


\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 5 (Last revised \number\month/\number\day/\number\year)}
\bigskip
\item{7.1}{Show that the definition of the function field of an irreducible 
affine or projective variety is independent of the open affine set chosen
to form the fraction field. Show that for a projective variety that the function
field is the degree zero elements of the function field of the homogeneous coordinate ring.}

\itemitem{Proof:}{It is enough to show that if $U' \subset U \subset X$, are affine open
subsets of an irreducible
affine variety X, then the surjective map $\psi:K[U] \to K[U']$ induces an isomorphism of fraction
fields.  Suppose that $f/g \in K(U')$. Then there are elements $F,G \in K[x_1,
\ldots, x_n]/I(U)$ such that the image of $F,G$ modulo $I(U')$ is $f,g$.
Hence the map on fraction fields is surjective.  Further, if
$f/g \not = 0$ in the fraction field $K(U')$ it is not zero in $K(U)$ since
$U' \subset U$ is dense.  Thus the field of fractions is independent of the dense open
subset chosen.  For projective varieties X, let $S(X)^0$ be the degree zero elements
in the fraction field of the homogeneous coordinate ring.  There is  a
map $S(X)^0 \to K(X \cap U_0)$ obtained by restricting to the open subset where $X_i \not =0$.
Given $G(X_1,\ldots,X_n)/H(X_1,\ldots,X_N) \in K(X \cap U_0)$, consider 
the image of $X_0^kG(X_1/X_0,\ldots,X_n/X_0),X_0^lH(X_1/X_0,\ldots,X_n/X_0)$
in the homogeneous coordinate ring. With proper choice of k,l their ratio is in $S(X)^0$ and maps to
$G(X_1,\ldots,X_n)/H(X_1,\ldots,X_N) \in K(X \cap U_0)$. Thus the fraction fields
are isomorphic.




\item{7.5}{Show that the the graph $\Gamma_\phi$ of a rational
map $\phi$ is independent of which open set U is utilized in
the definition.  Show that the image of a rational map $\phi$ is the
closure
of the set $\phi(U)$ for any dense open subset $U \subset X$ where it is regular.}


\itemitem{Proof:}
{Let $\phi: X --\to \bP^n$ be a rational map. Suppose that $U'\subset U$ are open
dense subsets of $X$ where $\phi$ is regular.  Since the graph of a regular
map $\phi:U \to \bP^n$ is closed in $U\times \bP^n$ there is a closed set
$W \subset X\times \bP^n$ such that $W \cap (U\times \bP^n)= \{(u,\phi(u))|u \in U\}$.
This shows that the closure of $\{(u,\phi(u))|u \in U\}$ in $X\times \bP^n$
has 1-element fibers over
$u\in U$. Consider the closure of $\{(u',\phi(u'))|u' \in U'\}$ in $U \times \bP^n$.
Its projection to U is a closed set containing the dense set $U'$, so for each 
$u \in U$ the closure contains a point with first coordinate $u$. By the remark on fibers,
the point must be $(u,\phi(u))$, so that the graph computed
with $U'$ contains the graph computed with $U$, hence they are equal. Thus given 
open sets $U_1,U_2$ dense in X, the graph we compute is the same as that associated
to $U_1\cap U_2$, so it is independent of the choice of open dense subset.
}
\itemitem{}{\indent For the last statement of the problem note that
for $U \subset X$ open and $\phi $ regular on $U$ we have $\phi(U)$ contained
in the image of $\phi$, hence its closure is.  Further 
$\pi_2^{-1}(\overline{\phi(U)})$ is closed and contains 
all points of form $(u,\phi(u))$ for $u \in U$. Thus it contains the graph of
$\phi$, so that the image of $\phi$ is the closure of $\phi(U)$.}
\medskip

