\magnification=\magstep1
\def \bA {{\bf A}}
\def \bP {{\bf P}}
\def \bZ {{\bf Z}}
\def \bG {{\bf G}}
\def \bQ {{\bf Q}}
\def \bO {{\bf O}}
\def \bF {{\bf F}}
\def \bT {{\bf T}}
\def \bP {{\bf P}}


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

\bigskip


\centerline{ Problem Set 6 (Last revised \number\month/\number\day/\number\year)}
\bigskip

\item{11.6}{Using any characterization of dimension, show that if $X \subset \bP^n$
is any quasiprojective variety and $Z\subset \bP^n$ is any hypersurface
not containing an irreducible component of $X$, then
$${\rm dim}(Z\cap X)= {\rm dim}(X)-1.$$}



\itemitem{Proof:}{We will use the transcendence degree definition of
dimension. It will be enough to do this for $X$ irreducible, by writing
the quasiprojective variety as a union of irreducible components. Further, since the dimension of $X$ is the same as the dimension of the closure $\bar X$,
it will be enough to treat projective varieties $X$. Finally, by intersecting with
the standard affine subsets of projective space, we need only prove the result
for affine varieties intersected with hypersurfaces.  Let $f$ be a regular
function on $X$, which is not 0 or constant on $X$.  We first show that the
zero set of $f$ on $X$ has dimension strictly less than $\dim(X)$, by
showing that any proper closed subvariety $Y \subset X$ has $\dim(Y) < \dim(X)$. We have
$K[X] \to K[Y]$ is surjective. If $\dim(Y)=\dim(X)=r$
let $f_1,\ldots f_r$ be algebraically independent elements in the
affine coordinate ring $K[X]$, with algebraically independent images $\bar f_i$ in
$K[Y]$.  Choose $u \not = 0 \in K[X]$ which has image zero in $K[Y]$ (that is,
take a polynomial not vanishing on $X$ which vanishes on $Y$).  Then there
is a polynomial in $n+1$ variables such that $P(u,f_1,\ldots,f_r)=0$,
by the transcendence basis definition of dimension. Since $K[X]$ is a domain,
we may take $P$ irreducible. It is not a multiple of the first variable,
since $u \not = 0$.  But then $P(0,\bar f_1,\ldots,\bar f_r)=0$ is a nontrivial
relation contradicting the independence of $\bar f_i$. Thus a proper closed 
subset of an irreducible variety has strictly smaller dimension.  In particular, 
if $f \in K[X]$ is not constant, by the nullstellensatz its zero set defines
a proper subvariety, which by the above has smaller dimension.  
It only remains to show that the dimension of the zero set of $f$
is at least $\dim(X) -1$. Observe that  given a (possibly reducible) 
variety $X_i$ we can find non constant linear functions that do not
vanish identically on any component of $X_i$ (choose a point in each component,
and take a hyperplane missing these points).  Let $X_1$ be the zero set of
$f$ on $X$, and inductively apply the above to produce $f_2,f_3,\ldots, f_r$
of the same total degree as $f$ (take powers of the linear functions)
and $X_2\subset X_3\subset\cdots\subset X_r$ such that $f_j$ is nonzero
on each component of $X_j$, and $X_{j+1}$ is its zero set in $X_j$.
We will show that $f_2,\ldots f_r$ are algebraically independent on $X_1$,
which will prove that every component of the zero set of a nonconstant
regular function on $X$ has dimension $\dim(X)-1$.  Let $P \in K[T_2,\ldots , T_r]$.
We want to show that $R=P(f_2,\ldots,f_n)$ does not vanish identically
on any component of $X_1$.  For this it is  enough to show that if
$Q\in K[X], R\cdot Q=0$ on $X_1$, then $Q=0$ on $X_1$, for if $R$ vanishes on an irreducible
component take $Q$ to be any nonzero element of $K[X]$ vanishing on all
other components.  By the nullstellensatz the first vanishing condition may be
replaced by $Z(f) \subset Z(R\cdot Q)$  which is equivalent to $f$ divides a power
of $R\cdot Q$, and the second is equivalent to $f$ divides a power of $Q$.
We can reduce to a purely algebraic statement if we observe that projection from a
point in projective space not contained in a variety Z exhibits the
homogeneous coordinate ring of the projection as a subring of the homogeneous
coordinate ring of $Z$, 
and the latter ring is a finite module over the former (this is easy to see for the
projection from $[0,0,\ldots,0,1]$ by observing that since this point
is not in $Z$, there is a homogeneous polynomial in $I(Z)$ which is of the
form $Z_n^d+\ldots$, and thus the homogeneous coordinate ring of $Z$
is generated by $1,Z_n,\ldots,Z_n^{d-1}$ over the homogeneous coordinate
ring of the projection).  Thus projection from a linear space is finite in the
sense that the homogeneous coordinate ring of the variety is a finite
module over the homogeneous coordinate ring of the projection, and hence the same is
true for the coordinate rings of affine varieties which are dense in a projective variety.
Further, since a projective variety is isomorphic to its Veronese image,
if $f_0,\ldots f_s$ are linearly independent forms of the same degree which do not
vanish simultaneously on a variety $Z$ the map $[z]\mapsto [f_0(z),\ldots,f_s(z)]$
is a regular map which is given by projection on the Veronese image,
and thus is finite in the sense above, and preserves dimension.  Hence if we
projectivize our construction of  $X_1,\ldots,X_r$ earlier we see that
(since $\dim(X)=r, \dim(X_i) \leq r-i$) $K[X]$ is a finitely generated
module over $K[T_1,\ldots T_r]$ for some algebraically independent $T_i$.
We are thus reduced to the following Lemma.}

