\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}}
\def \bC {{\bf C}}

{Mathematics 535 \hfill Algebraic Geometry \hfill Fall 2016}

\bigskip


\centerline{ Problem Set 13 (Last revised \number\month/\number\day/\number\year)}
\bigskip


\item{1.}{(Harris 14.2) Show that the dimension of the Zariski tangent space $T_pX$ is an upper-semicontinuous
function of p in the Zariski topology.}

\medskip

\item{2.}{(Harris 14.3) (i) Show that if $X$ is a hypersurface then the locus of singular
points is a proper subvariety of $X$. (ii) Use this together with the
fact that any variety is birational to a hypersurface (11.23) to prove that
the singular points of any variety form a proper subvariety.}

\medskip

\item{3.}{(Harris 14.5) Let $X\subset \bA^2$ be the curve $y-x^p=0$ and let $Y=\bA^1$
and let $f:X\to Y$ be given by $f(x,y)=y$. Show that if $K$ has characteristic
$p$ then $X,Y$ are smooth, $f$ is surjective, but $df_q$ is zero for all
$q\in X$.}



\medskip


\item{4.}{(Harris 14.6) Let $f:X\to Y$ be any regular map of affine varieties. Show that
the dimension of the kernel of the differential $df_p$ is an upper semicontinous
function of $p$.}


\medskip







\item{5.}{(Harris 14.12) Verify that the row space of this matrix is the projective
tangent space to $X$ at $p$.}

\medskip


\item{6.}{(Harris 14.13) Use Corollary 14.10 to show that the Veronese and Segre
varieties are smooth and verify this directly by looking at their defining
equations. (Alternately, deduce this from 14.3 and the transitive 
action of an algebraic group on these varieties).}



\medskip



\bye




















\bye


i1 : R=QQ[x_0..x_3]

o1 = R

o1 : PolynomialRing

i2 : S=QQ[s,t]

o2 = S

o2 : PolynomialRing

i3 : I=ker map(S,R,{s^3,s^2*t,s*t^2,t^3})

             2                       2
o3 = ideal (x  - x x , x x  - x x , x  - x x )
             2    1 3   1 2    0 3   1    0 2

o3 : Ideal of R

i4 : resolution I

      1      3      2
o4 = R  <-- R  <-- R  <-- 0

     0      1      2      3

o4 : ChainComplex
i5 : oo.dd

          1                                                   3
o5 = 0 : R  <----------------------------------------------- R  : 1
               | x_1^2-x_0x_2 x_1x_2-x_0x_3 x_2^2-x_1x_3 |

          3                         2
     1 : R  <--------------------- R  : 2
               {2} | -x_2 x_3  |
               {2} | x_1  -x_2 |
               {2} | -x_0 x_1  |

          2
     2 : R  <----- 0 : 3
               0

o5 : ChainComplexMap

i6 : bettin

o6 = betti o4

o6 : Symbol

i7 : betti o4

o7 = total: 1 3 2
         0: 1 . .
         1: . 3 2
i13 : hilbertPolynomial (I,Projective=>false)

o13 = 3$i + 1
















\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}}
\def \bC {{\bf C}}

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

\bigskip


\centerline{ Problem Set 7 (Last revised \number\month/\number\day/\number\year)}
\bigskip

\item{13.1}{Let $X \subset \bP^n$ be a finite set of $d$ points.
Show that for sufficiently large values of $m$ (say $m \geq d-1$) the Hilbert
function $h_X(m)=d$.}

\itemitem{Proof:}{The Hilbert function depends only on the
homogeneous coordinate ring, which is unchanged by projective
transformations, so we may assume all points are in the open set
$z_0\not = 0$ so the points are of the form $p_i=[1,a_{i1},\ldots,a_{in}]$.
Consider the map of vector spaces from homogeneous polynomials of
degree m to $K^d$ given  by $F(z_0,\ldots,z_n)\to
(F(p_1),\ldots,F(p_n))$, where the polynomial is evaluated on the
chosen representatives
above for $p_i$.  By choosing a linear polynomial $L_i$ which is
zero on $p_i$ and nonzero on all other $p_j$ and forming the degree
$d-1$ polynomials $F_j=\prod_{i \not = j}L_i$ we see that by
multiplying these by powers of $z_0$ that the vector space map above
is onto, and kernel is the degree m polynomials in the ideal of
functions vanishing
on $X$. Thus $h_X(m)=d$ for $m \geq d-1$.}



\medskip


\item{13.5}{Find the Hilbert function of a hypersurface of degree d
in $\bP^n$ and verify that the dimension of the hypersurface is $n-1$.}

\itemitem{Proof:}{Let $X=Z(F)$ be a hypersurface, where $F$ is a
square free nonconstant polynomial of degree $d$, so that $I(X)=(F)$. The vector
space of polynomials in $I(X)$ of degree $m$ is zero when $m < d$, and
is
isomorphic to the space of all polynomials of degree $m-d$ (via
multiplication by $F$) when $m \geq d$. Thus for $m \geq d$ we have
$h_X(m)={m+n \choose n} - {m-d+n \choose n}=dm^{n-1}/(n-1)! + \cdots$.
This agrees with the dimension being $n-1$.