\item{7.6}{Show that if $\phi:X -- \to \bP^n$ is a rational map and
$Z \subset X$ is a subvariety such that the restriction $\eta$ of $\phi$ to
Z is defined , then the graph $\Gamma_\eta \subset Z \times \bP^n$
is contained in the set $\pi_1^{-1}(Z)=\Gamma_\phi\cap (Z\times\bP^n)
\subset X \times \bP^n$  but may not equal it. Thus the proper transform of Z can be 
properly contained in the total transform of Z.}

\itemitem{Proof:}{By definition $\Gamma_\eta$ is the closure of $\{(u,\eta(u))\}$
in $Z \times \bP^n$ for $u$ in some open set in $\bP^n$ intersected with Z.  
This is clearly contained in $Z \times \bP^n$ and also
in $\Gamma_\phi$ (we can take the open in Z to be the intersection
of Z and an open in X and $\eta$ to be the restriction of $\phi$ to the smaller open).
For an example where it is strictly larger, consider $X =\bP^3$ and let
Z be the plane $X_0=0$ and let $\phi:\bP^3 --\to \bP^2$ the projection
$[x_0,x_1,x_2,x_3] \to [x_0,x_1,x_2]$. Then $\Gamma_\phi$ is the
zero set of $x_0y_1-x_1y_0,x_0y_2-x_2y_0,x_1y_2-x_2y_1$ in $\bP^3 \times \bP^2$,
the image of Z is $\bP^2$ (in fact  any set containing [0,0,0,1] has image $\bP^2$),
while $\Gamma_\eta$ is the zero set of $x_1y_2-x_2y_1,x_0,y_0$ in $Z \times\bP^n
\subset \bP^3 \times \bP^2$. Note the image of $\eta$ is $\{[0,*,*]\}$, so the proper transform
of Z is just $\bP^1$, while its image is $\bP^2$.
\medskip


\item{7.7}{Show that the image of a rational normal curve $C \subset \bP^n$
under projection from a point $p \in C$ is a rational normal curve in $\bP^{n-1}$.
Is this true if we project from a point not in C?}


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

\item{7.8}{Let $X \subset \bP^m$ be an irreducible quasiprojective variety.
Assume that the characteristic of K is 0. Show that a map of sets
$\phi:X \to \bP^n$ is rational if and only if its graph is closed in
$X \times \bP^n$.  Alternately, a subvariety $\Gamma \subset X\times \bP^n$
that meets the general fiber $\{p\}\times \bP^n$ in one point determines a rational mapping
from $X --\to \bP^n$.}

\itemitem{Proof:}{}
\medskip

\item{7.9}{Let $\phi:X --\to\bP^n$ be a rational map with graph
$\Gamma_\phi$.  Show that there is a unique
maximal
open subset $U \subset X$ such that projection on the 
first fiber induces an isomorphism from
$\pi_1^{-1}(U)\subset \Gamma_\phi$ to U.  Show that there is a pair $(U,f)$
in the equivalence class $\phi$ such that for all other $(V,g)$ in
the class, $V \subset U$.}

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

\item{7.10}{Show that two varieties $X,Y$ are birational if and only if
they have isomorphic function fields if and only if they contain nonempty
dense open subsets isomorphic to one another.}

\itemitem{Proof:}{Let (U,f) and (V,g) determine rational maps $\phi: X --\to Y,
\psi:Y--\to X$ which have composition the identity map. Then the open set
$f^{-1}(V)\cap U$ is isomorphic to $V \cap f(U)$ via $f$ and its inverse $g$.
So birational varieties have isomorphic open subsets. The converse is clear
from the definition of a rational map as a regular map on an open dense
subset.  Clearly then birational varieties have isomorphic function fields,
since the function field depends only on an open dense subset.  Conversely,
any isomorphism $K(X) \to K(Y)$ corresponds to a rational map $X --\to Y$
which clearly is birational.}


