M231: Exercise sheet 1
1. Let V be a vector space.
(a) Let U1 , U2 , W ≤ V with U1 , U2 ≤ W . Show that U1 + U2 ≤ W .
(b) Deduce that if X1 , X2 , Y ≤ P(V ) with X1 , X2 ≤ Y then the join X1 X2 ≤ Y .
(c) Show that X1 X2 is the intersection of all projective subspaces of P(V ) that contain
X1 and X2 .
2. Let P(V ) be a projective space of dimension at least 2 (thus dim V ≥ 3). Prove:
(a) Distinct projective lines in P(V ) intersect in at most one point.
(b) Distinct projective lines in P(V ) intersect if and only if they lie in some (unique)
projective plane.
(c) Give an example of two non-intersecting projective lines in P(R4 ).
3. Let P(V ) be a projective space and X1 , X2 ≤ P(V ) projective subspaces of complementary dimension: dim X1 + dim X2 = dim P(V ).
(a) Prove that the Xi intersect.
What does this tell us when dim P(V ) = 3?
(b) Deduce that if L is a line and X is a hyperplane then either L lies on X or L
intersects X in a single point.
4. Let V be a vector space over a field F.
(a) Let v, w ∈ V \ {0}. Show that [v] = [w] if and only if w = λv, for some λ ∈ F× .
(b) Show that the relation ∼ defined on V \ {0} by v1 ∼ v2 if and only if v2 = λv1 ,for
some λ ∈ F× , is indeed an equivalence relation.
(c) Let V \ {0}/ ∼ be the set of equivalence classes and define a map Φ : P(V ) →
V \ {0}/ ∼ be letting Φ([v]) be the equivalence class of v. Show that Φ is a
well-defined bijection.
Thus we may view P(V ) as the set of equivalence classes under ∼. This can be used,
for example, to give it a (quotient) topology.
Maybe more challenging
5. Let F = Z/2Z be the field with two elements.
(a) How many points are there in P(F2 )?
(b) How many points are there in the plane P(F3 ) and how many lines?
(c) How many lines go through a point of P(F3 )?
(d) Repeat the exercise with F = Z/3Z, the field with three elements.
Are any patterns emerging?
February 14, 2017
Home page: http://go.bath.ac.uk/ma30231
M231: Exercise sheet 1—Solutions
1.
(a) Any element of U1 + U2 is of the form u1 + u2 with ui ∈ U1 , i = 1, 2. Thus both ui ∈ W
and W is closed under addition so that u1 + u2 ∈ W .
To see that U1 + U2 is a vector subspace, first note that it is non-empty since it contains
0 = 0 + 0. Second, if u1 + u2 , v1 + v2 ∈ U1 + U2 and λ ∈ F, then
(u1 + u2 ) + λ(v1 + v2 ) = (u1 + λv1 ) + (u2 + λv2 ) ∈ U1 + U2 .
(b) Write Xi = P(Ui ), i = 1, 2 and Y = P(W ). We are now in the situation of the last part
and so X1 X2 = P(U1 + U2 ) ≤ P(W ) = Y .
(c) Let Z be the intersection of all projective subspaces containing X1 and X2 . Now X1 X2 is
such a subspace so that Z ⊂ X1 X2 . Conversely, if Y is any such a subspace, we have just
shown that X1 X2 ≤ Y so that X1 X2 ⊂ Z. This bakes the cake.
2.
(a) Theorem 1.5 says there is a unique line through two distinct points. Thus distinct lines
can intersect in at most one point.
(b) Let Li = P(Ui ), i = 1, 2, be distinct lines in P(V ) and consider the join L1 L2 . The
dimension formula gives:
dim(L1 L2 ) = dim L1 + dim L2 − dim(L1 ∩ L2 ) = 2 − dim(L1 ∩ L2 )
and dim(L1 ∩L2 ) < 1 since L1 6= L2 . Thus L1 and L2 intersect if and only if dim(L1 ∩L2 ) =
0 if and only if L1 L2 is a projective plane. In particular, if the Li intersect, they lie in the
plane L1 L2 . The converse is the second part of Theorem 1.5 which follows at once from our
analysis: if P(V ) is a projective plane, then dim(L1 L2 ) ≤ 2 so that 1 > dim(L1 ∩ L2 ) ≥ 0.
Finally, for uniqueness, any projective plane P that contains L1 and L2 must also contain
L1 L2 whence P = L1 L2 since both have dimension 2.