\medskip


\item{13.6}{Find the Hilbert function of the Segre variety $\Sigma_{n,m}=
\sigma(\bP^n\times \bP^m) \subset \bP^{nm+n+m}$ and verify that the dimension
is $n+m$.}

\itemitem{Proof:}{(T. Pham)
Denote $d=(n+1)(m+1)$. Under the map 
$$\sigma : P^n\times P^m \to P^{d-1},$$
$$([x_0,\ldots, x_n], [y_0,\ldots, y_m]) \mapsto (\ldots, x_iy_j,
\ldots),$$
the homogeneous polynomials of degree $k$ in the coordinates
$[z_0,\ldots, z_{(n+1)(m+1)}]$ on $P^{d-1}$ pull back to give all 
bihomogeneous polynomials of bidegree $(k, k)$ in $x_i$ and $y_j$,
respectively.
If $X\subset P^{d-1}$ is the image of this map, then
$$h_X(k)= {k+n \choose n}{k+m \choose m}=h_{\bP^n}(k)h_{\bP^m}(k)$$
Therefore, the degree of $h_X(k)$ is $m+n$, so $\dim X = m+n$.

}


\medskip

\item{13.8}{Determine the arithmetic genus of (i) a pair of skew lines in $\bP^3$
(i) a pair of incident lines in either $\bP^2$ or $\bP^3$ (iii) three concurrent
but not coplanar lines in $\bP^3$ and (iv) three concurrent coplanar lines 
in either $\bP^2$ or $\bP^3$.}


\itemitem{Proof:}{(S. Ganguly)
(i)By a change of coordinates, we can assume WLoG that the two lines
$L_{1}$ and $L_{2}$ are given by : $L_{1}: \{[t_{0},t_{1},0,0]\}$
and $L_{2}: \{[0,0,t_{2},t_{3}]\}$. Now let us find the dimension of
the vector space of homogeneous polynomials of degree m (for large
enough m) which vanish on the variety $X= L_{1}\bigcup L_{2} \subset
\bP^{3}$.  If x,y,z and w are the coordinates then a degree m
monomial vanish on $X$ iff that monomial contains at least one of x
and y and at least one of z and w.By some simple combinatorial
arguments, the number of such monomials is  $4{m-2+3 \choose
m-2}-4{m-3+3 \choose m-3}-2{m-4+3 \choose m-4}+4{m-4+3 \choose
m-4}-{m-4+3 \choose m-4}= (m^{3}+6m^{2}-m-6)/6$ so, the
Hilbert polynomial is ${m+3 \choose
3}-(m^{3}+m^{2}-m-1)/6=2m+2$.So the Arithmetic Genus$=
(-1)^{1}(2-1)=-1$}
\itemitem{(ii)}{
In $\bP^{2}$, let our lines be $\{[0,s_{1},s_{2}]\}$ and
$\{[t_{0},t_{1},0]\}$.The monomials vanishing on the two lines are the
ones containing both x and z and their number is ${m \choose 2}$.So
the Hilbert polynomial is $p_{X}(m)={m+2 \choose m}-{m \choose 2} =
2m+1$.  So the arithmetic genus is$(-1)^{1}(1-1)=0$.  In $\bP^{3}$,
let our lines be be $\{[0,s_{1},s_{2},0]\}$ and
$\{[t_{0},t_{1},0,0]\}$. The monomials vanishing on the two lines are
the ones containing either w or both x and z and their number is $$
{m-1+3 \choose m-1}+{m-2+3 \choose m-2}-{m-3+3 \choose
m-3}=(m^{3}+6m^{2}-m)/6$$ So the Hilbert polynomial is ${m+3
\choose 3}-(m^{3}+6m^{2}-m)/6=2m+1$,  so the arithmetic
genus$=(-1)^{1}(1-1)=0$.}