\itemitem{}{Lemma: Let $A\supset K[T_1,\ldots,T_r]$ be a finitely generated
$K[T_1,\ldots,T_r]$ module which is a domain, $x=T_1,y=P(T_2,\ldots,T_r)\not = 0, u \in A$.
If $x$ divides a power of $yu$ in $A$, the $x$ divides a power of $u$.}

\itemitem{}{Proof: Note that by replacing $y,u$ by powers, it is enough to show
that if $x,z \in K[T_1,\ldots,T_r]$ are relatively prime, then $x$ divides
$zv$ in $A$ implies that $x$ divides a power of $v$.  Let $zv=xw$, and
let $F(T)=T^l+b_1T^{l-1}+\ldots + b_l$ be the minimal polynomial of
$w \in A$ considered in the fraction field.  Since $A$ is finite over
$K[t_1,\ldots,T_r]$ there exist monic polynomials with coefficients in 
$K[T_1,\ldots,T_r]$ which w satisfies, and since $A$ is a domain it is
easy to see that in the minimal polynomial $b_i \in K[T_1,\ldots,T_r]$.
The minimal polynomial of $v=xw/z$ is then $$(x/z)^lF((z/x)T)=
T^l+(xb_1/z)T^{l-1}=\ldots +(x^lb_l/z^l)$$
which has coefficients in $K[T_1,\ldots, T_r]$ since $v$ satisfies a 
monic polynomial with coefficients in this ring (by finiteness). Thus
$z^i$ divides $b_i$, since $z,x$ are relatively prime. Hence from
the minmial polynomial we have $x$ divides $v^l$ as desired.}



\medskip

\item{11.9}{Let $\Gamma \subset \bP^n$ be a general collection of d points.
It is known  that $\Gamma$ is a complete intersection if and only if $n=1,d=1,2$ 
or $d=4,n=2$.  Show this for $n=2, d\leq 7$.}

\itemitem{Proof:}{(B. Kennedy)}

\medskip



\item{11.10}{Let $\Gamma \subset \bP^2$ be a general collection of d points. Show that
$\Gamma$ is a set-theoretic complete intersection.}

\itemitem{Proof:}{(see Fact of the day \#10)
}


\medskip

\item{11.11}{Find a homogeneous quadratic polynomial $Q(Z_0,\ldots,Z_3)$
and a homogeneous cubic polynomial $P(Z_0,\ldots,Z_3)$ whose common zero locus
is a rational normal curve $C \subset \bP^3$.}

