Problem Set 10
Math 205, Fall 2016
1. Let k be a field, and let f be a homogeneous polynomial of positive degree in X0 , · · · , Xn .
Give a precise definition of an open subscheme D(f ) of Pnk , analogous to the definition of
principal open subsets of affine schemes. Prove that
p
∼
D(f ) = Spec k ` : ` ∈ N, p homogeneous with deg p = ` deg f .
f
In particular, D(f ) is affine. Give a criterion for one such open set to contain another, and
for a collection of such open sets to be an open cover of another.
2. Let k be a field, and let S be the ring k[X0 , X1 , · · · , Xn ].
(a) Let I ⊂ S be a homogeneous ideal. Show that there exists an ideal sheaf s(I) on Pnk
such the sections of s(I) on D(f ) are given by
a
: ` ∈ N, a ∈ I, a homogeneous with deg a = ` deg f .
f`
(b) For any ideal sheaf I on Pnk , let t(I) denote the subset of S spanned over k by homop
geneous polynomials p such that deg
p ∈ I(D(Xi )) for all i ∈ {0, 1, · · · , n}. Show that
xi
t(I) is a homogeneous ideal.
(c) Show by example that if I is a homogeneous ideal, then t(s(I)) need not be equal to I.
(d) Show, however, that I and t(s(I)) agree in sufficiently high degree, i.e. there exists an
integer m0 such that for all m ≥ m0 , Im = t(s(I))m (here Im denotes the degree-m
elements of I).
Note. The ideal t(s(I)) is called the saturation of the ideal I. Two homogeneous ideals define
the same subscheme of Pnk if and only if they have the same saturation.
Remark. In the following exercise, you may use the following fact without proof (see e.g.
Atiyah-MacDonald Corollary 11.2): If I is any homogeneous ideal in S = k[X0 , · · · , Xn ],
then the function hI (m) = dimk (S/I)m is a polynomial for sufficiently large m (i.e. there
exists an integer m0 and a polynomial pI (m) such that hI (m) = pI (m) for all m ≥ m0 ). If
X = V (I) is the closed subscheme defined by I, this polynomial will also be denoted pX (m),
and called the Hilbert polynomial of X. By part (d) of problem 2, there is no ambiguity about
pX , even though different homogeneous ideals may define the same scheme X.
3. (a) Let k, S be as in the previous problem. Prove that if f ∈ S is a homogeneous polynomial
of degree d, whose class in S/I is not a zero-divisor, then for all m,
pI+(f ) (m) = pI (m) − pI (m − d).
(b) Let f, g ∈ k[X, Y, Z] be homogeneous polynomials in three variables, with no common
factors. Let C = V (f ) and D = V (g). Show that pC∩D (m) = (deg f )(deg g) for all
integers m (the Hilbert polynomial of the intersection is constant).
(c) Deduce that if k is algebraically closed and C ∩ D is a disjoint union of reduced points,
then there are exactly (deg f )(deg g) such points.
4. (Examples of Bézout’s theorem, over R)
due Wednesday, December 7.
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Problem Set 10
Math 205, Fall 2016
(a) Let r be a positive real number. Consider the following two subchemes of P2R , whose
R-points consist of circles of radius r centered at (±1, 0).
C1 = V ((X + Z)2 + Y 2 − r2 Z 2 )
C2 = V ((X − Z)2 + Y 2 − r2 Z 2 )
Show that the schematic intersection C1 ∩ C2 is isomorphic to one of the following three
schemes: Spec C t Spec R t Spec R, Spec C t Spec C, or Spec C t Spec R[ε]/ε2 .
(b) For i = 1 and i = 2, let ai , bi , ci be real numbers with ai 6= 0. Let Pi be the scheme
V (ai X 2 + bi XZ + ci Z 2 − Y Z) ⊂ P2R . Describe the R-points of these schemes. List all
possible isomorphism classes for the schematic intersection P1 ∩ P2 . Show in particular
that this intersection is never reduced.
5. Let f, g ∈ C[x, y] be irreducible polynomials in two variables, with neither dividing the other.
Suppose that f (0, 0) = g(0, 0) = 0. Let C = V (f ) and D = V (g), considered as subschemes
of P2C . Let Z denote the schematic intersection C ∩ D. Let p denote the closed point of P2C
corresponding to the origin (0, 0). Show that the following conditions are equivalent.
(a) Z is reduced at p.
(b) dimC OZ,p = 1 (i.e. the curves meet with multiplicity 1 at the origin).
(c) The determinant of fgxx fgyy is nonzero when evaluated at (x, y) = (0, 0). (Geometrically,
this condition means that the tangent lines to C and D are distinct at the origin, i.e.
the curves meet transversely.)
6. Find an R-scheme X such that XC = Spec C ×R X is isomorphic to P1C (i.e. X becomes the
projective line after base change), but such that X ∼
6 P1R .
=
(More generally, a scheme over a field k that becomes isomorphic to PnK after base change to
some field extension K/k is called a Brauer-Severi variety.)
due Wednesday, December 7.
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