\itemitem{(iii)}{We take the lines to be $\{[t_{0},t_{1},0,0]\}, \{[t_{0},0,t_{2},0]\},
\{[t_{0},0,0,t_{3}]\}$.  A monomial vanishing on all three must contain
one of z and w, one of y and w and one of y and z.  The number of such
monomials is $3{m-2+3 \choose m-2}+{m-3+3 \choose m-3}-6{m-3+3
\choose m-3}+4{m-3+3 \choose m-3}-{m-3+3 \choose
m-3}=(m^{3}+6m^{2}-7m)/6$  So the Hilbert polynomial is ${m+3
\choose 3}-(m^{3}+6m^{2}-7m)/6=3m+1$,  so the arithmetic genus is
$(-1)^{1}(1-1)=0$}

\itemitem{(iv)}{In $\bP^{2}$ let's take the three lines to be  
$\{[0,s_{1},s_{2}]\}$ and $\{[t_{0},t_{1},0]\}$ and
$\{[r_{0},r_{1},r_{0}]\}$.
The monomials vanishing on the two lines are the ones containing some
positive powers of x, z and $(x-z)$ and their number is ${m-3+2
\choose m-3}=(m^{2}-3m+2)/2$.  So the Hilbert polynomial is 
${m+2 \choose m}-(m^{2}-3m+2)/2=3m$, so the Arithmetic
genus $=1$.  In $\bP^{3}$ let's take the three lines to be
$\{[r_{0},r_{1},0,0]\}$, $\{[s_{0},0,s_{2},0]\}$ and
$\{[t_{0},t_{1},t_{1},0]\}$. The vector space of all degree m
homogeneous polynomials that vanish on this three lines is spanned by
the monomials that have either w as a factor or $yz(y-z)$ as a factor,
and the number of such monomials is $${m-1+3 \choose m-1}+{m-3+3
\choose m-3}-{m-4+3 \choose m-4}=(m^{3}+6m^{2}-7m+6)/6$$  So
the Hilbert polynomial is: ${m+3 \choose m}- 
(m^{3}+6m^{2}-7m+6)/6=3m$  and the Arithmetic Genus$=1$.
}

\medskip

\item{13.9}{Consider a plane curve $X \subset \bP^2$ of degree d and its
image $Y=\nu_2(X) \subset \bP^5$ under the quadratic Veronese map.  Compare the
Hilbert polynomials of the two and observe in particular that the arithmetic genus is
the same.}


\itemitem{Proof:}{%not done by(B. Kennedy)
Let $X \subset {\bf P}^2$ be a plane curve,
$Y = \nu_2(X) \subset {\bf P}^5$, and $Z = \nu_2({\bf P}^2)$.  The Veronese
furnishes us with the following exact sequence:
$$0 \longrightarrow I(Y)_r \cap I(Z)_r \longrightarrow I(Y)_r \longrightarrow
I(X)_{2r} \longrightarrow 0,$$
where the second map is given by pullback via the Veronese, and the kernel
$I(Y)_r \cap I(Z)_r = I(Z)_r$ arises from the relations determining the
Veronese variety.  That the pullback map is surjective is established in
Harris's discussion of this construction (p. 24).  Thus
$$
\dim I(Y)_r = \dim I(X)_{2r} + \dim\  I(Z)_r, {\rm\ and}$$ $$
{5 + r \choose r} - h_Y(r) =
{2+2r \choose 2r} - h_X(2r) +
{5 + r \choose r} -h_Z(r).
$$
But $h_Z(r) = {2+2r \choose 2}$
(see Harris's ex. 13.4), so $h_X(2r) = h_Y(r)$ for every $r$ and
consequently $p_X(2r) = p_Y(r)$ also.  Since the arithmetic genus is
manufactured from the constant term of the Hilbert polynomial, it follows
that the Veronese preserves the arithmetic genus, while multiplying the
degree of the curve by 2.  It is clear that general principles are at work
here, and probably the significant point to make is that for any variety, the
Hilbert polynomial for its image under the Veronese map of degree $d$ is found
by replacing the argument $r$ by $d\cdot r$.


