then
 \[
 h_i(K_{\cdot}(f_1,\dots,f_r,M)) = \left\{\begin{array}{ll}
 M/(f_1,\dots,f_r)M & \text{if } i=0\\
 0 & \text{if } i>0
 \end{array} \right.
 \]
 \end{proposition}
 ($i=0$ is trivial ... you don't even need the regular sequence.)
 Thus, if $M$ is free, the Kozul complex is a free resolution of
 $M/(f_1,\dots, f_r)M$.

 \begin{theorem}
 Let $X$ be a locally complete intersection closed subscheme of
 $P=\P^N_k$, with ideal sheaf $\I$.  Then
 \[
     \omega_X^{\circ} = \omega_P \otimes
     \wedge^r(\I/\I^2)\check{}.
 \]
 In particular, the dualizing sheaf is a line sheaf on $X$.
 \end{theorem}
 \begin{proof}
 in the works
 \end{proof}

 \begin{remark} This works for any field.
 \end{remark}

 \underline{Next:} Compare $\omega_X^{\circ}$ with $\omega_X$ when
 $X$ is a non-singular variety.

 We prove half of a theorem from chapter II:
 \begin{theorem}[II.8.17] Let $X$ be a non-singular variety over
 an algebraically closed field $k$.  Let $Y\subseteq X$ be an
 irreducible closed subscheme, and let $\I$ be its sheaf of
 ideals.  Then $Y$ is non-singular if and only if
 \begin{itemize}
  \item[(1)] $\Omega_{X/Y}$ is locally free, and
  \item[(2)] the second exact sequence is a short exact sequence
  \[
    0\to \I/\I^2 \to \Omega_{X/k}\otimes \O_Y\to \Omega{Y/k} \to
    0.
  \]
  \end{itemize}
  In this case, $\I/\I^2$ is locally free (on $Y$) of rank
  $r=\codim Y$ and $Y$ is a locally complete intersection in $X$.
  \end{theorem}
  \begin{proof}[Half Proof]
  in the works
  \end{proof}