Oral exam practice problems: Algebraic Geometry
Alberto Garcı́a Raboso
TP1. Let Q1 and Q2 be the quadric hypersurfaces in Pn given by the equations
f1 ≡ x20 + · · · + x2n
=0
f2 ≡ a0 x20 + · · · + an x2n = 0
respectively, where all the coefficients ai are nonzero and different from one another. How many
singular elements are there in the pencil generated by Q1 and Q2 ?
Denote by Q[λ:µ] ⊂ Pn the element of the pencil over [λ : µ] ∈ P1 , given by the equation
n
X
(λ + µai )x2i = 0
i=0
A quadric hypersurface is singular if and only if it is not of maximal rank, so the singular elements
are located at λ + µai = 0, i = 0, . . . n; since the coefficients ai are distinct, there are n + 1 of them.
Moreover, all of the singular fibers are quadric hypersurfaces of rank one less than the maximal, and
thus cones over a nonsingular quadric hypersurface in Pn−1 with vertex a point.
RD1. a) Calculate the Betti numbers of the intersection of the quadrics Q1 and Q2 of problem TP1.
b) In the case of odd n, calculate the Hodge numbers of Q1 ∩ Q2 .
a) First of all, notice that Q1 ∩ Q2 is a complete intersection. By an iteration of Krull’s principal
ideal theorem, a subscheme defined by two equations has codimension at most two. It is clearly not
of codimension zero, for f1 and f2 are nonzerodivisors. To see that it is not of codimension one,
notice that f1 and f2 are irreducible and of the same degree; since they do not differ by a unit, they
define different prime ideals.
Q1 ∩ Q2 is smooth: the projective tangent plane at a point p = [x0 : . . . : xn ] ∈ Q1 ∩ Q2 is the
intersection of the tangent planes of Q1 and Q2 at p, i.e., it is given by the kernel of the matrix
!
!
∂f1 /∂x0 . . . ∂f1 /∂xn
2x0 . . . 2xn
=
∂f2 /∂x0 . . . ∂f2 /∂xn
2a0 x0 . . . 2an xn
Hence p is a singular point if the 2 × 2 minors of this matrix vanish at p. The equations of this
singular locus are (ai − aj )xi xj = 0 for 0 ≤ i < j ≤ n, whose only solutions are the coordinate points
[1 : 0 : . . . : 0], [0 : 1 : . . . : 0], . . . [0 : 0 : . . . : 1], which are not contained in Q1 nor in Q2 .
1
Using the Lefschetz hyperplane theorem and Poincaré duality, we have