}

\medskip

\item{13.13}{Verify the possible free resolutions of a configuration of
four points in the plane detailed in example 13.12}

\itemitem{Proof:}{()}

\medskip

\item{13.14}{Find the minimal free resolutions of $\Gamma$ in $\bP^3$ in
case (i) $\Gamma$ consists of the four coordinate points (ii) $\Gamma$
consists of five points, no four coplanar.}

\itemitem{Proof:}{(
%o31 = total: 1 6 8 3
%          0: 1 . . .
%          1: . 6 8 3
% 0\to S(-4)^3 \to S(-3)^8 \to S(-2)^6 \to S \to S(\Gamma)
%                 x_0x_1 x_0x_2 x_1x_2 x_0x_3 x_1x_3 x_2x_3 
%
%

)}
\medskip
\item{13.15}{Find the minimal free resolution of the twisted cubic curve.}

\itemitem{Proof:}{%not done by(K. Dalili)  SEE MACAULAY session at end
Let $X$ be the twisted cubic with $I(X)=(x_1^2-x_0*x_2,x_1*x_2-x_0*x_3,x_2^2-x_1*x_3)$
and let $S=K[x_0,\ldots,x_3]$ and $S(x)=S/I(X)$ be the homogeneous coordingate ring.
The kernel of $S \rightarrow S(X)$ is generated by three elements $F_i$ of degree 2, so there
is a map of $S(-2)^3\rightarrow S \rightarrow S(X)$ which has  $-x_2F_1+x_1F_2-x_0F_3,
x_3F_1-x_2F_2+x_1F_3$ in the kernel.  Any other relation $p_1F_1+p_2F_2+p_3F_3=0$
with $p_i$ homogeneous  could be written modulo the two preceding as $ax_1^kF_1+\cdots=0$,
which implies that a=0, that is that modulo known relations $p_2F_2+p_3F_3=0$.  Since $x_0$ appears
in $F_2$ but not $F_3$ no such relation is possible.  Thus we have the sequence
$0\rightarrow S(-3)^2 \rightarrow S(-2)^3\rightarrow S \rightarrow S(X)$  as the minimal free
resolution of the twisted cubic curve.  Note that this means that the Hilbert polynomial
is 3m+1, as we expect for a degree 3 curve which is isomorphic to the projective line.

}
\medskip

\item{13.17}{Prove a weak form of the Bezout theorem in $\bP^2$: if $F,G$ are
polynomials of degree $d,e$ on $\bP^2$ without common factors such that $F,G$
generate the ideal of their intersection, then the intersection consists of
$d\cdot e$ points.  Similarly show that if $\Gamma \subset \bP^3$ is a complete
intersection of surfaces of degrees $d,e,f$ then $\Gamma$ consists of $d\cdot e\cdot f$
points.}

\itemitem{Proof:}{(F. Louro)
Since $F,G$ generate the ideal of $X=\{F=G=0\}$, the intersection
is by definition complete. As described in page~$172$, the Betti
numbers for
a complete intersection in ${\b P}^2$ of two
curves~$F=0$ and~$G=0$ of degrees~$d,e$ are:
$a_{11}=d$, $a_{12}=e$, $a_{22}=d+e$.
By the discussion preceding the Exercise,
the Hilbert function of~$X$ is~$h_X(m)={m + 2 \choose 2}
- {m-d+2 \choose 2} - {m-e+2\choose 2} + {m-{d+e}-2 \choose 2}
= \ldots \ldots = de$, a polynomial of degree~$0$, so that
the intersection of $F,G$ has dimension~$0$ (obviously
of pure dimension~$0$). The desired result then follows
from Exercise $13.1$.
For the case of $3$ hypersurfaces intersecting completely
in~${\bP}^3$, one has $a_{11}=d, a_{12}=e,a_{13}=f$,
$a_{21}=d+e, a_{22}=d+f,a_{33}=e+f$, $a_{31}=d+e+f$
and $h_{\Gamma}(m)= {m + 3 \choose 3}
- {m-d+3 \choose 3} - {m-e+3\choose 3} - {m-f+3\choose 3} 
+ {m-d-e+3 \choose 3} + {m-d-f+3\choose 3} + {m-e-f+3\choose 3} 
-  {m-{d+e+f}-3 \choose 3}
= \ldots \ldots \ldots= def$.




}


\medskip

\item{13.18}{Find the Hilbert polynomial of a complete intersection in
$\bP^3$ of surfaces of degrees $d,e$.  In case this intersection is a
curve, what is the arithmetic genus of this curve?}

