Algebraic Geometry
Prof. Pandharipande
D-MATH, FS 2014
Solutions Sheet 3
1. Most of the problem is immediate from the case of projective varieties. For a
discussion of the Segre map see for example Harris, Example 2.11.
2. Let us index the points starting by 0, so we have n+3 points p0 , . . . , pn+2 . As the
points are in general linear position, we can find a coordinate transformation of
CPn , such that for 0 ≤ i ≤ n we have pi = [0, . . . , 0, 1, 0, . . . , 0], where the 1 is
at the i-th position. Consider now n + 1 (pairwise distinct) points [ai , bi ] in CP1 ,
0 ≤ i ≤ n for some ai , bi ∈ C with ai , bi 6= 0. We use homogeneous coordinates
x, y on CP1 . Let us consider the linear polynomials λi = bi x − ai y and define a
function f : CP1 −→CPn of degree n by
" n
#
n
n−1
Y
Y
Y
f ([x, y]) =
λj (x, y), . . . ,
λj , . . . ,
λj
j=1
j=0
j6=i
As λj (ai , bi ) = 0 for i = j and 6= 0 otherwise, we have f ([ai , bi ]) = pi for
i = 0, . . . , n. Furthermore, we have


!
Y
f ([0, 1]) = 
(−aj )
j6=i
Q
1
1
1
=
, ,...,
,
a0 a1
an
(1)
i=0,...,n
n−1
since j6=i (−aj ) = (−1) a0 · · · an /ai and all the ai are non-zero. Let pn+1 =
[c0 , . . . , cn ]. As pn+1 is in general linear position, ci 6= 0 for all i (otherwise
it would be in the span of n of the points). By setting, ai = 1/ci we have
f ([0, 1]) = pn+1 . With the same argument for [1, 0] and the bi , we are done.
3. This is discussed e.g. in Harris Example 2.4.
4. V (x20 + x21 + x22 ) is projectively equivalent to V (x0 x2 − x21 ) which isomorphic to
CP1 via the Veronese map ν1 : CP1 −→CP2 .
Alternatively, pick a point P on the the quadric X = V (x20 + x21 + x22 ). Then
each line through P intersects the quadric in exactly one other point. Conversely
any other point A on X gives a unique line through P and Q. This defines an
bijection X−→{ lines through P } ∼
= CP1 and one can show that this gives an
isomorphism of varieties.
5. Let C = V (xn + y n + z n ) ⊂ CP2 be the Fermat curve. Consider the projection
π : C−→P1
given by [x, y, z] 7→ [y, z]. This is well defined on C as [1, 0, 0] ∈
/ C. The map
π is of degree n, that is a general point has n preimages. The only points that
do not have this property, are the n points of the form [1, ω1 ], . . . , [1, ωn ], where
ω1 , . . . , ωn are the n solutions to the equation 1+z n . The points Pi = [1, ωi ] have
only a single preimage given by [0, 1, ωi ]. They are branch points of branching
index n − 1.
One may now argue by cut-and-paste arguments. This uses that the topological
Euler characteristic is additive under cut and paste into ’nice’ pieces and compatible with pullback by purely degree n maps (as it is simply # points − #
edges + # faces of a triangulation).
X
χ(C) = χ(π −1 (P1 \ {P1 , . . . , Pn })) +
χ(π −1 (Pi ))
i
1
= deg(π)χ(P \ {P1 , . . . , Pn }) + n
X
= n(χ(P1 ) −
χ(Pi )) + n
= n(2 − n) + n
= (3 − n)n
Alternatively, one may use the abstract Hurwitz formula which says that a map
f : C−→C 0 between smooth projective curves of degree deg(f ), satisfies
X
branch index,
χ(C) = deg(f )χ(C 0 ) −
branch points
see e.g. Hartshorne page 301. In fact, in the complex case this can be proved
precisely by the above cut and paste argument.
For the genus use χ(C) = 2 − 2g, to get
g=
(n − 2)(n − 1)
.
2
6. Consider the vector space V of homogeneous polynomials of degree 2 in the
variables x, y, z, w. This is a 10 dimensional space and its projectivization parametrizes quadrics in P3 . Let L1 , . . . , L4 be 4 general skew lines in P1 . Pick
3 generic points on each of the lines L1 , L2 , L3 . The condition for a quadric
Q ∈ P(V ) to contain a point is a linear condition on V . As the lines are choosen
general and there are three points on each line, there will be a 1-dimensional
subspace of V of homogeneous polynomials vanishing at all the points and therefore a unique quadric Q that contains the choosen points (This needs a check,
that the conditions are independent!). The restriction of the polynomial that
defines Q to the lines Li , i = 1, 2, 3 is a polynomial of degree 2 that vanishes at
three distinct points, hence must be zero. This shows that Q contains the three
lines L1 , L2 , L3 .
As the lines were choosen generic, Q is a smooth quadric and after a change of
coordinates we may assume it is given by V (xw − yz) which is isomorphic to
P1 × P1 via the Segre map (see Problem 1). The lines L1 , L2 , L3 are then given
P1 × Pi for Pi ∈ P1 , i = 1, 2, 3 some points (This follows as they are distinct
and of degree 1). Any line that meets the three lines L1 , L2 , L3 must meet the
quadric Q in three distinct points and with the same argument as above is contained in Q. It must hence given by MR = R × P1 for R ∈ P1 a point.
Consider now the 4th general line L4 . L4 meets the quadric Q in precisely 2
points (R1 , S1 ), (R2 , S2 ) ∈ P1 × P1 . This shows that there are precisely two lines
(namely MR1 and MR2 ) incident to all four lines L1 , . . . , L4 .
7. Let ωi = [ai , bi ] ∈ P1 , i = 1, 2, 3 be the three solutions to the equation a3 +b3 = 0
on P1 (with homogeneous coordinates a, b on P1 ). For each 1 ≤ i, j ≤ 3, define
the line
n
o
1
Lij = [sai , sbi , taj , tbj ] | [s, t] ∈ P .
It is immediate that Lij is contained in X = V (X 3 +Y 3 +Z 3 +W 3 ). In total this
are 9 different lines on X. The same argument works as well for the pairing of
the coordinates into the pairs (13|24) and (14|23) (the case above was (12|34)).
This makes 3 × 9 = 27 lines.