\magnification=\magstep1
\def \bA {{\bf A}}
\def \bP {{\bf P}}
\def \bZ {{\bf Z}}
\def \bG {{\bf G}}
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\def \bF {{\bf F}}
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\def \bP {{\bf P}}


{Mathematics 535 \hfill Algebraic Geometry \hfill Fall 2002}

\bigskip


\centerline{ Problem Set 8 (Last revised \number\month/\number\day/\number\year)}
\bigskip
\item{18.2}{Assuming that the characteristic of $K$ is 0, use exercises 7.16 and 11.44 to
show that part (iii) of Definition 18.1 is well-defined, and use this to show that
parts (i) and (ii) are also well-defined.}

\itemitem{Proof:}{In 11.44 we saw that the dimension of the universal $(n-k)$-plane
section $\Omega^k$ of an irreducible variety $X$ is $(n-k)k+k$.  Further, the map
from $\Omega^k$ to the Grassmannian $\bG(n-k,n)$ which takes a pair consisting of a 
$n-k$-plane and a point of $X$ on it to the plane is clearly onto (every $n-k$ plane
intersects a $k$-dimensional variety in $\bP^n$ by one of our characterizations of
dimension). Since $\Omega^k$ and $\bG(n-k,n)$ have the same dimension, the general fiber dimension
is zero, and by 7.16 (in characteristic 0) the general fiber always has the same number of points.
Thus (iii) is well defined. Now let $\Gamma $ be a general $(n-k-2)$-plane in $\bP^n$.
The projection $\pi_\Lambda: X \to \bP^{k+1}$ has image which is a dimension k
irreducible variety in $\bP^{k+1}$, so it is a hypersurface.  The general
$n-k$-plane containing $\Gamma$ hits $X$ in deg(X) points, and after projection 
this gives a line in $\bP^{k+1}$ hitting the hypersurface in deg(X) points. Further,
the general line in $\bP^{k+1}$ is the projection of a general $n-k$-plane in 
$\bP^n$ under $\pi_\Lambda$. Hence the degree of the hyperplane is the same as the
number of points determined in (iii) of the definition. So (ii) is well defined
and agrees with (iii). Similarly the general $(n-k-1)$-plane $\Lambda$ in $\bP^n$
is spanned by a general $(n-k-2)$-plane $\Gamma$ and a general point $P$ in $\bP^n$.
The map $\pi_\Lambda$ factors as $\pi_\Gamma$ followed by $\pi_P$. Computing the
fibers in these maps gives the degree of $\pi_\Lambda$.  We have seen that the general 
projection to a hyperplane is birational, while the projection of a hypersurface
of degree d from a point is clearly d to 1. Thus the degree of $\pi_\Lambda$
for general $\Lambda$ is well defined , and equal to the degree defined in (iii)
or (i).}

\medskip

\item{18.7}{Let $X \subset \bP^n$ be any subvariety of pure dimension of degree 1. 
Show that $X$ is a linear subspace of $\bP^n$.}

\itemitem{Proof:}{We can use 18.5 to conclude that the intersection of
X with any hyperplane not containing $X$ has degree 1. Further, since
the degree of a reducible variety is the sum of degrees of its components,
$X$ is irredcuible, as well as its intersection with a hyperplane.  Suppose
$X$ has dimension 1 and is not linear, so that there are point $p,q,r\in X$
not on a line. Choose a hyperplane $H$ containing $p,q$ but not $r$.  Then
$X\cap H$ has degree 1, hence is a finite set, but contains $p,q$. This contradiction
shows that $X$ is linear if it has degree 1 and is pure of dimension 1.  By induction
assume that all pure varieties of dimension at most $k$ which have degree 1 are linear.
Let $X$ have dimension $k+1$ and degree 1 be a proper variety in $\bP^n$. 
Then given any two points $p,q$ in $X$ we may find a hyperplane containing $p,q$ and
not containing some point of $X$. Then $X \cap H$ has dimension $k$, and degree 1
by the generalized Bezout formula, so by induction it is linear. In particular,
the line from $p$ to $q$ is contained in $X$. Thus $X$ contains all lines joining any
two of its points, so it is linear.}

%This is clear if $X$ is a hypersurface. If not, 
%choose a point p not in $X$ and consider the cone $\overline{pX}$.  It has degree 1 as well,
%so that by induction on the codimension it is linear.  Thus if $x_1,x_2$ are in
%$X$ the line $\overline{x_1x_2}$ is in $X$ or contians $p$.
%
%there is a projection
%of $X$ which preserves degree and has image of codimension one less, so by induction
%the projection is linear. }
%
\medskip

\item{18.11}{Give a proof of the nondegeneracy of the general hyperplane
section of the variety $X$ without invoking the notion of degree or Bezout's theorem
as follows. First show that if the general hyperplane section of $X$ spans a $k$-plane we have a rational map $\phi:\bP^{n*}--\to \bG(k,n)$ defined by sending a general hyperplane
$H\in\bP^{n*}$  to the span of $H\cap X$.  Next use the fact
that the universal hyperplane section of $X$ is irreducible to deduce that
for any $H\in\bP^{n*} $ and any point $\Lambda \in \phi(H)$ the hyperplane section
$H\cap X$ lies on the $k$-plane $\Lambda$.  It follows that
if the general hyperplane section is degenerate then all are. But any $n$
independent points of $X$ will span a hyperplane $H$ with $H \cap X$ nondegenerate.}

\medskip

\item{18.14}{Let $Y\subset \bP^n$ be a variety of dimension $k$
and degree e, and $Z=\nu_d(Y)$ its image in $\bP^N$ under
the Veronese map.
Use the various degree computing techniques to compute the degree of $Z$.
}


\medskip


\item{18.18}{Let K have characteristic 0. Let $X,Y\subset \bP^n$ be disjoint
subvarieties of dimensions $k$ and $l$, and $\Gamma \subset \bP^n$ a general
plane of dimension $n-k-l-1$. Show that the projections $\pi_\Lambda(X)$
and $\pi_\Gamma(X)$ intersect transversely.}

\medskip

\item{18.21}{Show that a general tangent line to $X$ at a general point $p\in L$
has contact of order 2 with $X$ at $p$.}
\medskip



\item{18.22}{Show by the same argument that a smooth cubic hypersurface
in $\bP^n$ is unirational for all $n \geq 3$.}

\medskip


\item{18.23}{Show that for any fixed $d$ there exists an $n_0=n_0(d)$ such that
for all $n \geq n_0$ a smooth hypersurface of degree $d$ in $\bP^n$ is unirational.}

\medskip




\bye


