\item{Proof:}{(J. Burdges)
Let $R = k[x_0,x_1,x_2,x_3]$ be the homogeneous coordinate ring of
$\bP^3$.
Since $X \subset \bP^3$ is a complete intersection of two surfaces,
$X = Z(I)$ where $I = (F,G)$ for some relatively prime $F,G \in R$
which describe the surfaces and satisfy $\deg F = d$ and $\deg G = e$.
We construct the Koszul complex for $X$,
$$ 0 \to R(-e-d) \buildrel\psi\over\to
         R(-d) \oplus R(-e) \buildrel\phi\over\to
         R \to R/I \to 0 $$
by taking $\phi(\alpha,\beta) = \alpha F + \beta G$ and $\psi(\gamma)
= (-G\gamma,F\gamma)$.  We only really need to worry about exactness
between the interesting maps, i.e. $\hbox{im}~\psi = \ker \phi$.  This
follows from the fact that $F$ and $G$ are relatively prime, via the
Chinese remainder theorem.  The twistings have no effect on the maps,
except to make them degree zero.
This exact sequence of modules is an exact sequence of vector spaces
if we restrict to the common degree pieces, so we can use the usual
rank plus nullity theorem to compute the dimension of $S = R/I$ to get
$$ h_X(m) = \dim S_m = \dim R_m - \dim (R(-d) \oplus R(-e))_m + \dim
R(-e-d)_m $$
and we observe that $\dim (R(-d) \oplus R(-e))_m = \dim R(-d)_m + \dim
R(-e)_m$.
We should recall that $\dim R(j)_m = {m+j+3 \choose m+j}_0$, so this
will not usually be a polynomial, but for $m+j > 0$ we find that
$$ \dim R(j)_m = {1\over6} m^3 + ({j\over 2} + 1) m^2 + ({j^2\over 2}
+ 2j + {11\over 6})m +({j^3\over 6} +j^2+{11\over 6}j +1) $$
Thus $p_X(m) = de m + de(-e-d+4)/2$.
Hence the degree of the complete intersection is the product of the degrees.
Finally. we don't need really to worry about Riemann surfaces since
Example 13.7 explains that we are just looking for the arithmetic
genus.
Since $\dim X = 1$, this works out to be $1+de(d+e-4)/2$.

}

\medskip

\item{14.1}{Let $K \simeq \bC$ be the complex field. Prove that $X$ is smooth at $p$ if and only if
$X$ is a smooth complex manifold at $p$ if and only if $X$ is a real
$C^\infty$ submanifold of $\bC^n$ at $p$.}

\itemitem{Proof:}{We may assume that X is an affine variety $\subset
\bC^n$ and $I(x)=(f_1,...,f_m)$ of complex dimension k.  Let $J=[\partial f_i/\partial
x_j(p)]$ be the Jacobian matrix at a point p in X.  X is smooth at p
if and only if the rank of J is n-dim(X).  For a subset of complex n
space which is the zero set of functions $f_i$ holomorphic at p to be a
smooth complex manifold exactly the same condition on the Jacobian is
to be checked.  When we consider $X \subset {\bf R}^{2n}$ as the zero set
of the real functions $u_i,v_i$ (where $f_j=u_j+iv_j$) the test to see
if X is a real $C^\infty$ submanifold of $\bC^n$ at $p$ is to check if
the rank of the real matrix 
$$\pmatrix{ \partial u_i/\partial x_j  &\partial v_i/\partial x_j\cr
             \partial u_i/\partial y_j  &\partial v_i/\partial y_j\cr}$$
evaluated at p has rank 2n-2k.  By the Cauchy--Riemann equations the
matrix J gives a linear map of the real 2n dimensional vector space
${\bC^n} $ to itself which has matrix the real matrix above.  Thus
the rank of that real matrix is twice the rank of the complex matrix J.
So X is smooth at p as a variety if and only if the the set of points
of X in ${\bf R}^{2n}$ is a smooth real manifold at p.  Note of course, we
are not saying anything about the possible real coordinate points on X.
 }
\medskip

\item{14.2}{Show that the dimension of the Zariski tangent space $T_pX$ is an upper-semicontinuous
function of p in the Zariski topology.}

\itemitem{Proof:}{The dimension of the Zariski tangent space at $p$ to $X \subset \bA^n$
is 
just the n minus the rank of the $l \times n$ Jacobian matrix $\partial f_i /\partial x_j$ 
if $I(X)=(f_1,\ldots,f_l)$.
The dimension of the tangent space is greater than
or equal to a positive integer $m$ if and only if the Jacobian matrix has
rank less than or equal to $n-m$ if and only if all $(n-m+1)\times(n-m+1)$
minors of the Jacobian have determinant zero. This is clearly a set of equations
describing a closed subset of $X$, proving the upper semicontinuity.}

\medskip

\item{14.3}{(i) Show that if $X$ is a hypersurface then the locus of singular
points is a proper subvariety of $X$. (ii) Use this together with the
fact that any variety is birational to a hypersurface (11.23) to prove that
the singular points of any variety form a proper subvariety.}