\itemitem{Proof:}{(K. Dalili)
}

\medskip

\item{11.15}{Let $X, Y$ be irreducible projective varieties, $f:X \to Z$
and
$g:Y\to Z$ surjective maps. Show that the dimension of the fiber product
satisfies 
$${\dim}(X\times_Z Y) \geq \dim(X)+\dim(Y)-\dim(Z).$$
Find an example where strict inequality holds. Show by example that some irreducible components
of $X\times_Z Y$ may have strictly smaller dimension.}


\itemitem{Proof:}{(T. Pham)
}

\medskip

\item{11.16}{a) Show  by example that in general the function $\mu$ is not the
sum of the pullbacks of the functions $\mu_k$ (so that we cannot deduce upper
semicontinuity of $\mu$ directly from the corresponding statement for the maps $\pi_k$).\hfill\break
b)Justify the third reduction step: that in order to show that $\mu$ is upper semicontinous in general it is enough to show that for
any map $\pi$ there is an open subset $U\subset X$ where it attains its minimum.}

\itemitem{Proof:}{(a) Consider $X=Z(x-yz,yw-z)\subset \bA^4$ and the projection
$\pi_1:X \to \bA^3$ given by $(x,y,z,w)\mapsto(x,y,z)$. The image $\pi_1(X)$
consists of $(yz,y,z)$ such that if $y=0$ then $z=0$. The closure $X_1$ of
the image is given by $x-yz=0$. Let $\pi_2$ be the projection $(x,y,z) \mapsto (x,y)$.
Then the fiber of $\pi$ over $(0,0)$ is $X_{(0,0)}=\{(0,0,0,w)\}$ so that
$\mu((0,0,0,1))=1$. The fiber of $\pi_2$ over $(0,0)$ is $\{(0,0,z)\}$,
which is larger than the closure of the image of the fiber $X_{(0,0)}$.
Then $\mu_1(0,0,0,1)+\mu(0,0,0)=2 > \mu(0,0,0,1)$.  Thus $\mu$ is not the
sum of the pullbacks of $\mu_k$ so we don't get upper semicontinuity directly
from the simple case of projection to affine space of dimension one less}
\itemitem{}{(b)
Suppose that it is proved that there is a nonempty open subset $U$ of $X$
where $\mu$ takes on its minimum $\mu_0(X)$. Now apply the result to
the variety $X-U$, and continue to produce a decreasing sequence of
closed subsets $X \supset X_1 \supset X_2 \supset \cdots$. Since algebraic
varieties are Noetherian topological spaces, this sequence must terminate after
a finite number of steps with the empty set, that is $\mu$ takes
on constant value $\mu_n$ on some $X_n$.  On each set $X_i$ we have that the
minimum value $\mu_i$  of $\mu$ on $X_i$ satisfies $\mu_0\leq\mu_1\leq\ldots
\leq \mu_n$. Then $\{p \in X:\mu(p)\geq m\}= \cup X_i$ taken over all
$i$ such that $\mu_i \geq m$, so it is a closed set.
}

\medskip

\item{11.19}{Consider the variety $\Phi$ of lines meeting a given
line $l_0$. Let $l_0^*\subset {\bP^3}^*$ be the locus of planes $H \subset \bP^3$ containing $l_0$ and consider the incidence corespondence
$$\Omega=\{(l,H):l \subset H\}\subset \Phi \times l_0^*.$$
Show that the fibers of $\Omega$ over $l_0^*$ are isomorphic to $\bP^2$, while the map
$\pi_1:\Omega \to \Phi$ is one-to-one except over $l_0$ where the
fiber is $\bP^1$.  Show that the incidence correspondences $\Psi$ and
$\Omega$ are isomorphic (they are each isomorphic to the scroll $X_{2,2,1}$),
but not via an isomorphism commuting with the maps to $\Phi$.} 


\itemitem{Proof:}{(S. Ganguly)
}

\medskip

\item{11.20}{Let $X\subset \bP^n$ be an irreducible non-degenerate $k$-dimensional variety.
For $l<n-k$ find the dimension of the closure of the locus of $l$-planes
meeting $X$ in at least two points. Show by example that no analogous formula 
exists if we replace two by three, even if we require $l \geq 2$. }


\itemitem{Proof:}{()
}


\medskip

\item{11.21}{Let $X\subset \bP^n$ be an irreducible non-degenerate $k$-dimensional 
variety. Show that the plane spanned by $n-k+1$ general points of $X$ meets $X$ in
only finitely many points and use this to compute the dimension of the closure of the
locus of $l$-planes spanned by their intersections with $X$ when $l \leq n-k$.
}