\medskip

\item{7.12}{Show that projection from $p=[0,0,1]$ in $\bP^2$
gives a birational isomorphism of the curve $C$ given by the
zeros of $ZY^2-X^3-X^2Z$ with $\bP^1$.}
\itemitem{Proof:}{The rational map $\gamma:\bP^1 --\to C$ can be given by
sending $[x_0,x_1] $ to $[x_0^3+x_0x_1^2,x_1^3+x_0^2x_1,x_0^3]$ which is
the analytic formula determining the intersection of a line of slope $x_1/x_0$
through the origin with the affine curve $y^2=x^3+x^2$.}
\medskip

\item{7.13}{Show that for all m,n the product $\bP^m\times \bP^n$ is 
rational.  Give an explicit birational isomorphism of 
$\bP^m\times \bP^n$ and $\bP^{m+n}$ (see also 7.28).}

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

\item{7.14}{Let $Q \subset \bP^n$ be a quadric hypersurface and
let $p\in Q$ be any point not on the vertex of Q.  Show that projection
from p is a birational isomorphism of Q with $\bP^{n-1}$ (see also 7.27).}

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

\medskip
\item{7.19}{(a)Show that the construction of $Bl_Y(X)$ is local, that is
that for X a variety, Y a subvariety of X and $U\subset X$ an affine open set the inverse image of
U in $Bl_Y(X)$ is $Bl_{Y\cap U}(U)$. (b) Show that the construction is
independent of choice of generators for the ideal of Y.}

\medskip

\item{7.20}{Verify the naturality property of blowups and use it to
show that the construction of $BL_N(M)$ makes sense. (b) Show that if
a complex manifold is a subvariety of ${\bf C}^n=\bA_{\bf C}^n$ the two 
constructions of blow-ups $BL_N(M)$ coincide.}

\medskip

\item{7.23}{Show that the proper transforms in $\Gamma$ of the two families 
of lines on Q are the proper transforms of the families
of lines in $\bP^2$ through q and r.}

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

\item{7.25}{Consider the image $X \subset \bP^4$ of  the rational map
 $\phi([Z_0,Z_1,Z_2])=$ $[Z_0^2,Z_1^2,Z_0Z_1,Z_1Z_2]$, the so called
cubic scroll.  Let $\Gamma \subset
\bP^2\times \bP^1$ be the graph of projection from $q=[0,0,1]$.  Show that
the image of the exceptional divisor $E\subset \Gamma$ is a line $L$ in $\bP^4$.
Show that the images of proper transforms in $\Gamma$ of lines in $\bP^2$ through
q are lines $M_\lambda$ in $\bP^4$ which meet L once.  Finally, show that the
images under $\phi$ of lines
in $\bP^2$ not passing through q are conics in $\bP^4$ disjoint from L, and
meeting each $M_\lambda$ in one point.
 }