\itemitem{Proof:}{(i) It is enough to do this for an irreducible hypersurfaces
which is the zero set of a nonconstant  irreducible polynomial $F(x_1,\ldots,x_n)$.
The locus of singular points $X_{sing}$ is the closed set of points 
$(x_1,\ldots,x_n) \in X$
such that $\partial F /\partial x_i$ simultaneously vanish.  If $X_{sing}=X$
then the ideals $(F), (F, \partial F/\partial x_1,\ldots,\partial F/\partial x_n)$
define the same variety so that $\partial F/\partial x_i$ is a multiple of
$F$. The total degree of $\partial F/\partial x_i$ is less than that of $F$,
so it must be true that $\partial F/\partial x_i=0$ as a polynomial.
In characteristic zero it is easy to see that the partial derivative of a
monomial is nonzero if it is not a constant, so this is impossible in
characteristic 0.  In characteristic $p$ the only monomials with all
partial derivatives zero are $p$th powers, and since a $p$th power of
a sum in characteristic $p$ is a sum of $p$th powers we conclude that $F$ is
a $p$th power contradicting that it is irreducible. So in each case $X_{sing}$
is a proper subvariety of $X$, so that its complement $X_{sm}$,
the set of smooth points, is a 
nonempty open subset (which is dense if X is irreducible).
(ii) Since any variety is birational to a hypersurface, it has a a nonempty open subset isomorphic
to an open subset in a hypersurface. By (i), not every point in a nonempty
 open subset in a hypersurface is singular. Hence in an arbitrary variety there
are always smooth points, and the set of singular points is a proper subvariety
of strictly smaller dimension.}
\medskip

\item{14.5}{Let $X\subset \bA^2$ be the curve $y-x^p=0$ and let $Y=\bA^1$
and let $f:X\to Y$ be given by $f(x,y)=y$. Show that if $K$ has characteristic
$p$ then $X,Y$ are smooth, $f$ is surjective, but $df_q$ is zero for all
$q\in X$.}

\itemitem{Proof:}{Since the partial of $y-x^p$ with respect to y is 1,
X is smooth. For each $y$ in $K$ there is a p-th root z and
we have $f(z,y)=y$ so $f$ is
surjective.  The map induced on contangent spaces is $y-y_0 /to
x^p-x_0^p$
and the latter is in the square of the maximal ideal, so the map on
cotangent spaces is 0, thus the map $df_q$ is identiacally 0 in this
case.  Hence in characteristic p the differential may never be surjective.



 


}
\medskip


\item{14.6}{Let $f:X\to Y$ be any regular map of affine varieties. Show that
the dimension of the kernel of the differential $df_p$ is an upper semicontinous
function of $p$.}


\itemitem{Proof:}{Suppose that $X \subset A^m, Y \subset A^n,
f=(h_1,...,h_n)$.  Then the map $df_p$ is induced by the  n x m matrix
$[\partial h_i/ \partial x_j(p)]$ which gives a linear map $K^m /to K^n$
which restricted to $T_pX$ is the map desired.  Thus the kernel of
$df_p$ on $T_pX$ is the simultaneous kernel of the linear map given
by this matrix on $K^m$ and the kernel of $[\partial g_k/\partial x_j
(p)]$ where I(X) is generated by $g_k$.  Thus the kernel of $df_p$ is
the kernel of a n' x m matrix.  The dimension of the kernel is greater
than or equal to t if and only if the rank is less than or equal to
n'-t, which is given by the vanishing of suitable minors.  Hence the
set of points where the kernel dimension is at least t is closed,
proving that this dimension is an upper seim-continuous function on X.



}

\medskip

\item{14.7}{Let $X \subset \bA^4$ be the hypersurface $ax^2+y^2=b$, and let $Y=\bA^2$ with coordinates $a,b$ and let $f$ be the 
projection. Show that $X$ is smooth of dimension 3, and that every fiber of $f$ has dimension 1. Find the set of points $p\in X$ such that
$df_p$ has rank $<2$ and verify 14.6 for this map.  At the same time, find the locus of
$p\in X$ such that the Zariski tangent space to $X_{f(p)}$ at $p$ has dimension greater that
1 and show that it is not closed.}


