Algebraic Geometry, PS 6

Bryden Cais

November 26, 2002

1 Lecture 12

1. We have seen that lines in ℙ^{3}(K) are identified with points of G(2, 4). Let x ∈ ℙ^{3}(K) be any point. By a projective change of coordinates, we may assume without loss of generality that x is the point (1, 0, 0, 0) = e_{1}, where {e_{i}} is the standard basis of K^{4}. We can write any element of G(2, 4) as f_{1} ∧ f_{2} for f_{1}, f_{2} ∈ K^{4}. The condition that the line fiven by f_{1}, f_{2} pass through x is just the condition that

f_{1} ∧ f_{2} ∧ e_{1} = 0.

Writing f_{1} ∧ f_{2} in Plücker coordinates as

f_{1} ∧ f_{2} = ∑_{i<j} p_{ij}e_{i} ∧ e_{j}

we see that the condition that f_{1} ∧ f_{2} pass through x is given by

p_{23} = p_{24} = p_{34} = 0.

Since we have in addition the condition

p_{12}p_{34} − p_{13}p_{24} + p_{14}p_{23} = 0,

we see that the set of lines passing through x, π_{x}, is a closed subset of G(2, 4) given by the equations above. Viewing π_{x} via the Plücker embedding, we define the morphism

φ : π_{x} → ℙ^{2}(K)

by φ({p_{ij}}) = (p_{12}, p_{13}, p_{14}). From the defining equations (in Plücker coordinates) of π_{x} it is clear that φ is an isomorphism with inverse given by

φ^{−1}(x, y, z) = (x, y, z, 0, 0, 0) = (p_{12}, p_{13}, p_{14}, p_{23}, p_{24}, p_{34}) ∈ G(2, 4).

Now let π_{P} be the set of lines contained in a given plane P ⊂ ℙ^{3}(K). Without loss of generality, as before, we may suppose that P is the span of e_{1}, e_{2}, e_{3}. Then the condition that a line f_{1} ∧ f_{2} ∈ G(2, 4) in ℙ^{3}(K) be contained in P is just the condition

e_{1} ∧ e_{2} ∧ e_{3} ∧ f_{1} = e_{1} ∧ e_{2} ∧ e_{3} ∧ f_{2} = 0,

and writing f_{i} = ∑_{j} a_{ij}e_{j} we see that f_{1} ∧ f_{2} ∈ π_{P} if and only if a_{14} = a_{24} = 0.

5. (a) We first show that the set of flags

F (k, l; n + 1) = {(U, W) : U ⊂ W ⊂ K^{n+1}} ⊂ G(k, n + 1) × G(l, n + 1)

is a closed set. Write

U = [v_{1} ∧ v_{2} … ∧ v_{k}] ∈ G(k, n + 1)
W = [w_{1} ∧ w_{2} … ∧ w_{l}] ∈ G(l, n + 1).

Then the condition that U ⊂ W is precisely the condition that v_{i} ∈ W for all i, or equivalently, that

v_{i} ∧ w_{1} ∧ w_{2} … ∧ w_{l} = 0

for 0 ≤ i ≤ k. This gives a system of algebraic equations in plücker coordinates, thus showing that F (k, l) is closed. Now let

F (m_{1}, …, m_{r}; n + 1) ⊂ G(m_{1}, n + 1) × ⋯ × G(m_{r}, n + 1)

be a flag of type (m_{1}, …, m_{r}) and for i < j let

π_{i,j} : F (m_{1}, …, m_{r}; n + 1) → F (m_{i}, m_{j}; n + 1)

be the natural projection map given by π_{i,j}(V_{0}, V_{1}, …, V_{r}) = (V_{i}, V_{j}). Then it is clear that

F (m_{1}, …, m_{r}; n + 1) = ⋂_{i<j} π_{i,j}^{−1}(F (m_{i}, m_{j}; n + 1)).

Obviously, π_{i,j} is continuous for each i < j so that π_{i,j}^{−1}(F (m_{i}, m_{j}; n + 1)) is closed since F (m_{i}, m_{j}; n + 1) is. Thus, F (m_{1}, …, m_{r}; n + 1) is a finite intersection of closed sets and hence closed.