\itemitem{Proof:}{Consider the rational map from the product $X^l$ of l copies of $X$
to the Grassmannian $\bG(l,n)$ which is defined on the open set of points 
$(x_1,\ldots,x_l)$ such that $x_1\wedge\ldots\wedge x_l \not = 0$ by taking the
span of the l points.  If the fibers of this map are generally finite, then
the dimension of the closure of the locus of $l$-planes spanned by their intersections with $X$
will be $l\dim(X)$.}

\medskip

\item{11.23}{Counting dimensions, show that the variety ${\cal S}(X) \subset 
\bG(1,n)$ of secant lines to a variety $X \subset \bP^n$ of dimension $k<n-1$ is a proper 
subvariety of the variety ${\cal C}_1(X)$ of incident lines. Use this to deduce
that (in characteristic 0) the general projection $\pi_\Lambda:X \to \bP^{k+1}$ from
an
$(n-k-1)$-plane $\Lambda\subset \bP^n$ is
birational onto its image.  Use this to give another proof that every variety is birational to
a hypersurface.}

\itemitem{Proof:}{The variety of secant lines is the image of a rational
map from $X\times X$ with finite fibers so it is irreducible of dimension
$2k$.  The variety of incident lines is the projection of
$\Sigma=\{(\ell,p)|p \in \ell\}\subset \bG(1,n)\times X$ onto $\bG(1,n)$,
and this is a generically finite to one map since a general line hits a proper
subvariety in finitely many points.
The projection of $\Sigma$ onto X is surjective, and fibers are all irreducible
of dimension $n-1$ (the fiber is the set of all lines through a point),
so that $\Sigma$ is irreducible of dimension $k+n-1$.  Clearly 
${\cal S}(X)\subset {\cal C}_1(X)$ is a closed subset of dimension $2k$ in
an irreducible variety of dimension $k+n-1$, so if $k<n-1$ it must be
a proper subvariety.  Thus the general incident line  is not a secant line,
so it can hit $X$ only one time.  Hence if $\dim(X)<n-1$ projection
from any point $p \in \bP^n$ not on $X$ is generically one to one.  This dominant map
corresponds to an extension of function fields. In characteristic zero the degree of the function field extension
is the number of points in the general fiber, which is 1. So $X$ is birational to
$\pi_p(X)$.  We may repeat this to obtain that $X$ is birational to
its projection on $\bP^{k+1}$, which must be a hyperplane since it is
irreducible of
dimension one less than the surrounding projective space.  Hence (if the characteristic
of K is 0) every variety is birational to its projection on a hypersurface.}


\medskip

\item{11.25}{Show that if $X\subset \bP^n$ is an irreducible curve then the chordal variety
$S(X)$ is three-dimensional unless $X$ is contained in a plane.}


\itemitem{Proof:}{(J. Burdges)
}


\medskip

\item{11.26}{Let $\Sigma=\sigma(\bP^2\times\bP^2) \subset \bP^8$ be the Segre
 variety.  Show that the chordal variety of $\Sigma$ has dimension 7, so that $\Sigma$ has deficiency $\delta(\Sigma)=4$.}


\itemitem{Proof:}{()
}

\medskip

\item{11.28}{Let $X\subset \bP^n$ be an irreducible nondegenerate variety with deficient
secant variety.}
\item{}{1) Show that the cone $\overline X=\overline{p,X}\subset \bP^{n+1}$ over $X$ has deficiency $\delta(\bar X)=\delta(X)+1$.}
\item{}{ii) Show that the general hyperplane section $Y=X\cap H$ of $X$ has deficiency
$\delta(Y)=\delta(X)-1$.}