if i even, 0 ≤ i ≤ 2(n − 2)
1

bi = bn−2 if i = n − 2


0
otherwise
χ = n − 1 − bn−2
for n odd, and
bi =



1
if i even, 0 ≤ i ≤ 2(n − 2), i 6= n − 2
bn−2


0
if i = n − 2
otherwise
χ = n − 2 + bn−2
for n even. For another way of calculating the Euler characteristic of Q1 ∩ Q2 , consider the exact
sequence of vector bundles
0 → TQ1 ∩Q2 → TPn |Q1 ∩Q2 → NQ1 ∩Q2 → 0
with NQ1 ∩Q2 ∼
= OPn (2)⊕2 Q1 ∩Q2 , which yields an equation on total Chern classes:
c TPn |Q1 ∩Q2 = c TQ1 ∩Q2 c NQ1 ∩Q2
(1 + ω)n+1 = (1 + c1 + · · · cn−2 )(1 + 2ω)(1 + 2ω)
where ci are the Chern classes of the tangent bundle of Q1 ∩ Q2 and ω = c1 (OPn (1))|Q1 ∩Q2 . The top
Chern class cn−2 is then the coefficient of ω n−2 in (1 + ω)n+1 /(1 + 2ω)2 ; equivalently,
(1 + ω)n+1
dω
cn−2 = Resω=0
ω n−1 (1 + 2ω)2
(1 + ω)n+1
(1 + ω)n+1
= −Resω=−1/2
dω − Resω=∞
dω
ω n−1 (1 + 2ω)2
ω n−1 (1 + 2ω)2
(1 + z)n+1
(−1)n
(1 + z)n+1
=
Resz=0 2
dz + Resz=0 2
dz
n−1
8
z (1 − z)
z (2 + z)2
(
0
if n odd
(−1)n n n
=
+ =
4
4
n/2 if n even
Hence
Z
χ=
cn−2 =
Q1 ∩Q2
(
0
2n
if n odd
=⇒
if n even
2
bn−2 =
(
n−1
n+2
if n odd
if n even
b) Since we want to consider the case of n odd, let n = 2g + 1, so that bn−2 = b2g−1 = 2g. Denote
π
by P2g+1 × P1 ⊃ X −→ P1 the family of quadrics given by the pencil in problem TP1; we have seen
that it is the vanishing locus of the bihomogeneous polynomial
!
!
n
n
n
X
X
X
2
2
2
(λ + µai )xi = λ
xi + µ
ai xi
i=0
i=0
i=0
The expression on the right hand side exhibits X as the blowup of P2g+1 along the base locus of the
pencil, B := Q1 ∩ Q2 ; in particular, since B is smooth, X is too.
By the Lefschetz hyperplane theorem we know then that hp,q (X) = hp,q (P2g+1 ×P1 ) for p+q ≤ 2g.
Together with Poincaré duality, this gives all the Hodge numbers of X outside of the middle dimension
p+q = 2g +1: the only nonzero ones are h0,0 (X) = h2g+1,2g+1 (X) = 1 and hp,p (X) = 2 for 1 ≤ p ≤ 2g.
We can also find the Betti number b2g+1 (X) from the topological Euler characteristic of X, which
π
in turn can be calculated from the knowledge of the projection X −→ P1 . The general fiber of this
map is a smooth quadric Q(2g) in P2g+1 , whose Euler characteristic can be calculated as above to
be χ Q(2g) = 2g + 2. Using once more the Lefschetz hyperplane theorem and Poincaré duality we