(c) We just need 2-dimensional linear subspaces Ui ≤ R4 with U1 ∩ U2 = {0}. For example,
take U1 = {(λ0 , λ1 , 0, 0) : λ0 , λ1 ∈ R} and U2 = {(0, 0, λ2 , λ3 ) : λ2 , λ3 ∈ R}.
3.
(a) We argue as in Theorem 1.5 and wheel out the dimension formula: let dim P(V ) = n and
dim X1 = k so that dim X2 = n − k. Then
n = dim P(V ) ≥ dim(X1 X2 ) = dim X1 + dim X2 − dim(X1 ∩ X2 )
= k + (n − k) − dim(X1 ∩ X2 )
and rearranging this gives dim(X1 ∩ X2 ) ≥ 0 so that the Xi intersect.
When n = 3, the only interesting possibility is k = 1, 2 and the result asserts that any
projective line must intersect any projective plane. As we see below, either this intersection
point is unique dim(X1 ∩ X2 ) = 0 or the line lies in the plane dim(X1 ∩ X2 ) = 1.
(b) L and X have complementary dimension and so must intersect by the first part. Now
X ∩ L ≤ L so that either dim X ∩ L = 0 and X ∩ L is a point or dim X ∩ L = 1 in which
case X ∩ L = L so that L ≤ X.
4.
(a) If [v] = [w] then w ∈ [v] which means there is λ ∈ F× with w = λv. Conversely, if w = λv,
then for any µ ∈ F× , µw = (µλ)v ∈ [v] so that [w] ⊂ [v]. Applying the same argument
starting from v = λ−1 w gives [v] ⊂ [w] so that [v] = [w].
(b) Reflexive: v ∼ v since v = 1v.
Symmetric: v1 ∼ v2 means v2 = λv1 , some λ ∈ F, if and only if v1 = λ−1 v2 so that
v2 ∼ v1 .
Transitive: if v1 ∼ v2 and v2 ∼ v3 , we have v2 = λv1 and v3 = µv2 whence v3 = (µλ)v1
and v1 ∼ v3 .
(c) First we show that Φ is well-defined: if [v] = [w], by part (a), we have that v ∼ w so
that v and w define the same equivalence class. Moreover, if Φ([v]) = Φ([w]) then v ∼ w
so, again by part (a), [v] = [w] and Φ is injective. Finally, the equivalence class of any
v ∈ V \ {0} is Φ([v]) so Φ surjects also.
5.
(a) The same argument as that given in lectures establishes that, for any field F, P(F2 ) ∼
=
F t {∞} so that |P(F2 )| = |F| + 1. When F = Z/2Z, this yields |P(F2 )| = 2 + 1 = 3.
Alternatively, view P(F2 ) as F2 \ {0}/ ∼ and observe that all equivalence classes have
the same size which is |F× | because, for any v ∈ F2 \ {0}, λ 7→ λv is a bijection from
F× to the equivalence class of v. Since equivalence classes partition, we have |P(F2 )| =
|F2 \ {0}|/|F× | = (22 − 1)/(2 − 1) = 3.
(b) Either argue that P(F3 ) ∼
= F2 t F t {∞} so that |P(F3 )| = 22 + 2 + 1 = 7 or that
3
3
×
|P(F )| = |F \ {0}|/|F | = (23 − 1)/(2 − 1) = 7.
As for the lines in P(F3 ), a line is determined by any pair of distinct points lying on it.
There are 7 × 6/2 such pairs and any particular line is equally well determined by any of
the 3 pairs of the 3 points on it. Thus there are 7 = 7 × 6/(2 × 3) lines in the plane P(F2 ).
(c) Fix a point, then there are 6 pairs that include that point and each determines a line
which is counted twice over: once for each of the other two points on the line. Thus there
are 6/2 = 3 lines through each point.
(d) The same arguments apply for the field with 3 elements: |P(F2 )| = 3 + 1 = 4 = (32 −
1)/(3 − 1). Similarly, |P(F3 )| = 13. Meanwhile, each line has 4 points and is determined
by any of the 6 pairs of point that lie thereon. In all, there are 13 × 12/2 = 13 × 6 pairs of
points in P(F3 ) and thus 13 lines. Finally, there are 12 pairs that include a given point
and each determines a line through that point which is counted 3 times over, giving 4
lines through a given point.
This suggests two facts:
1.
Projective planes contain the same number of points as lines.
2.
The number of points on a line in a projective plane is the same as the number of lines
through a point of a projective plane.
All of this is true, as we shall see later. Meanwhile, I invite you to prove these facts now for an
arbitrary finite field.