(b) Consider the projection map

π : F (m_{1}, …, m_{r}; n + 1) → F (m_{1}, m_{2}, …, m_{r−1}; n + 1).

The fibre over a point

(V_{1}, V_{2}, …, V_{r−1}) ∈ F (m_{1}, m_{2}, …, m_{r−1}; n + 1)

consists of all points V_{r} ∈ G(m_{r}, n+1) with V_{r−1} ⊂ V_{r}. Choosing such a V_{r} corresponds to choosing a subspace W of the orthogonal complement of V_{r−1} in V = K^{n+1} of dimension m_{r} − m_{r−1}, so that V_{r} := W ⊕ V_{r−1}. But the set of all such spaces is identified with

G(m_{r} − m_{r−1}, n + 1 − m_{r−1}),

and this is therefore the fibre over any point. Since our projection map is surjective and all fibres clearly have the same dimension, namely

dim G(m_{r} − m_{r−1}, n + 1 − m_{r−1}) = (m_{r} − m_{r−1})(n + 1 − m_{r}),

it follows that we have

dim F (m_{1}, …, m_{r}; n + 1) = dim F (m_{1}, m_{2}, …, m_{r−1}; n + 1) + (m_{r} − m_{r−1})(n + 1 − m_{r}).

Repeated application of this formula gives

dim F (m_{1}, …, m_{r}; n + 1) = ∑_{i=0}^{r−1}(m_{r−i} − m_{r−i−1})(n + 1 − m_{r−i}),

where for convenience we set m_{0} = 0.

10. Set k = k_{0}(t) where k_{0} is an algebraically closed field and let F ∈ k[T_{0}, …, T_{n}] be a homogenous polynomial of degree d ≤ n. We claim that V (F)(k) ≠ ∅. By clearing denominators, we may assume without loss of generality that the coefficients of F are all polynomials in t. Let m be the maximum of the degrees of these polynomials and for each i write

T_{i} = ∑_{j=0}^{r} a_{ij}t^{j}

for unknowns a_{ij} ∈ k_{0}. Substituting in the T_{i} to the equation

F (T_{0}, T_{1}, …, T_{n}) = 0

we get a polynomial in t all of whose coefficients must be equal to 0. Observe that the degree of this polynomial is at most

m + rd,

and hence there are at most m + rd + 1 equations. However, there are rn unknowns, and since d ≤ n we can choose r large enough so that 1 + m + rd < (r + 1)(n + 1) so that there are more unknowns than equations, whence there exists a nonzero solution, as claimed.

2 Lecture 13

1. Let k be an algebraically closed field of characteristic 0 and let V be an irreducible plane cubic. We know that V (k) has an inflection point, and without loss of generality (by a projective change of coordinates) we may suppose that the point (1, 0, 0) ∈ ℙ^{2}(k) is an inflection point. It follows that this is the only point in V (k) on the line at infinity (since it is a triple point and bezout’s theorem tells us that the line at infinity has exactly 3 intersection points with V (k), counting multiplicity). We then see that X is given by an equation of the form

a_{0}X^{3} + a_{1}X^{2}Z + a_{2}XZ^{2} + a_{3}Z^{3} + b_{1}Y^{2}Z + b_{2}Y Z^{2} + c_{0}XY Z = 0.

Observe that we must have a_{0} ≠ 0 (since otherwise our cubic is reducible). Moreover, since the only point on the line at infinity is smooth, if V (k) has a singular point, it lies in the affine space Z = 1. Therefore, scaling X and dehomogenizing, our cubic takes the form

X^{3} + a_{1}X^{2} + a_{2}X + a_{3} + b_{1}Y^{2} + b_{2}Y + c_{0}XY = 0.

Replacing X by X − a_{1}/3 brings this to the form

X^{3} + aX + b + cY^{2} + dY + eXY = 0.

If c = 0 and e = 0 then our cubic is obviously smooth since it is of the form Y = X^{3} + aX + b (or possibly X^{3} + aX + b = 0). So we may assume that e ≠ 0, and then we change Y to (Y − a)/e to bring our cubic to the form

X^{3} + XY + u + wY = 0.