thus have the Betti numbers of Q(2g) :


2 if k = 2g

(2g)
bk Q
= 1 if k even, 0 ≤ k ≤ 4g, k 6= 2g


0 otherwise
The 2g + 2 singular elements of the pencil are cones over a nonsingular quadric Q(2g−1) in P2g with
vertex a point, hence isomorphic to the Thom space of a trivial (real) vector bundle of rank 2 over
Q(2g−1) , which in turn is isomorphic to Σ2 Q(2g−1) + . Proceeding as with Q(2g) , we obtain
bk Q(2g−1) =
bk
(
1
0
Σ2 Q(2g−1) + =
if k even, 0 ≤ k ≤ 4g − 2
otherwise
(
1
0
if k even, 0 ≤ k ≤ 4g
otherwise
i.e., the topological Euler characteristic of the singular elements of the pencil is one less than that
of the general ones. Since there are 2g + 2 of them, we arrive at
h i
χ(X) = χ(P1 )χ Q(2g) + (2g + 2) χ Σ2 Q(2g−1) + − χ Q(2g)
= 2(2g + 2) − (2g + 2) = 2g + 2
Collecting all the previous results, we have b2g+1 (X) = 2g and hence
3
p,q
p+q=2g+1 h (X)
P
= 2g.
π
We now consider the Leray spectral sequences associated to the map X −→ P1 and the sheaves
p
ΩX :
E2r,s (ΩpX ) = H r (P1 , Rs π∗ ΩpX ) V H r+s (X, ΩpX )
Notice that, since dim P1 = 1, these spectral sequences have only two columns, and they actually
degenerate at the E2 -step. Hence, for p, q ≥ 1,
hp,q (X) = hq (X, ΩpX ) = h0 (P1 , Rq p∗ ΩpX ) + h1 (P1 , Rq−1 p∗ ΩpX )
Moreover h0,q (X) = hq (X, OX ) = 0, the last equality coming from the long exact sequence in cohomology associated to the exact sequence of sheaves
0 → OP2g+1 ×P1 (−2, −1) → OP2g+1 ×P1 → OX → 0
We now investigate the sheaves Rs π∗ ΩpX for p > 0. Let U ⊂ P1 the complement of the 2g + 2
points over which the singular fibers are located; then π −1 (U ) ∼
= U × Q(2g) , so that
p
(Rs π∗ ΩpX )(U ) ∼
= H s U × Q(2g) , ΩU ×Q(2g)
M M
∼
H a U, ΩiU ⊗ H b Q(2g) , ΩjQ(2g)
=
a+b=s i+j=p
i
h p−1
p
∼
= H 0 (U, OU ) ⊗ H s Q(2g) , ΩQ(2g) ⊕ H s Q(2g) , ΩQ(2g)
From the Hodge numbers of Q(2g) , we see that Rs π∗ ΩpX = 0 unless p = s or p = s + 1. For the middle
dimension p + q = 2g + 1, we have h0,2g+1 (X) = h2g+1,0 (X); if 1 ≤ p ≤ 2g, then
hp,2g+1−p (X) = h0 (P1 , R2g+1−p p∗ ΩpX ) + h1 (P1 , R2g−p p∗ ΩpX )
The first term vanishes unless p = g + 1, while the second is nonzero only if p = g, that is, the
only nonzero Hodge numbers in the middle dimension are hg,g+1 (X) and hg+1,g (X), which have to
be equal because of complex conjugation. In sum,


