Finite Projective Geometry 2 nd year group project. B. Doyle, B. Voce, W.C Lim, C.H Lo Mathematics Department - Imperial College London Supervisor: Ambrus P´ al June 7, 2015 Abstract The Fano plane has a strong claim on being the simplest symmetrical object with inbuilt mathematical structure in the universe. This is due to the fact that it is the smallest possible projective plane; a set of points with a subsets of lines satisfying just three axioms. We will begin by developing some theory direct from the axioms and uncovering some of the hidden (and not so hidden) symmetries of the Fano plane. Alternatively, some projective planes can be derived from vector space theory and we shall also explore this and the associated linear maps on these spaces. Finally, with the help of some theory of quadratic forms we will give a proof of the surprising Bruck-Ryser theorem, which shows that if a projective plane has order n congruent to 1 or 2 mod 4, then n is the sum of two squares. Thus we will have demonstrated fascinating links between pure mathematical disciplines by incorporating the use of linear algebra, group the- ory and number theory to explain the geometric world of projective planes. 1
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Finite Projective Geometry
2nd year group project.
B. Doyle, B. Voce, W.C Lim, C.H Lo
Mathematics Department - Imperial College London
Supervisor: Ambrus Pal
June 7, 2015
Abstract
The Fano plane has a strong claim on being the simplest symmetrical object with inbuilt
mathematical structure in the universe. This is due to the fact that it is the smallest possible
projective plane; a set of points with a subsets of lines satisfying just three axioms. We will
begin by developing some theory direct from the axioms and uncovering some of the hidden
(and not so hidden) symmetries of the Fano plane. Alternatively, some projective planes can
be derived from vector space theory and we shall also explore this and the associated linear
maps on these spaces.
Finally, with the help of some theory of quadratic forms we will give a proof of the surprising
Bruck-Ryser theorem, which shows that if a projective plane has order n congruent to 1 or 2
mod 4, then n is the sum of two squares. Thus we will have demonstrated fascinating links
between pure mathematical disciplines by incorporating the use of linear algebra, group the-
ory and number theory to explain the geometric world of projective planes.
Now by definition, the zi are also linear combinations of the xi. We must have that there is a zi
and a yj such that both contain some multiple of x1. Without loss of generality, assume they are
z1, y1. Set z1 = y1 if the coefficient of x1 is different in z1 and y1, if not set z1 = −y1. Then we get
x1 in terms of the other xi, and the z21 and y21 terms cancel out. Continue doing this, each time
expressing xj in terms of the xk, where k > j, in such a way that without loss of generality z2j
cancels out y2j . We get in the end,
nx2N+1 = w2 + y2N+1 (3)
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with w, yN+1 being some rational multiple of xN+1. Now there were no conditions on xN+1 so
we can pick it in such a way that both yN+1 and w are integers. (xN+1 is the product of the two
respective denominators of w, yN+1). Therefore (3) has integer solutions, and by 5.6, n is the
sum of two squares.
The orders of projective planes which Bruck-Ryser deals with are 2, 5, 6, 9, 10, 13, 14, 17, 18, 21, 22,
25, 26 . . . and of those 6, 14, 21, . . . are not the sum of two squares. Therefore there is no projec-
tive plane of order 6, 14, 21. However Bruck-Ryser does not prove that if n is the sum of two
squares then there is a projective plane, in fact it has been proved that there is no projective
plane of order 10 - see [8].
It has been conjectured that all finite projective planes have prime power order, currently the
first possible counter-example is 12.
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A Appendix - Rings and Fields
In this subsection we will develop the tools needed to prove that for a field F and a polynomial
p(x), the splitting field of p(x) always exists. The following proofs are influenced by [9] and
[12].
All rings considered in this section are assumed to be commutative and with unity.
Theorem A.1. Let R be a ring and I be an ideal. Note that (R,+) is an abelian group and (I,+) is a
subgroup of (R,+). Any subgroup of an abelian group is normal, hence R/I = {a + I : a ∈ R} is a
well-defined group under addition. Define multiplication on R/I by (a+ I)(b+ I) = ab+ I . This gives
R/I a ring structure.
Proof. We check that this multiplication is well-defined. It then follows from the ring axioms
that this muliplication satisfies the axioms for a ring.
Suppose a+I = a′+I ,b+I = b′+I , then we have a−a′ ∈ I and b−b′ ∈ I . We need to show that
ab+I = a′b′+I , i.e. ab−a′b′ ∈ I . Note that ab−a′b′ = ab−a′b+a′b−a′b′ = b(a−a′)+a′(b−b′).
Since a− a′, b− b′ ∈ I , b(a− a′) ∈ I and a′(b− b′) ∈ I . Hence ab− a′b′ ∈ I .
Proposition A.2. Let R be a ring, I be an ideal of R. Then there is a one-to-one correspondence between
ideals in R that contain I and ideals in R/I .
Proof. Let π : R→ R/I be the canonical projection, i.e. π(a) = a+ I . Let
I = {J : J is an ideal in R containing I } and I ′ = {J : J is an ideal in R/I}. π induces a map
π′ : I → I ′ defined by π(J) = {a ∈ R/I : a = π(r)for some r ∈ J}.
We first check that this is a well-defined map from I to I ′. Let J be an ideal in R that contains I ,
we show that π(J) is indeed an ideal of R/I . Let a, b ∈ π(J). Then a = r + I, b = s+ I for some
r, s ∈ J . We have r + s ∈ J , hence r + s+ I ∈ π(J) and this addition is clearly abelian as (R,+)
is abelian. Hence (π(J),+) forms an abelian group. Next, let c ∈ R/I , i.e. c = t + I for some
t ∈ R. Then ca = (t + I)(r + I) = tr + I where tr ∈ J as r ∈ J . Hence ca ∈ π(J). Therefore,
π(J) is an ideal.
