Algebraic Geometry, PS 4

Bryden Cais

October 30, 2002

1 Lecture 7

Exercise 1: Let X be a projective d subspace in ℙ_{k}^{n}. By definition, X is given by n − d linearly independent linear equations. By a projective change of coordinates, X is isomorphic to the projective subspace of ℙ_{k}^{n} given by the linear equations

{T_{0} = 0, T_{1} = 0, …, T_{n−d} = 0}.

Then projection onto the last d + 1 coordinates gives an isomorphism of X with ℙ_{k}^{d}, as required. Exercise 2: Consider the Segre embedding s(1, 1)_{K}(ℙ_{k}^{1}(K) × ℙ_{k}^{1}(K)) → ℙ_{k}^{3}(K) for any K. It is defined by

s(1, 1)_{K}([X_{0}, X_{1}], [Y_{0}, Y_{1}]) = [X_{0}Y_{0}, X_{0}Y_{1}, X_{1}Y_{0}, X_{1}Y_{1}] = [Z_{0}, Z_{1}, Z_{2}, Z_{3}].

Let Q be the hypersurface in ℙ_{k}^{3} given by

Z_{0}Z_{3} − Z_{1}Z_{2} = 0  (1)

It is clear that the image of ℙ_{k}^{1} × ℙ_{k}^{1} under s(1, 1)_{K} is contained in Q since any point in the image satisfies 1. We show that the two are in fact isomorphic. Define f : Q → ℙ_{k}^{1} × ℙ_{k}^{1} by

f([Z_{0}, Z_{1}, Z_{2}, Z_{3}]) = \begin{cases}(1, Z_{2}/Z_{0}), (1, Z_{1}/Z_{0}) & if Z_{0} ≠ 0 \\ (1, Z_{3}/Z_{1}), (Z_{0}/Z_{1}, 1) & if Z_{1} ≠ 0 \\ (Z_{0}/Z_{2}, 1), (1, Z_{3}/Z_{2}) & if Z_{2} ≠ 0 \\ (Z_{1}/Z_{3}, 1), (Z_{2}/Z_{3}, 1) & if Z_{3} ≠ 0 \end{cases} .

It is not difficult to check that f is well defined: one need only observe that on the patches of ℙ_{k}^{3} where more than one Z_{i} is nonzero the different definitions of f all agree. For example, if Z_{0}, Z_{1} ≠ 0 we have

[1, Z_{2}/Z_{0}], [1, Z_{1}/Z_{0}] = [Z_{0}, Z_{2}], [Z_{0}, Z_{1}]
= [Z_{0}Z_{1}, Z_{2}Z_{1}], [Z_{0}, Z_{1}]
= [Z_{0}Z_{1}, Z_{0}Z_{3}], [Z_{0}, Z_{1}]
= [Z_{1}, Z_{3}], [Z_{0}, Z_{1}]
= [1, Z_{3}/Z_{1}], [Z_{0}/Z_{1}, 1].

The other compatibility computations are identical, and we omit them. Observe that

s(1, 1)_{K} ∘ f = id_{f}

and

f ∘ s(1, 1)_{K} = id_{ℙ_{k}^{1}×ℙ_{k}^{1}} .

Since both f, s(1, 1)_{K} are morphisms, this shows that they are isomorphisms and ℙ_{k}^{1} × ℙ_{k}^{1} ≃ Q. Now suppose that k is an algebraically closed field and let

F (T_{0}, T_{1}, T_{2}, T_{3}) = (T_{0}, T_{1}, T_{2}, T_{3}) (\begin{matrix}a_{00} & a_{01} & a_{02} & a_{03} \\ a_{10} & a_{11} & a_{12} & a_{13} \\ a_{20} & a_{21} & a_{22} & a_{23} \\ a_{30} & a_{31} & a_{32} & a_{33} \end{matrix}) (\begin{matrix}T_{0} \\ T_{1} \\ T_{2} \\ T_{3} \end{matrix})

be a symmetric, nondegenerate quadratic form (that is, a_{ij} = a_{ji} and the matrix (a_{ij}) is nonsingular). Since k is algebraically closed and not of characteristic 2, we know that every symmetric, nonsingular matrix is equivalent (by elementary row and column operations) to the identity matrix. Moreover, by a projective change of coordinates (projective automorphism) on the form F, we can perform any row and column operations on the matrix (a_{ij}). It follows that the hypersurface defined by F is (via an appropriate projective change of coordinates) isomorphic to Q, which is given by the symmetric, nonsingular matrix

(\begin{matrix}0 & 0 & 0 & 1/2 \\ 0 & 0 & −1/2 & 0 \\ 0 & −1/2 & 0 & 0 \\ 1/2 & 0 & 0 & 0 \end{matrix}) .