Then change X to X − w to obtain

(X − w)^{3} + XY + u = 0

Any singular point (x, y) of a cubic of this form must satisfy

3(x − w)^{2} + y = x = (x − w)^{3} + xy + u = 0,

so that we must have x = 0, w^{2} = −y and w^{3} = u. However, if w^{3} = u then our cubic has the form

(X − w)^{3} + w^{3} + XY,

which is reducible. Thus we conclude that in this case, there can be no singular cubics, so that any singular cubic has c ≠ 0 above. Thus we can scale Y and bring the cubic to the form

X^{3} + aX + b + Y^{2} + uY + vXY = 0.

Now change Y to Y − u/2 − vX/2 to bring it to the form

Y^{2} = X^{3} + AX^{2} + BX + C,

and change X to X − A/3 to finally get our cubic in the form

Y^{2} = X^{3} + pX + q.

A point (x, y) of this cubic is singular if and only if

y^{2} = x^{3} + px + q
y = 0
3x^{2} = −p.

These equations imply that we must have

2px + 3q = 0,

and hence that

27q^{2} + 4p^{3} = 0.

There are then two possibilities: either p = q = 0 and we obtain the cubic

Y^{2} = X^{3}

which has a cusp at the point (0, 0), or p, q ≠ 0 but 27q^{2}+4p^{3} = 0 so that the cubic X^{3}+pX+q has a double (but not a triple) root. Again by a projective change of coordinates we may assume that this double root is X = 0, whence (after scaling X, Y) we get the cubic

Y^{2} = X^{3} + X^{2},

which has a node at (0, 0). Our analysis shows that up to projective changes of coordinates, these are the only two singular cubics in ℙ^{2}(k).

5. Any singular point of the quartic V (T_{0}^{2}T_{1}^{2} + T_{1}^{2}T_{2}^{2} + T_{0}^{2}T_{2}^{2} − T_{0}T_{1}T_{2}T_{3}) in ℙ^{3}(K) must satisfy

T_{0}^{2}T_{1}^{2} + T_{1}^{2}T_{2}^{2} + T_{0}^{2}T_{2}^{2} − T_{0}T_{1}T_{2}T_{3} = 0  (1)
2T_{0}(T_{1}^{2} + T_{2}^{2}) − T_{1}T_{2}T_{3} = 0  (2)
2T_{1}(T_{0}^{2} + T_{2}^{2}) − T_{0}T_{2}T_{3} = 0  (3)
2T_{2}(T_{0}^{2} + T_{1}^{2}) − T_{0}T_{1}T_{3} = 0  (4)
T_{0}T_{1}T_{2} = 0.  (5)

The last equation gives us three possibilities:

Case 1: T_{0} = 0. Then by (4) we have T_{1}T_{2} = 0 (and it is seen that (3),(2),(1) give no further information). It is therefore easily checked that the locus of points

(0, 0, 1, t), (0, 1, 0, t), (0, 0, 0, 1)

for any t ∈ K are singular points.

Case 2: T_{1} = 0. Clearly, the equation under consideration is invariant under any permutation of T_{0}, T_{1}, T_{2}. We can thus immediately write down the locus of singular points in this case:

(0, 0, 1, t), (1, 0, 0, t), (0, 0, 0, 1)

for any t ∈ K.

Case 3: T_{2} = 0. Again, we find that the points

(1, 0, 0, t), (0, 1, 0, t), (0, 0, 0, 1)

for any t ∈ K are singular points.

In summary, we have found all singular points:

(0, 0, 1, t), (0, 1, 0, t), (1, 0, 0, t), (0, 0, 0, 1)

for any t ∈ K.

6. Suppose that X is a surface in ℙ^{3}(K) that contains three non-coplanar lines passing through

a point x. Then clearly, we have

dim T (X)_{x} ≥ 3.

On the other hand, since X is a surface in ℙ^{3}(K), it has dimension 2 so that

dim T (X)_{x} > dim X.

Since one of the characterizations of a smooth point is

dim T (X)_{x} = dim X,

we see that x must be singular.