\itemitem{Proof:}{()
}

\medskip

\item{11.29}{Let $M$ be the projective space of $m\times n$ matrices, and let
$M_k\subset M$ be the subvariety of matrices of rank at most $k$. Take $k$
so that $2k < \min(m,n)$. Show that the chordal variety $S(M_k)$ is equal
to the subvariety $M_{2k} \subset M$ of matrices of rank at most $2k$.  Indeed. show that for a map $A:K^m \to K^n$
of rank $2k$ and any pair of complementary $k$-dimensiaaonal subspaces $\Lambda,
\Xi\subset {\rm \ image}M(A)$ the composition of A with projections of $\Lambda$ and $\Xi$ give an expression of
A as a sum of two matrices of rank k. Deduce that in the case $\bP^n=M, X=M_k$
the fiber of the incidence correspondence $\Sigma$ in 11.22 over a general point $S(X)$ has dimension $2k^2$ and hence
that $\dim(S(M_k)) \leq 2 \dim(M_k)+1-2k^2$.  }


\itemitem{Proof:}{()
}

\medskip

\item{11.31}{Let $X\subset \bP^n$ be an irreducible nondegenerate variety of dimension k. use 11.21
to deduce that for any $l \leq n-k$ the general fiber of the secant plane map
$$s_l:X^{l+1} \to \bG(l,n)$$
is finite.}

\itemitem{Proof:}{()
}

\medskip

\item{11.33}{Give the proof of Proposition 11.32 in case $2l=n$ either
by bounding the dimension of the family  of secant $l$-planes containing a general point of
$S_l(X)$ or by using the ideas in the  cases of 11.32 that were proved and 
arguing
that $S_l(X)$ strictly contains $S_{l-1}(X)$ and thus has higher dimension.
Do the case $2l > n$ similarly, or deduce it from the above.}

\itemitem{Proof:}{()
}


\medskip

\item{11.35}{i) Verify that in characteristic 0 if a general polynomial of degree $n$
is the
sum of $d$ $n$th powers of linear polynomials, with $2d-1=n$, then it is uniuely expressible in this way.\hfill\break
ii) What is the smallest number $m$ such that every polynomial of degree $n$ is expressible
as a sum of $m$ $n$th powers?}

\itemitem{Proof:}{}
\medskip

\item{11.38}{Show that any two closed subvarieties $X,Y \subset \bP^n$
whose dimensions sum to at least n must intersect.}
\itemitem{Proof:}{Let $\Lambda_0,\Lambda_1$ be the n-planes in $\bP^{2n+1}$ given
by $z_0=\cdots=z_n=0$ and $z_{n+1}=\cdots=z_{2n+1}=0$ respectively.
Let $\tilde X\subset \Lambda_0,\tilde Y\subset \Lambda_1$ be isomorphic to $X,Y$ respectively
(that is defined by the equations giving $X$ and the plane $\Lambda_0$, and similarly
for $Y$).  Consider the $n$-plane $L$ given by $z_0=z_{n+1},\ldots, z_n=z_{2n+1}$.
Let $\tilde J(X,Y)$ be the join, which clearly has dimension $\dim(X)+\dim(Y)+1$
(consider the rational map $\tilde X \times \tilde Y --\to \bG(1,2n+1)$ which has
image of dimension $\dim X +\dim Y$, and the map of this image to $\tilde J$ which
has fibers of dimension 1). Thus $\tilde J(X,Y)\cap L$  is nonempty since the sum of
dimensions is at least $2n$.  Let $q$ be in this intersection. It must be of the form
$(q',q')$ since it is in $L$ , and since it is on a line joining a point of $\tilde X$
to $\tilde Y$ there are points $(p_1;0) \in \tilde X, (0,p_2) \in \tilde Y$ such
that $(q',q')=\lambda (p_1,0)+\mu (0,p_2)$ . This  implies that $p_1=p_2 \in X \cap Y \subset \bP^n$.
In fact, using the map $X\times Y \to \bP^n\times \bP^n$ and considering the
diagonal in the target space, which is defined by $n$ equations, we see that the
inverse image of the diagonal is just $X\cap Y$ which is defined as the zero set
of $n$ functions in $X \times Y$. Thus $\dim(X\cap Y)\geq \dim(X)+\dim(Y) -n$.}