\itemitem{Proof:}{The graph $\Gamma \subset \bP^2 \times \bP^1$ is described
by $\{[z_0,z_1,z_2]\times[w_0,w_1]| z_0w_1=z_1w_0\}$ and the exceptional divisor
E is given by the set $\{[0,0,1]\times[w_0,w_1]\} \simeq \bP^1$.
The image under the Segre imbedding is isomorphic to $X$, in fact the
map from $\Gamma$ to $X$ is given by $[z_0,z_1,z_2]\times [w_0,w_1]$
goes to $[z_0w_0,z_1w_1,z_1w_0,z_2w_0,z_2w_1]$.  Hence the image of
E is the line $L:\{[0,0,0,a,b]\}$.  Let $\ell_{[a,b]}:bx_0-ax_1=0$ be
a line through $q=[0,0,1]$ in $\bP^2$.  Then the total transform of
$\ell_{[a,b]}$ in $\Gamma$ is $\{[z_0,z_1,z_2]\times [w_0,w_1]|
z_0w_1=z_1w_0,bz_0=az_1,bw_0=aw_1\}$. This is the union of the exceptional
divisor E and the proper transform $\{[\mu a, \mu b,\lambda]\times [\mu a, \mu b]\}$.
  The image of this proper transform under the
map to $\bP^4$ is seen to be $\{[\mu^2a^2,\mu^2b^2,\mu^2ab,\lambda \mu a,\lambda\mu b]\}$
which is just the line $M_{[a,b]}:av_3-bv_4=0$ in $\bP^4$, which intersects $L$
only at the point $[0,0,0,b,a]$.  The lines in $\bP^2$ missing $[0,0,1]$
are given by equations $Az_0+Bz_1+Cz_2=0$ with $C \not = 0$.  The total transform of
this in $\Gamma$ is just the set $\{[z_0,z_1,z_2]\times[w_0,w_1]|
Az_0+Bz_1+Cz_2=0,z_0w_1=z_1w_0\}$. Since $(Az_0+Bz_1+Cz_2)w_0=(Az_0+Bz_1+Cz_2)w_0=0$
The image of this in $\bP^4$ lies in the plane $Av_0+Bv_2+Cv_3=0=Av_2+Bv_1+Cv_4$
which clearly misses the line L.  Further the image of this total
transform is clearly a rational normal curve in this plane, 
that is a plane conic, which meets the line $M_{[a,b]}$ only at the image
of the point where $Az_0+Bz_1+Cz_2=0=bz_0-az_1$ that is the image of 
$[a,b,-(a+b)/C]$.  Hence $X$ is the union of lines joining a point on the directrix
to a corresponding point on a plane conic in $\bP^4$.


\medskip

\item{7.26}{Let X be as in 7.25. Let $r\in X$ be a point and let $\pi_r:X --\to\bP^3$
be projection from r, and let $Q\subset \bP^3$ be the image of $\pi_r$.  Show that if r is not
on the line L defined in 7.25 (the so called directrix of X) then Q
is a quadric of rank 4, while if r is on L, Q is a rank 3 quadric.}

\itemitem{Proof:}{If $r\in X$ is not on the directrix $L:v_0=v_1=v_2=0$,
then $r$ is on some plane conic contained in X by the discussion above (it
is the image of a point on a line not through $[0,0,1]$ in $\bP^2$).
Under projection from $r$ this conic goes to a line in $\bP^3$
and the directrix goes to a disjoint line, so that the projection of the
scroll is the union of lines joining a point on one line to a corresponding
point on a disjoint line which is a smooth quadric (8.15). 
If $r$ is on the directrix under projection to a hyperplane distinct
from $r$ the directrix $L$ projects to a point.  The points in 
one of the  plane conics lying on $X$ map to a plane conic, so $X$ projects to
the cone over a smooth conic, which is a rank 3 quadric in $\bP^3$.
}
\medskip



\item{7.27}{Consider the projection of a quadric hypersurface in $Q\subset \bP^n$
from a point $p\in Q$ and assume that the rank of $Q$ is $n+1$.  Show that
this projection is a birational isomorphism of $Q$ with $\bP^{n-1}$ and
describe this map in terms of blowing up and blowing down.}

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

\item{7.28}{Generalize 7.11 by showing that $\bP^m\times \bP^n$ is birational
with $\bP^{m+n}$ via the map 
$$\phi([Z_0,\ldots,Z_n],[W_0,\ldots,W_n])=[Z_0W_0,Z_1W_0,\ldots,Z_mW_0,Z_0W_1,\ldots, Z_0W_n].$$
What is the graph of this map? Describe the map in terms of blowing up and blowing down.} 

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

\medskip

\item{7.29}{Consider the group law for addition $A:K^n\times K^n\to K^n$.
This gives a rational map of $\bP^n\times \bP^n$ to $\bP^n$.
For n=1,2 describe the graph of this map and describe the map
in terms of blowing up and down.}



\itemitem{Proof:}{}

\medskip



\bye


