Since Q is isomorphic to ℙ_{k}^{1} × ℙ_{k}^{1} by the above, we see that every hypersurface given by an equation of the form F is isomorphic to ℙ_{k}^{1} × ℙ_{k}^{1}. The projection maps p_{i} : Q → ℙ_{k}^{1} for i = 1, 2 are given as the first and second components (respectively) of our function f above. Exercise 3: Recall that Ver_{1}^{n} is the image of ℙ_{k}^{1} under the map

v_{1,n}([X_{0}, X_{1}]) = [X_{0}^{n}, X_{0}^{n−1}X_{1}, …, X_{1}^{n}] := [Z_{0}, Z_{1}, …, Z_{n}] ∈ ℙ_{k}^{n}.

Since Z_{i} = X_{0}^{n−i}X_{1}^{i}, we observe that any point of Ver_{1}^{n} satisfies Z_{i}Z_{j} = Z_{k}Z_{l} for i + j = k + l. Moreover, since v_{1,n} is a morphism, we know that the image Ver_{1}^{n} is closed in ℙ^{n}. Now let C be the affine curve given by

{Z_{i} − Z_{1}^{i} = 0}_{i=2}^{n} .

Observe that for any point (1, Z_{1}, Z_{2}, …, Z_{n}) of C we have

Z_{i}Z_{j} = Z_{1}^{i+j},

which only depends on i + j. Therefore, C is contained (via the usual inclusion of affine space into projective space) in Ver_{1}^{n} since the defining ideal of Ver_{1}^{n} vanishes on C. Observe that C is not a closed subset of ℙ_{k}^{n}. Indeed, any point of C satisfied the equations

Z_{i}Z_{j} = Z_{k}Z_{l}

for i + j = k + l. However, (0, 0, …, 1) also satisfies these equations but is not a point of C. On the other hand, any point of Ver_{1}^{n}(K) with Z_{0} ≠ 0 has the form

(1, t, t^{2}, …, t^{n})

for t ∈ K, while there is only one point with Z_{0} = 0, the point (0, 0, …, 1). It is clear that any affine point of Ver_{1}^{n}(K) is also a point of C. It follows that Ver_{1}^{n}(K) = C ∪ (0, 0, …, 1). Since C is not closed in ℙ_{k}^{n} and Ver_{1}^{n} is, it follows that the projective closure of C is Ver_{1}^{n}. Exercise 6:

1. Let M be the set of (n+1)×(n+1) rank 1 matrices, up to proportionality. We know that the image S of ℙ_{k}^{n}(K)′′ × ℙ_{k}^{n}(K)′′ under the Segre embedding is isomorphic to ℙ_{k}^{n}(K)′′ × ℙ_{k}^{n}(K)′′. Any point of S has the form

(X_{i}Y_{j})_{i,j=0}^{n}

with (X_{0}, …, X_{n}), (Y_{0}, …, Y_{n}) ∈ ℙ_{k}^{n}. Therefore, we define

f : S → M

by

f(X_{0}Y_{0}, …, X_{n}Y_{n}) = (X_{i}Y_{j}).