1 if p = q = 0 or p = q = 2g + 1




2 if 1 ≤ p = q ≤ 2g
hp,q (X) =

g if p = g, q = g + 1 or p = g + 1, q = g




0 otherwise
We can finally compute the Hodge numbers of B:
have



1
hp,q (B) = hp+1,q+1 (X) − hp+1,q+1 (P2g+1 ) = g


0
4
since X is the blowup of P2g+1 along B, we
if 0 ≤ p = q ≤ 2g − 1
if p = g − 1, q = g or p = g, q = g − 1
otherwise
TP2. a) Let Q be the singular quadric hypersurface in P4 given by the equation x20 +x21 +x22 +x23 = 0.
Calculate its divisor class group, its Picard group and its cohomology with coefficients in O and C.
b) Compute the Picard group of Spec Op,Q , where p ∈ Q is the singular point. Use this to show that
p is not a quotient singularity.
e of Q along P is smooth and
c) Show that Q contains a plane P through p. Prove that the blow-up Q
answer the same questions.
d) Show that the blow-up of Q at p is a P1 -bundle over P1 × P1 and identify which one it is.
a) First of all, change coordinates in P4 :
y0 =
1
(x0 + ix1 ),
2
y1 =
1
(x0 − ix1 ),
2
1
(x2 + ix3 ),
2
y2 =
y3 =
1
(−x2 + ix3 ),
2
y4 = x4
Now, Q is the cone over X := V+ (y0 y1 − y2 y3 ) ⊂ P3 with vertex the point p = [0 : 0 : 0 : 0 : 1]. The
singular point is of codimension 3 in Q, so that Cl(Q) ∼
= Cl(Q − {p}). But Q − {p} ∼
= X × A1 , which
yields
Cl(Q) ∼
= Cl(Q − {p}) ∼
= Cl(X × A1 ) ∼
= Cl(X) ∼
=Z⊕Z
since X ∼
= P1 × P1 . Following the chain of isomorphisms, we find generators for Cl(Q):
P1 := V+ (y1 , y2 ) ⊂ Q
P2 := V+ (y1 , y3 ) ⊂ Q
Neither of these Weil divisors is Cartier, for Q is normal and they are not locally principal at
p. Indeed, mp,Q /m2p,Q ∼
= mp,P4 /m2p,P4 is a vector space of dimension 4 with basis the classes of
y0 , y1 , y2 , y3 ; the image of the prime ideal associated to P1 contains the classes of both y1 and y2 , so
it could not be generated by a single element, and similarly for P2 . However, P1 ∪ P2 = V+ (y1 ), so
that Z · (P1 + P2 ) ⊆ CaCl(Q). Suppose there were another Weil divisor in CaCl(Q)\Z · (P1 + P2 ).
Then, by taking an appropriate linear combination, we would have kP1 ∈ CaCl(Q) for some k ≥ 1.
To show that kP1 is not locally principal for any such k, look at
mkp,Q
mk+1
p,Q
∼
=
(mkp,P4 + q)/q
(mk+1
+ q)/q
p,P4
∼
=
mkp,P4
mkp,P4 ∩ (mk+1
+ q)
p,P4
∼
=
mkp,P4
mk+1
+ mkp,P4 ∩ q
p,P4
where q ⊂ Op,P4 is the ideal generated by y0 y1 − y2 y3 . It is clear that this is a vector space with
basis
{y0a y2b y3c | a, b, c ≥ 0, a + b + c = k} ∪ {y1a y2b y3c | a, b, c ≥ 0, a + b + c = k}
Again the image of the ideal associated to kP1 is of dimension strictly greater than one, proving
that CaCl(Q) = Z · (P1 + P2 ). Finally, since Q is integral, we have Pic(Q) ∼
= CaCl(Q) ∼
= Z.
The cohomology with coefficients in O can be easily deduced from the exact sequence
0 → OP4 (−2) → OP4 → OQ → 0
5
We obtain H 0 (Q, OQ ) ∼
= C and H i (Q, OQ ) ∼
= 0 for i > 0.
Since Q − {p} ∼
= X × A1 , we can identify Q with the Thom space of a trivial (real) bundle of
1
rank 2 over P × P1 , so that Q ' Σ2 (P1 × P1 )+ . From this the Betti numbers of Q follow: b0 = 1,
b1 = 0, b2 = 1, b3 = 0, b4 = 2, b5 = 0 and b6 = 1.
b) First of all, observe that a prime divisor in Spec Op,Q , i.e., a height one prime ideal of Op,Q ,
corresponds to a prime divisor on Q that passes through p. Let ι : Spec Op,Q → Q be the canonical
inclusion. If P is a prime divisor in Q, then define
(
ι−1 (P ) if p ∈ P
∗
ι (P ) :=
0
if p 6∈ P
and extend linearly to a map ι∗ : Div(Q) → Div(Spec Op,Q ). Moreover, if f ∈ K(Q)∗ , then