\medskip

\item{11.39}{Show that if $m > n$ there does not exist a nonconstant regular map
$\phi:\bP^m\to \bP^n$ by using 11.38.}
\itemitem{Proof:}{The image $\phi(\bP^m) $ is a closed subset of $\bP^n$.
Choose a point $p \in \phi(\bP^m)$ with $\dim(\phi^{-1}(p))$ minimal and a
 hyperplane $H \subset \bP^n$ which does not contain $p$ and does not 
contain $\phi(\bP^m)$ but does contain points $q$ with minimal fiber dimension
(Since $\phi $ is not constant there are points in the image besides $p$. A general hyperplane will do, since those containing p or
the image form a closed subset and the set of points with minimal fiber dimension is open).
Then $\dim(\phi^{-1}(p))=m-\dim(\phi(\bP^m))\geq m-n$ and the dimension of
$\phi^{-1}(H)$ is $\dim(H\cap \phi(\bP^m))+m -\dim(\phi(\bP^m))=
m-1$.  Since the sum of these two dimensions is at least $m-n+m-1 \geq m$
the varieties $\phi^{-1}(p)$ and $\phi^{-1}(H)$ intersect, contradicting that
we chose $p \not \in H$.

}
\medskip

\item{11.41}{Find the dimension of the flag variety $\bF(a_1,\ldots,a_n)$ in general.}

\itemitem{Proof:}{()
}


\medskip

\item{11.43}{Find the dimension     of the more general incidence correspondence
$$\Psi=\{(\Gamma,\Lambda):\dim(\Gamma \cap \Lambda) \geq m\} \subset \bG(k,n)\times \bG(l,n).$$
}

\itemitem{Proof:}{()
}


\medskip

\item{11.44}{Let $X \subset \bP^n$ be an irreducible $k$-dimensional 
variety. Find the dimension
of the universal $(n-l)$-plane section $\Omega^{(l)}(X)$ 
and show that it is irreducible.}

\itemitem{Proof:}{Consider the universal $(n-l)$-plane
section $\Omega^{(l)}(X)=\{(\Lambda,x)| x \in \Lambda\} \subset \bG(n-l,n)\times X$.
The projection to $X$ has fibers consisting of all $(n-l)$-planes 
in $\bP^n$ containing
a fixed point, which is isomorphic to the Grassmanian $\bG(n-l-1,n-1)$.
Thus $\Omega^{(l)}(X)$ is irreducible and has dimension $k+ (n-l)l$. 


\medskip

\item{12.6}{i)Verify that the surface S described in the middle of page 153 is
indeed a cubic surface in $\bP^3$.\hfill\break
ii) Verify that a birational map of surfaces $X \to S$ can fail to be birational
on only finitely many curves in $X$.}

\medskip

\item{12.7}{Consider general values of $n,d,k$ such that 
$\phi(n,d,k)=(k+1)(n-k)- {k+d \choose d}$ is zero, that is for which we expect
the general hypersurface of degree $d$ in $\bP^n$ to contain a finite number of
$k$-planes.  In cases with $d \geq 3$ find a hypersurface $X_0 \subset \bP^n$ of degree $d$
with Fano variety $F_k(X)$ containing an isolated point $\Lambda_0$ and deduce
that every such hypersurface does indeed contain a $k-plane$.}
\medskip

\item{12.11}{Find again the dimension of the parameter space of rational
normal curves $C \subset \bP^n$ using the construction of the rational normal
scroll in 9.15.}

\medskip

\item{12.13}{Find the dimension of the space of plane conic curves $C \subset \bP^3$
and show that for $d \geq 4$ not every surface of degree $d$ in $\bP^3$ contains a
conic. what do you expect hapeens when $d=3$? Show directly that it does.}
\medskip


\item{12.16}{In case both $a, b$ are $\leq 2$ set up an incidence correspondence
between curves and quadrics to compute the dimension of
${\cal V}_{a,b}$.  In particular, check that you get the same answer as before for
the space ${\cal V}_{2,1}$ of twisted cubics.}

\medskip



\item{12.17}{Observe that a curve of type $1,d-1)$ on a smooth quadric
$Q \subset \bP^3$ is a rational curve of degree $d$, that is the image of $\bP^1$ under a map
to $\bP^3$ given by a quadruple of polynomials of degree $d$ on $\bP^1$, or
equaivalently a projection of the rational normal curve in $\bP^d$. Use a count of parameters
to show that for $d \geq 5$ the general rational curve of degree $d$ is not a 
curve of type $(1,d-1)$ on a quadric.}

\medskip



\bye


