We must check that matrix (X_{i}Y_{J}) has rank 1. Certainly, the rank of (X_{i}Y_{j}) is at least 1 since not all X_{i}Y_{j} are 0 (this would imply that X_{i} = 0, Y_{i} for all i, which is impossible since (X_{0}, …, X_{n}), (Y_{0}, …, Y_{n}) are points of projective space). However, (X_{i}Y_{j}) has at most rank 1 since the determinant of any 2 × 2 minor is

X_{i}Y_{j} · X_{(}i + r)Y_{(}j + s) − X_{(}i + r)Y_{j} · X_{i}Y_{(}j + s) = 0,

for any 1 ≤ r ≤ n − i and 1 ≤ s ≤ n − j. Therefore, the matrix (X_{i}Y_{j}) indeed has rank 1. Our map f is thus well defined. It is clear that f is injective, for X_{i}Y_{j} = X′_{i}Y′_{j} for all i, j implies that X_{i} = X′_{i}, Y_{j} = Y′_{j}. Surjectivity is similarly clear: any 2 × 2 minor of a rank 1 (n+1)×(n+1) matrix has determinant 0, so that the entries of our matrix (when considered as a lexographically ordered tuple) satisfy the defining equations of S and hence gives a point of S.

2. We know that any point of Ver_{n}^{2}(K)′′ can be written

(Z_{ij}) := (X_{i}X_{j})

for i, j = 0 … n and (X_{0}, …, X_{n}) ∈ ℙ_{k}^{n}(K)′′. We therefore define

f : Ver_{n}^{2}(K)′′ → M

(where M is the set of rank 1 symmetric square matrices with coefficients in K of size m = \binom{n+2}{2} up to proportionality) by

f(Z_{ij}) = (Z_{ij})

(where we clearly mean the matrix with entries Z_{ij} on the right and the point Z_{ij} on the left). This is well defined since both the point (Z_{ij}) and the matrix (Z_{ij}) are defined up to proportionality and the symmetric matrix (Z_{ij}) has rank 1 since it certainly does not have rank 0 (since not all entries are 0) but we have

Z_{ij}Z_{(i+r)(j+s)} − Z_{(i+r)j}Z_{i(j+s)} = X_{i}X_{j}X_{i+r}X_{j+s} − X_{i+r}X_{j}X_{i}X_{j+s} = 0

so that every 2 × 2 minor has determinant 0. As in part a), it is clear that f is injective. Surjectivity follows similarly since any 2 × 2 minor of a rank 1 symmetric matrix must have determinant 0, that is, any such matrix, when considered as a lexographically ordered tuple (that is, (a_{ij}) ↦ (a_{i}j) with 0 ≤ i ≤ j ≤ n) satisfies the defining equations of Ver_{n}^{2}(K)′′.

2 Lecture 8

Exercise 1: Let S = {(a, b, c) ∈ ℙ^{2}(K) : a ≠ 0, b ≠ 0}∪{(1, 0, 0)}. If S = Z_{1}\Z_{2} for some projective algebraic sets Z_{1}, Z_{2}, then for any copy of affine space 𝔸^{2}(K) ⊂ ℙ^{2}(K) we have S ∩ 𝔸^{2}(K) = Z_{1} ∩ 𝔸^{2}(K)\Z_{2}∩𝔸^{2}(K), and Z_{i}∩𝔸^{2}(K) is a closed (in the subspace topology) set in 𝔸^{2}(K). Therefore, it is enough to show that S∩𝔸^{2}(K) is not of the form Z_{1}\Z_{2} for affine algebraic sets Z_{1}, Z_{2}. We choose the copy of 𝔸^{2}(K) corresponding to a = 1. Then T := S ∩𝔸^{2}(K) = {(b, c) ∈ 𝔸^{2}(K) : b ≠ 0}∪(0, 0). Since (by a previous problem set) T is dense in 𝔸^{2}(K), we must have Z_{1} = 𝔸^{2}(K). But this requires that Z_{2} = {(b, c) ∈ 𝔸^{2}(K) : b = 0} \ (0, 0), which is not closed since any polynomial vanishing on the line b = 0 must also vanish at the origin (a polynomial restricted to a line is a polynomial in one variable and hence has finitely many roots unless it is identically 0 on the line). Thus, T is not locally closed, and it follows that S is not quasi projective. Exercise 4: The union of quasi projective algebraic sets is not necessarily quasi projective. For example, the sets S_{1} = {(a, b, c) ∈ ℙ^{2}(K) : a ≠ 0, b ≠ 0} and S_{2} = (1, 0, 0) are both quasi projective, but their union is, by problem 1, not quasi projective. Any finite intersection of quasi projective sets is again quasi projective. For if S_{1}, S_{2} are quasi projective, we may write S_{i} = Z_{1i} \ Z_{2i} for closed sets Z_{ij} and i, j = 1, 2. It follows that