X
X
ι∗ (f ) = ι∗ 
vP (f ) · P  =
vP (f ) · ι−1 (P )
P prime
P prime
p∈P
is the principal divisor associated to f considered as an element of K(Spec Op,Q )∗ = K(Q). Hence
ι∗ descends to a map of divisor class groups, which we still denote ι∗ . The surjectivity of the these
maps is clear: if
X
D=
nP · ι−1 (P )
P prime
p∈P
e = D for
then ι∗ (D)
e=
D
X
nP · P
P prime
p∈P
Since P4 = V+ (y4 ) ∩ Q ∈ Div(Q) does not pass trough the singular point, we have that
ι∗ (y1 /y4 ) = ι∗ (P1 + P2 − P4 ) = ι−1 (P1 ) + ι−1 (P2 )
is a principal divisor in Div(Spec Op,Q ). Furthermore, if (f ) = k · ι−1 (P1 ) then f ∈ OQ (U ) for some
open neighborhood U of p in Q, in contradiction with the fact that kP1 is not locally principal at
the singular point. In conclusion, Cl(Spec Op,Q ) = Z · P1 .
We now study the divisor class group of a quotient singularity. Let G be a finite subgroup
of SU (3) acting on C3 ; the C-algebra homomorphism C[x, y, z]G ,→ C[x, y, z] corresponds to the
canonical quotient map π : C3 → C3 /G. We can mimic the above construction to define a map
π ∗ : Div(C3 /G) → Div(C3 ) which also descends to a homomorphism of divisor class groups: if P be
a prime divisor in C3 /G passing through the origin, then π ∗ (P ) is a Weil divisor which is invariant
under the action of G. Since Cl(C3 ) = 0 (C[x, y, z] is a UFD), there exists f ∈ K(C3 ) such that
6
Q
(f ) = π ∗ (P ). But then Gf := g∈G gf is G-invariant and descends to a rational function on the
quotient; hence (Gf ) = |G| · P ∈ Div(C3 /G) and Cl(C3 /G) is torsion. Reasoning as above, we can
also show that the divisor class group of the local scheme at the singularity is a quotient of the latter
group, hence also torsion. This proves that the singular point of Q is not a quotient singularity.
c) We have already seen that Q contains two planes through the singular point, namely P1 and P2 .
e of Q along P1 can be constructed as the proper transform of Q with respect to the
The blow-up Q
4
blow-up of P along P1 . The latter can be constructed as follows: there is a surjection of OP4 [u, v]
L
onto the blow-up algebra k≥0 I k , where I = y1 · OP4 (−1) + y2 · OP4 (−1) and the map takes u and
v to y1 and y2 , respectively, considered as elements of degree one; the kernel of this map is generated
by the element uy2 − vy1 , so that
M
BlP1 P4 := Proj
Ik ∼
= Proj OP4 [u, v]/(uy2 − vy1 )
k≥0
= V+ (uy2 − vy1 ) ⊂ P4 × P1
The total transform of Q is then V+ (y0 y1 − y2 y3 , uy2 − vy1 ) ⊂ P4 × P1 . Looking at the inverse
image of D+ (y0 ) ∩ Q ⊂ P4 under the blow-up map π we can differentiate the component along the
e of Q:
exceptional divisor E ∼
= P2 × P1 from the proper transform Q
π −1 D+ (y0 ) ∩ Q = V (y10 − y20 y30 , uy20 − vy10 ) = V y10 − y20 y30 , y20 (u − vy30 ) ⊂ D+ (y0 ) × P1
Thus,
e = V+ (y0 y1 − y2 y3 , uy2 − vy1 , uy0 − vy3 ) ⊂ P4 × P1
Q
The smoothness can be checked affine-locally.
e with coefficients in O, notice that E ∪ Q
e is a complete
In order to calculate the cohomology of Q
4
1
intersection in P × P , so that we have the exact sequence
0 → OP4 ×P1 (−3, −1) → OP4 ×P1 (−2, −0) ⊕ OP4 ×P1 (−1, −1) → OP4 ×P1 → OE∪Qe → 0
The cohomology of P4 × P1 with coefficients in OP4 ×P1 (a, b) follows immediately from that of P4
e O e) ∼
and P1 and the Künneth formula. A short calculation yields H 0 (E ∪ Q,
C and H i (E ∪
E∪Q =
e O e ) = 0. Similarly, E ∩ Q
e is the vanishing locus of uy0 − vy3 in P2 × P1 with coordinates
Q,
E∪Q
[y0 : y3 : y4 ] × [u : v], so that
0 → OP2 ×P1 (−1, −1) → OP2 ×P1 → OE∩Qe → 0
e O e) ∼
e O e ) = 0. Finally E ∼
and H 0 (E ∩ Q,
C and H i (E ∩ Q,
= P2 × P1 gives H 0 (E, OE ) ∼
=C
E∩Q =
E∩Q
i