Define π−1 : I ′ → I by π−1(J) = {a ∈ R : π(a) ∈ J}, where J is an ideal in R/I . We
check that this is also a well-defined map. Let a, b ∈ π−1(J). Then a + I, b + I ∈ J . Hence
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a + b + I ∈ J , i.e. a + b ∈ π−1(J). This shows that (π−1(J),+) is an abelian group. Let r ∈ R.
ra + I = (r + I)(a + I) ∈ J , hence ra ∈ π−1(J). Hence π−1(J) is an ideal. Let i ∈ I , then
π(i) = I ∈ J . Hence I ⊂ π−1(J).
Since these two maps are inverses of each other, this shows that there is a one-to-one correspon-
dence between ideals of R containing I and ideals in R/I .
Proposition A.3. Let R be a ring. The only ideals of R are {0} and R if and only if R is a field.
Proof. SupposeR is a field. Let I be an ideal inR.Suppose I contains a non-zero element a, then
a−1a ∈ I , i.e. 1 ∈ I . Let x be any element in F , then x · 1 ∈ I . Hence I = R.
Now suppose the only ideals of R are {0} and R. Let r be a non-zero element in R and
consider the set {cr : c ∈ R}. This is clearly an ideal which contains a non-zero element, hence
this is equal to R. In particular, there exists c ∈ R such that cr = 1. Hence every non-zero
element in R contains a multiplicative inverse. This shows that R is a field.
Theorem A.4. Let R be a ring and M be a maximal ideal. Then R/M is a field.
Proof. By A.2, The ideals of R/M correspond to the ideals in R containing M . Since M is a
maximal ideal, the only ideals in R containing M are M and R. Hence the only ideals in R/M
are {M} and R/M (note that M is the zero element in R/M ). Hence R/M is a field.
Proposition A.5. Let F be a field, p(x) ∈ F [x] be an irreducible polynomial.
Then (p(x)) = {f(x) ∈ F [x] : p(x)|f(x)} is a maximal ideal.
Proof. Let I be an ideal containing (p(x)). Suppose it contains an element f(x) not in (p(x)),
then since p(x) does not divide f(x) and p(x) is irreducible, we have that hcf(f(x), p(x)) =
1. Since F [x] is an Euclidean domain, we know that there exists a(x), b(x) ∈ F [x] such that
a(x)f(x) + b(x)p(x) = 1. Since I is an ideal, it follows that 1 ∈ I . Hence I = F [x].
Theorem A.6. Let F be any field and f(x) ∈ F [x]. Then the splitting field of f(x) always exists.
Proof. We prove by induction on the degree of f(x). If deg(f) = 1, then F is the splitting field
of f(x). Suppose for every polynomial g(x) ∈ F [x] where deg(g(x)) ≤ n− 1, a splitting field for
g(x) exists. Let deg(f(x)) = n. Suppose f(x) does not factor completely into linear factors, then
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there exists an irreducible factor p(x) of degree at least 2. From A.5 and A.4,E1 := F [x]/(p(x)) is
a field. Note that by considering the canonical projection φ : F [x]→ F [x]/(p(x)), i.e. φ(f(x)) =
f(x) + (p(x)), and restricting the domain to F , F can be viewed as a subfield of F [x]/(p(x)).
Now let x = φ(x). Then p(x) = φp(x) = 0 in F [x]/(p(x)). Hence E1 is a field containing F [x]
that contains a root of p(x). Denoting this root by α, we see that f(x) = (x−α)p1(x) in E1[x]. By
the inductive hypothesis, there exists a field E containing E1 such that f(x) factors completely
into linear factors. Let K be the intersection of all subfields Ei of E containing F such that f(x)
factors completely into linear factors in Ei, then K is a splitting field of f(x).
24
References
[1] Simeon Ball and Zsuzsa Weiner, An introduction to finite geometry,http://www-ma4.upc.es/∼simeon/IFG.pdf (accessed May 27, 2015) - page 9
[2] A. Heyting, Axiomatic Projective Geometry - second edition, North-Holland,Wolters-Noordhoff, (1980)
[3] Johan Kahrstrom, On Projective Planes, Mid Sweden Universityhttp://kahrstrom.com/mathematics/documents/OnProjectivePlanes.pdf (accessed June1, 2015)
[4] Danny Kalmanovich, Finite Projective Planes, http://www.math.bgu.ac.il/∼dannykal/research/fpp%20revised.pdf (accessed May 27, 2015) - pages 8-9
[5] Peter J. Cameron, Combinatorics: Topics, Techniques, Algorithms, Cambridge UniversityPress - pages 141-143
[6] Albrecht Beutelspacher and Ute Rosenbaum, Projective Geometry: From Foundations toApplications, http://www.maths.ed.ac.uk/∼aar/papers/beutel.pdf (accessed June 1,2015)
[7] C.W.H. Lam, G. Kolesova and L. Thiel, A computer search for finite projective planes of order9, http://ac.els-cdn.com/0012365X9190280F/1-s2.0-0012365X9190280F-main.pdf? tid=a0202660-0874-11e5-b074-00000aacb35f&acdnat=1433173450 3af42e5f247ed30c8bd207b6048f9d8f (accessed June1,2015)
[8] C. W. H. Lam and L. Thiel and S. Swiercz, The Non-existence of Finite Projective Planes ofOrder 10,http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.39.8684&rep=rep1&type=pdf(accessed June 5, 2015)