S_{1} ∩ S_{2} = {Z_{11} \ Z_{21}} ∩ {Z_{12} \ Z_{22}} = (Z_{11} ∩ Z_{21}) \ (Z_{21} ∪ Z_{22})

is quasi projective since the intersection and union of closed (algebraic) sets is again closed. However, an arbitrary intersection of quasi projective varieties is not necessarily quasi projective. For example,

⋂_{x∈𝔸^{1}(K)\0}{(b, c) ∈ 𝔸^{2}(K) : c ≠ x}

is an arbitrary intersection of quasi projective sets that is (by problem 1) not quasi projective. Exercise 5: Let us find the irreducible components of the projective subset S of ℙ^{3}(K)given by T_{2}T_{0} − T_{1}^{2} = 0, T_{1}T_{3} − T_{2}^{2} = 0. Suppose first that T_{0} ≠ 0. Then if T_{1} = 0, we must also have T_{2} = 0, while there are no restrictions on T_{3}. This gives the set of points (1, 0, 0, t) for t ∈ K. If T_{1} ≠ 0, then we have T_{2} = T_{1}^{2} and T_{1}T_{3} = T_{1}^{4}, so we get the points (1, t, t^{2}, t^{3}) for t ∈ K. If T_{0} = 0, we get the point (0, 0, 0, 1). It is clear from this that S is the union of two irreducible curves given by

C_{1} :{(1, 0, 0, t) : t ∈ K} ∪ {(0, 0, 0, 1)}
C_{2} :{(1, t, t^{2}, t^{3}) : t ∈ K} ∪ {(0, 0, 0, 1)},

or given as projective curves by the equations

{Z_{1} = 0, Z_{2} = 0}
{Z_{1}^{2} − Z_{2}Z_{0} = 0, Z_{1}^{3} − Z_{3}Z_{0}^{2} = 0, Z_{2}^{3} − Z_{0}Z_{3}^{2}}.

Exercise 8: We first show that a regular map is continuous. To do this, we must show that the inverse image of a closed set is closed. We know that a regular map is given locally by polynomials. Let f : X → Y be a regular map of quasi projective varieties and let C be a closed set in Y given by the polynomials g_{s}. Let U_{i} be a finite covering of X by open sets. Then f|_{U_{i}} is just a polynomial map, for each i, so set f|_{U_{i}} = f_{i}. Set V_{i} = f^{−1}(C) ∩ U_{i}. Then it is clear that V_{i} is the quasi projective algebraic set defined by g_{s} ∘ f_{i}. For these polynomials certainly vanish on V_{i}. On the other hand, if they vanished at some point not in V_{i} then g_{s} would vanish at a point not in C, a contradiction. This shows that V_{i} is closed. The union of all V_{i} is f^{−1}(C), and since this is a finite union, we see that f is continuous.

Now let us show that the composition of regular maps is regular. Let f : X → Y and g : Y → Z be regular maps of quasi projective varieties. Let U_{i}, W_{j} be finite coverings of X, Y by open sets such that f|_{U_{i}} := f_{i}, g|_{W_{j}} := g_{j} are polynomial maps. Set

V_{ij} = f^{−1}(W_{j}) ∩ U_{i}.

Since f is continuous and W_{j} is open, we see that V_{ij} gives an open cover of X. Moreover, f|_{V_{ij}} = f_{i} and g|_{W_{j}∩f_{i}(U_{i})} = g_{j}. It follows that

g ∘ f|_{V_{i}j} = f_{i} ∘ g_{j}

is the composition of two polynomial maps and hence given by polynomials. It follows that g ∘ f is regular.