and H (E, OE ) = 0. Plugging this information in the exact sequence
0 → OE∪Qe → OE ⊕ OQe → OE∩Qe → 0
e O e) ∼
e O e ) = 0.
C and H i (Q,
we obtain H 0 (Q,
Q =
Q
7
e are all smooth, we have the following diagram of vector bundles on
Since P4 × P1 , BlP1 P4 and Q
e
Q:
0
0
/ Te
Q
TQe
/ 0
0
/ T
e
BlP1 P4 |Q
/ T 4 1| e
P ×P Q
/ N
e
BlP1 P4 \P4 ×P1 |Q
/ 0
/ Ne
Q\BlP1 P4
/ Ne 4 1
Q\P ×P
/ N
4
e
BlP1 P \P4 ×P1 |Q
/ 0
0
We know NBlP
1
0
P4 \P4 ×P1
0
0
0
∼
= OP4 ×P1 (1, 1)|BlP1 P4 and NQ\Bl
e
P
1
P4
∼
= OP4 ×P1 (2, 0)|Qe , so that
∼
NQ\P
e 4 ×P1 = OP4 ×P1 (2, 0) ⊕ OP4 ×P1 (1, 1) e
Q
From the middle column of the diagram above, we obtain
c TP4 ×P1 |Qe = c TQe c NQ\P
e 4 ×P1
(1 + ω)5 (1 + η)2 = (1 + c1 + c2 + c3 )(1 + 2ω)(1 + ω + η)
e and
where ci are the Chern classes of the tangent bundle of Q
η = c1 OP4 ×P1 (0, 1) ω = c1 OP4 ×P1 (1, 0) ,
e
Q
e
Q
We calculate
c1 = (2ω + η),
and χ =
R
e c3
Q
c2 = ω(2ω + 3η),
c3 = 3ω 2 η
= 6.
e = 1, b1 (E ∪ Q)
e = 0 and
The Lefschetz hyperplane theorem yields the Betti numbers b0 (E ∪ Q)
e
e
b2 (E ∪ Q) = 2. Moreover, we also know all of the Betti numbers of E and E ∩ Q. We will use a
e The following table contains all the
Mayer-Vietoris argument to compute the Betti numbers of Q.
information that we have so far.
8
i
e
bi (E ∪ Q)
0
1
2
3
4
5
6
1
0
2
x
·
·
·
e
bi (E q Q)
1
0
2
0
2
0
1
+
+
+
+
+
+
+
b0
b1
b2
b3
b2
b1
b0
e
bi (E ∩ Q)
1
0
2
0
1
0
0
e From the first two rows, it is immediate
Notice that we have also made use of Poincaré duality for Q.
that b0 = 1 and b1 = 0. Moreover, from the third and fourth rows we have b2 = 2 − x + b3 and x ≤ b3 .
Equating the value of the Euler characteristic that we calculated before with the one obtained from
the table, we arrive at the condition b3 = 2x. The only possible solution is then x = 0, b2 = 2 and
b3 = 0.
e Z) ∼
More precisely, the Lefschetz hyperplane theorem gives H 2 (E ∪ Q,
= Z ⊕ Z; since we also have
2
2
∼
e
e
H (E ∩ Q, Q) = Z ⊕ Z, it is clear from the table that H (Q, Z) is also torsion-free. We then deduce
e ∼
e is smooth, its divisor class
from the exponential sequence that Pic(Q)
= Z ⊕ Z. Moreover, since Q
group is isomorphic to its Picard group.
d) First consider the blow-up of C4 at the origin: it can be described as the set {(p, `) ∈ C4 × P3 :
p ∈ `}, i.e., as the total space of the tautological line bundle OP3 (−1) over P3 :
Bl0 C4 ∼
= Tot OP3 (−1) := Spec Sym OP3 (1)
The blow-up of P4 at the origin is then a fiberwise compactification of the latter, hence a P1 -bundle
over P3 . Any such bundle is the projectivization of a rank two vector bundle E on P3 , which always
splits as a sum of line bundles. To prove the latter fact, we may assume without loss of generality
that h0 (E) > 0: otherwise take a big enough twist, which can be undone at the end. If s : OP3 → E
is a non-zero global section, the image of its dual s∨ : E ∨ → OP3 is an ideal sheaf OP3 (−D) for some
effective divisor D and its kernel L is locally free of rank one; dualizing we obtain a short exact
sequence
0 → OP3 (D) → E → L∨ → 0
Since Ext1 (L∨ , OP3 (D)) ∼
= H 1 (P3 , L(D)) = 0, the extension is split.
Write then
π
∨
3
Bl0 P4 ∼
= P(L1 ⊕ L2 ) := Proj Sym(L∨
1 ⊕ L2 ) −−−−→ P
−1
−1
For 0 ≤ i ≤ 3, let f1,i
and f2,i
be local generators of L1 and L2 , respectively, as OP3 -modules over
3
the affine open D+ (zi ) ⊂ P . We then have isomorphisms
∼
=
∨
OP3 D+ (zi ) [ui , vi ] −−−−→ Sym(L∨
1 ⊕ L2 )(D+ (zi ))
9
sending
∨
ui 7→ f1,i ∈ Sym(L∨
1 ⊕ L2 )(D+ (zi ))
1
∨
∨
vi 7→ f2,i ∈ Sym(L1 ⊕ L2 )(D+ (zi ))
1
Suppose that the section of π given by the proper transform of the hyperplane at infinity V+ (y4 ) is
∨
∨
given by the canonical projection L∨