\itemitem{Proof:}{Since $X=Z(F), F=ax^2+y^2-b \in K[x,y,a,b]$ and the
partial of F with respect to the variable b is nonzero, X is a smooth
hypersurface.  The fibers of the map $(x,y,a,b) \to (a,b)$ are
isomrophic to the hypersurfaces  $X_{ab}$ in $A^2$ given by $ax^2+y^2=b$ and
since this is never the zero polynomial in x,y all fibers have
dimension 1.  The curve $X_{ab}$ is singular at points $(x,y)$ for
which $2ax=0,2y=0$.  If the characteristic of K is not 2, the singular
points of the fiber are all of form (0,0) if b=0, and if a=0 then
this is not a singular point .  Thus the
dimension of the tangent space to the fiber $X_{ab}$ at (x,y,a,b)
is greater than 1 if and only if (x,y,a,b) is of the form (0,0,a,0)
with $a \not = 0$, which is not a closed set.
The cotangent space at $(a_0,b_0)$ is generated by $a-a_0,b-b_0$.
The cotangent space to X at $(x_0,y_0,a_0,b_0)$ is generated by
$x-x_0,y-y_0,a-a_0$, and $b-b_0=2a_0(x-x_0)+x_0^2(a-a-0)+2y_0(y-y_0)$
so the transpose of $df_p$ is given by the matrix $\pmatrix{0 &2a_0\cr 0 &
2y_0\cr 1 & x_0^2}$ and thus $df_p $ has 
kernel dimension 
rank 3 - rank
of this matrix.  So $df_p$ has kernel dimension 2 if $a_0=y_0=0$ and
kernel dimension 1 otherwise (showing that it is upper semicontinuous).
Note the fiber is two lines if a=0, b not 0, and 1 line if a=b=0.

}

\medskip



\item{14.11}{(i) Verify directly that the map $h$ does not induce an
isomorphism between the tangent space to $Y$ and to $Z$ at the origin.
(ii) Show that $Y$ and $Z$ are not isomorphic.}


\itemitem{Proof:}{(i) The varieties are $X=\bA^1,Y=Z(y^2-x^3),Z=Z(y^3-x^4)$ and the maps are
$f(t)=(t^2,t^3)$, $g(t)=(t^3,t^4)$.  The map $h: Y \to Z$ is given by
$(x,y) \to (y,x^2)$.  The tangent space at (0,0) to Y or Z is 2
dimensional.  The map on cotangent spaces at the origin is 
$x->y, y -> x^2=0$ modulo squares is then a map of vector
spaces with kernel and image of dimension 1, so the same is true
of dh.  Thus h is not an isomorphism.  Note that the matrix of dh is
$\pmatrix{0 & 1 \cr 
         0 & 0 \cr}$ at the origin.
}
\itemitem{}{(ii) If Y and Z are isomorphic their affine coordinate
rings are isomorphic and the isomorphism must map the unique singular
points to each other.  Thus the local rings at (0,0) are isomorphic.
Let $m_Y,m_Z$ be the maximal ideals. Consider the vector space
$m^2/m^3$.  This has basis $x^2,xy$ for Y, but $x^2,xy,y^2$
are independent modulo $m_Z^3$ since no relation involves degree
2 elements.  Hence there is no possible isomorphism of these
two varieties.  Recall that the tangent cone is made from the sum of
$m^i/m^{i+1}$ so we are noting that the tangent cones are different
for these curves, which have multiplicity 2 and 3 at the origin.
Can also see this in the Hilbert polynomial of the algebra defining
the projective tangent cone.

}


\medskip


\item{14.12}{Verify that the row space of this matrix is the projective
tangent space to $X$ at $p$.}

\itemitem{Proof:}{We check this for the case of a hypersurface, the
general case is the intersection of hypersurfaces.  Let X be the zero
set of a homogeneous polynomial F.  We have that
$F(v_0(z),...,v_n(z))$ is identically 0.  The derivative with respect
to $z_j$ is $\sum \partial F /\partial x_i \partial v_i/ \partial z_j$.
Thus the row vector $(\partial v_0/\partial z_j (0), \ldots,\partial
v_n/\partial z_j (0))$ is in the projective tangent space.
Conversely, any line hitting X to order at least 2 at a point P has
direction expressible in terms of the rows of the matrix, by
approximating it by secant lines at P.  So the row space spans the
projective tangent space.