3 Lecture 9

Exercise 1: We show that the set of all degree m homogenous polynomials that split into the product of a degree k and a degree m − k homogenous polynomial is closed. It will then follow that the set of degree m reducible homogenous polynomials is closed since it is a finite union of closed sets. Under the Veronese embedding, v_{n}^{m} hypersurfaces of degree m in ℙ^{n} correspond to points of ℙ^{N_{m}}, where N_{m} = \binom{n+m}{m} − 1, since such a hypersurface is just a linear subspace (two homogenous degree m polynomials give the same hypersurface if and only if they are proportional) of 𝔸^{N_{m}}. Multiplication of polynomials gives a regular map (given by appropriate equations in the coefficients)

f : ℙ^{N_{k}} × ℙ^{N_{m−k}} → ℙ^{N_{m}}

where N_{i} = \binom{n+i}{i} − 1 as above. Clearly, the set of reducible homogenous degree m polynomials that split as a product of a degree m − k and a degree k homogenous polynomial is the image of f. By the Segre embedding, we have seen that ℙ^{N_{k}} × ℙ^{N_{m−k}} is a projective variety. By Theorem 1 of Lecture 9, the image of f is closed since f is regular and ℙ^{N_{k}} × ℙ^{N_{m−k}} is a projective variety and we are done.

Now let n = d = 2. We show that a degree 2 homogenous polynomial

F = ∑ a_{ij}T_{i}T_{j}

in three variables T_{0}, T_{1}, T_{2} is reducible if and only if det(a_{ij}) = 0. Since we are working over an algebraically closed field, we may apply a projective change of coordinates to put F in “diagonal form”. It is not difficult to see that F is irreducible if and only if the new, diagonalized form is irreducible. But there are only three possibilities for this form: T_{i}^{2}, T_{i}^{2} +T_{j}^{2}, T_{0}^{2} +T_{1}^{2} +T_{2}^{2}. Observe that since we are working over an algebraically closed field, the first two forms are reducible. Indeed, this shows that a quadratic form F is reducible if and only if it is equivalent to one of the first two forms above (since T_{0}^{2} + T_{1}^{2} + T_{2}^{2} is clearly not reducible). But F is equivalent to one of these forms if and only if the matrix (a_{ij}) does not have maximal rank; that is, if and only if the determinant of (a_{ij}) is 0. It follows that the projective variety of degree two reducible polynomials in three variables is given by

|\begin{matrix}a_{00} & a_{01} & a_{02} \\ a_{10} & a_{11} & a_{12} \\ a_{20} & a_{21} & a_{22} \end{matrix}| = 0.

Exercise 2: Let x ∈ K^{2}. If K^{2} \ {x} were an affine algebraic set, it would be given as the zero set of a system of equations, say F_{s} = 0. But any polynomial that vanishes on infinitely many points of a line vanishes on the entire line. We thus consider the restriction of the F_{s} to any line L passing through x. This shows that all the F_{s} vanish at x, a contradiction. If K^{2} \ {x} were a projective algebraic set then it’s image under a regular map would be closed. But the inclusion K^{2} \ {x} ↪ 𝔸^{2}(K) is certainly regular, but since 𝔸^{2} is irreducible, K^{2} \ {x} is not closed, a contradiction. Now let y ∈ ℙ^{n}(K). Similarly, if S = ℙ^{n}(K) \ {y} were a projective algebraic set, it would be closed. It would follow that ℙ^{n}(K) = S ∪ {y} is the union of two closed sets, a contradiction since we know that ℙ^{n}(K) is irreducible. It is false that ℙ^{n}(K) \ {y} is not an affine set. For example, let n = 1, y = (0, 1). Then ℙ^{1}(K) \ {y} ≃ 𝔸^{1}(K). Exercise 4: Let X ⊂ ℙ^{n} be a connected projective algebraic set and let Y be a projective set defined by F = 0, where F is a homogenous polynomial of degree d. Suppose that X ∩ Y = ∅ and let v_{n}^{d} be the Veronese map. Then v_{n}^{d}(X) ∩ v_{n}^{d}(Y) = ∅ since v_{n}^{d} is injective (and moreover v_{n}^{d}(X) is isomorphic to X). Since F is of degree d, we see that v_{n}^{d}(Y) is given by a linear equation. By a projective change of coordinates, we may assume that v_{n}^{d}(Y) is given by T_{0} = 0 and that v_{n}^{d}(X) ∩ v_{n}^{d}(Y) = ∅. But then it follows that v_{n}^{d}(X) is contained in an affine subset of ℙ^{N}. By Corollary 2 of Lecture 9, we conclude that v_{n}^{d}(X) is a single point, and hence, since v_{n}^{d} is injective, that X is a single point.