1 ⊕ L2 → L1 . In the notation just introduced, this corresponds
to the vanishing locus of the vi over each D+ (zi ). The complement of this section is given locally
∨
∼
by Spec OP3 D+ (zi ) [ui /vi ]. These pieces glue together to Spec Sym(L∨
1 ⊗ L2 ) , so that L1 ⊗ L2 =
4 ∼
OP3 (1). Since P(E) ∼
= P(E ⊗ L∨
2 ), we conclude that Bl0 P = P(OP3 (−1) ⊕ OP3 ). The blow-up of Q at
the origin is the proper transform of Q under π, i.e., Bl0 Q ∼
= P(OP1 ×P1 (−1, −1) ⊕ OP1 ×P1 ).
RD2. Analyze the plane curve C given by y 2 = f (x) =
arithmetic genus, singular points, . . . ).
Q10
i=1 (x
− i)i (calculate its geometric and
Q
i
Let C̄ be the projective closure of C, with equation Y 2 Z 53 − 10
i=1 (X − iZ) = 0. Its arithmetic
1
genus follows from the degree-genus formula: pa (C̄) = 2 (55 − 1)(55 − 2) = 1431.
The singular points of C are the solutions of the equations
0 = y 2 − f (x),
0=
∂ 2
y − f (x) = −f 0 (x),
∂x
0=
∂ 2
y − f (x) = 2y
∂y
These conditions are satisfied at the points (m, 0) such that f has a multiple root x = m; this happens
when m = 2, . . . 10. In an infinitesimal neighborhood of each of these points, we can redefine the x
coordinate to put the equation in the form y 2 = xm . Resolving these singularities, we see that the
normalization N of C has two points over each of the singular points for which m is even, and only
one for those for which m is odd.
The intersection of C̄ with the hyperplane at infinity Z = 0 is a highly singular point. Instead
of trying to resolve that singularity, we perform a clever change of coordinates in the affine plane,
valid for x 6= 0: we define z = x−1 and w = x−28 y; the equation of C then takes the form
w2 − z
10
Y
(1 − iz)i = 0
i=0
Adding the point z = w = 0 to C compactifies it; notice, in particular, that the point added is
nonsingular.
In summary, N is a 2-to-1 cover of P1 ramified over the points x = 1, 3, 5, 7, 9, ∞. By the
Riemann-Hurwitz formula,
2g(N ) − 2 = 2(−2) + 6
10
=⇒
g(N ) = 2
TP3. Classify all singular quadric hypersurfaces in Pn . Describe the singularities, resolve them and
describe the corresponding resolutions.
Since the only invariant of a quadratic form on Cn+1 is its rank, there are n + 1 isomorphism
classes of quadric hypersurfaces in Pn . A complete set of class representatives is given by
Qk,n = V+ (x20 + · · · + x2k−1 ) ⊂ Pn
where k = 1, . . . , n + 1 is the rank of the associated quadratic form. It is clear that a quadric
hypersurface is smooth if and only if its rank is maximal — in this case, if k = n + 1.
A singular quadric of rank 1 is a double hyperplane; in this case, all points are singular and a
resolution is provided by that same hyperplane with its reduced scheme structure. If k = 2, we have
the union of two hyperplanes intersecting transversely in a linear subvariety of dimension n − 2,
whose resolution consists of two disjoint copies of Pn−1 .
For 3 ≤ k ≤ n, Qk,n is a cone over a smooth quadric hypersurface in V+ (xk , . . . , xn ) ∼
= Pk−1 with
vertex S := V+ (x0 , . . . , xk−1 ) ∼
= Pn−k . To perform a resolution of singularities, we blow up Pn along
S:
BlS Pn = V+ {xi uj − xj ui }0≤i<j≤k−1 ⊂ Pn × Pk−1
TO DO: prove smoothness and describe these resolutions.
RD3. Look at the stratification of the theta divisor of a curve C by singularity type, and define a
stratification of the moduli space Mg by this data. Work this out for g ≤ 6.
TP4. Consider a generic pencil of plane curves of degree d. Show that its singular elements have a
unique singular point, and that it is an ordinary double point. Count how many of them there are.
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