 




}
\medskip


\item{14.13}{Use Corollary 14.10 to show that the Veronese and Segre
varieties are smooth and verify this directly by looking at their defining
equations. (Alternately, deduce this from 14.3 and the transitive 
action of an algebraic group on these varieties).}


\itemitem{Proof:}{The projective groups act transitively on the
Veronese of Segre varieties.  Thus all tangent spaces are isomorphic,
hence all have the same dimension as the variety.  Alternately,
one can reverify that the Veronese is isomorphic to a projective space
since it is bijective via the Veronese map, and the map on tangent
spaces is injective.  Further, using the equations $z_Iz_J=z_Kz_L$



}


\medskip


\item{14.14}{Show that the rational normal scroll $X_{a_1,\ldots,a_k} \subset
\bP^n$  is smooth if and only if all $a_i >0$.}

\itemitem{Proof:}{
If some $a_i=0$, then the scroll is a cone which will
be singular at points in its vertex. If all $a_i \not =0$, then we
must show that the tangent space has dimension k.  We can parameterize
the scroll as the set 
$$\{[\lambda_1 s^{a_1},\lambda_1 s^{a_1-1}t,\ldots,\lambda_1 t^a_1;
\lambda_2 s^{a_2},\ldots,\lambda_2 t^a_2;\dots;
\lambda_k s^{a_k},\lambda_1 s^{a_k-1}t,\ldots,\lambda_1 t^a_k\}$$.
The subset where s=1 is isomorphic to $A^1 \times P^{k-1}$ as is the
subset where t=1.  Thus the scroll $X_{a_1,\ldots,a_k}$ is a union of
open sets isomorphic to $A^1 \times A^{k-1}$ which intersect in open
sets isomorphic to affine space, so that the scroll is an irreducible
rational variety.  Each point has a neighborhood isomorphic to affine
space, so that every point is a smooth point.




%easiest to show scroll is locally a product of A^1 x P^k-1, hence
%rational, smooth, irreducible
%all rational normal scrolls are toric
%:http://www.google.com/url?sa=U&start=6&q=http://arxiv.org/pdf/alg-geom/9610018&e=912
%=sturmfels equations defining toric varieties





%The scroll is a
%union of {k-1}-planes by definition.  A point p on the scroll is on a 
%unique {k-1}-plane spanned by points on complementary planes.  Any line
%through p which is tangent to the scroll would also be tangent to
%the k-plane through p, hence it must lie in the k-plane.  Thus the
%tangent space is just the k-dimensional plane contained in the scroll
%through the point p, so that the scroll is smooth if it is not a cone.
} 
 

 
\medskip

\item{14.15}{Let $X\subset \bP^n$ be any irreducible nondegenerate variety (that is $X$ is not contained in any
hyperplane in $\bP^n$) and let $S(X)$ be its secant variety.
Show that the projective tangent space to $S(X)$ at a point of $X$ is all of $\bP^n$.}


\itemitem{Proof:}{Let p be a point of X.  Choose points $q_i$ of X
which span $\bP^n$ (this is possible since the span is a hyperplane
containing X).  The lines from p to $q_i$ are contained in the secant variety of
X.  Any line contained in a variety is in the tangent space to the
variety at any of its points.  Thus the tangent space contains lines
which span all of $\bP^n$, so it must be $\bP^n$.  As an application, the
secant variety must be singular along X if the secant variety is not
all of $\bP^n$, so for example the 4 dimensional secant variety to
the Veronese surface in $\bP^5$ is singular along the Veronese
surface.

% the det of symmetric 3 by 3 matrix is equation of secant,
% so z_0*z_1*z_2-z_2*z_3^2-z_1*z_4^2+2*z_3*z_4*z_5-z_0*z_5^2
%singular locus is precisely the veronese surface.


\medskip


\bye




















\bye


i1 : R=QQ[x_0..x_3]

o1 = R

o1 : PolynomialRing

i2 : S=QQ[s,t]

o2 = S

o2 : PolynomialRing

i3 : I=ker map(S,R,{s^3,s^2*t,s*t^2,t^3})

             2                       2
o3 = ideal (x  - x x , x x  - x x , x  - x x )
             2    1 3   1 2    0 3   1    0 2

o3 : Ideal of R

i4 : resolution I

      1      3      2
o4 = R  <-- R  <-- R  <-- 0

     0      1      2      3

o4 : ChainComplex
i5 : oo.dd

          1                                                   3
o5 = 0 : R  <----------------------------------------------- R  : 1
               | x_1^2-x_0x_2 x_1x_2-x_0x_3 x_2^2-x_1x_3 |

          3                         2
     1 : R  <--------------------- R  : 2
               {2} | -x_2 x_3  |
               {2} | x_1  -x_2 |
               {2} | -x_0 x_1  |

          2
     2 : R  <----- 0 : 3
               0

o5 : ChainComplexMap

i6 : bettin

o6 = betti o4

o6 : Symbol

i7 : betti o4

o7 = total: 1 3 2
         0: 1 . .
         1: . 3 2
i13 : hilbertPolynomial (I,Projective=>false)

o13 = 3$i + 1
















