# How do you create projective plane out of a finite field?

I have heard and read unclear mentions of links between projective planes and finite fields.

Is it possible to construct a projective plane (or a Steiner system) starting out with a field? Could you, for example, construct the Fano plane with help of a finite field?

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You can create a projective plane for any field $F$. Consider $F^n\setminus\{0\}$ up to scalar multiplication. This works for finite and infinite fields. – Grumpy Parsnip Aug 8 '13 at 13:51
That "[..] consider [..] multiplication [..]" is what I have heard about the relation between projective planes and fields, but never got the idea of actual relations. – Juris Aug 8 '13 at 14:02

There is a difference between the projective plane as an object of algebraic geometry and as a combinatorial construction (=Steiner system) that has not been adequately explained. The former makes sense over any field - the latter only over a finite field (together with the speculations that no field may be necessary for the required construction to exist).

Given any field $F$ we can construct the set $\mathbb{P}^2(F)$ in the usual way: it is the set of equivalence classes $S/\sim$ of $F^3\setminus\{(0,0,0)\}$, where two non-zero vectors $x=(x_1,x_2,x_3)$ and $y=(y_1,y_2,y_3)$ of $F^3$ are called equivalent, denoted $x\sim y$, if there is a scalar $\lambda\in F^*$ such that $y_i=\lambda x_i$ for all $i=1,2,3.$

The "points" of $\mathbb{P}^2(F)$ are equivalence classes, but to identify them we need to use coordinates. To that end we need a set of representatives - one from each equivalence class. A popular way of achieving that is to select $\lambda$ in such a way that the last non-zero coordinate is equal to $1$. After all, if $x_3\neq0$, we have (with $\lambda=1/x_3$) $$(x_1,x_2,x_3)\sim (\frac{x_1}{x_3},\frac{x_2}{x_3},1).$$ We can equally well choose to scale the first non-zero coordinate to be equal to one, if so desired.

If $F$ is finite, say $|F|=q$, this means that there are $q^2+q+1$ elements in $\mathbb{P}^2(F)$:

1. $q^2$ equivalence classes with representatives $(x,y,1)$, $x,y\in F$ arbitrary.
2. $q$ equivalence classes with representatives $(x,1,0)$, $x\in F$ arbitrary.
3. A single equivalence class with representative $(1,0,0)$.

So in a way $\mathbb{P}^2(F)$ is the union of a "usual" (affine) plane, a line and a point. If $F=\{0,1,2\}=\mathbb{Z}/3\mathbb{Z}$, then the $9+3+1=13$ elements are (the classes of) $P_1=(0,0,1)$, $P_2=(0,1,1)$, $P_3=(0,2,1)$, $P_4=(1,0,1)$, $P_5=(1,1,1)$, $P_6=(1,2,1)$, $P_7=(2,0,1)$, $P_8=(2,1,1)$, $P_9=(2,2,1)$, $P_{10}=(0,1,0)$, $P_{11}=(1,1,0)$ $P_{12}=(2,1,0)$ and $P_{13}=(1,0,0)$. The equivalence relation means that we equate, for example, the point $(2,1,2)$ with $2(2,1,2)=(1,2,1)=P_6$ and the point $(1,2,0)$ with the poin $2(1,2,0)=(2,1,0)=P_{12}$. In the field of $q$ elements there are $q-1$ non-zero constants, so in general $q-1$ points of $F^3$ form a single equivalence class. These are the non-zero points of a 1-dimensional subspace of $F^3$, and a useful point of view is to think of points of $\mathbb{P}^2(F)$ as lines through the origin in $F^3$.

To get the combinatorial design called the projective plane we need to also specify certain subsets of $\mathbb{P}^2(F)$, called "lines". These are constructed with the following recipe. Let $U\subset F^3$ be any 2-dimensional subspace. Then the line $L_U$ consists of the equivalence classes of the non-zero points of $U$. As $U$ is a subspace, it is closed under scalar multiplication. Therefore $U$ is a union of equivalence classes. If $|F|=q$, there are $q^2-1$ non-zero vectors in $U$, and these are divided into $(q^2-1)/(q-1)=q+1$ equivalence classes. Thus the line $L_U$ will have $q+1$ "points".

Continuing the example with $F=\mathbb{Z}/3\mathbb{Z}$ we see that, if $U$ is the span of the two (linearly independent) vectors $x=(1,2,1)$ and $y=(0,1,2)$ then the non-zero points in $U$ are $x,2x=(2,1,2)$ (both in $P_6$), $y,2y=(0,2,1)$ (both in $P_3$), $x+y=(1,0,0)$, $2x+2y=(2,0,0)$ (in $P_{13}$), $x+2y=(1,1,2)$ and $2x+y=(2,2,1)$ (in $P_9$). As promised, these fell into four equivalence classes, and thus $$L_U=\{P_6,P_3,P_{13},P_9\}.$$

You can do the same construction for all the $13$ 2-dimensional subspaces of $F^3$, and end up with $13$ lines, four points on each. Work out a few of those by hand to gain some fluidity.

The characteristic combinatorial properties of this construction are: 1) any pair of lines intersects at exactly one point, 2) given any two points there is a unique line containing them. These then follow from results of linear algebra: 1) any two 2-dimensional subspaces of $F^3$ intersect in a 1-dimensional subspace, 2) any two 1-dimensional subspaces of $F^3$ span a unique 2-dimensional subspace.

Hopefully this clears up some of the fog.

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The open questions are whether "combinatorial" projective planes, with $q^2+q+1$ points and lines, exist when $q$ is not a prime power. IIRC it is believed that they do not exist unless $q$ is a prime power. But it is open, when $q\equiv2\pmod4$ and $q>10$. The non-existence of a projective plane with $q=10$ was big news at some point. The proof required extensive computer checking (after several reductions to make complexity manageable). I am not up to speed with this, caveat reader. – Jyrki Lahtonen Aug 9 '13 at 6:02
This is a great answer and it clears quite a lot of fog. And I finally see where the $q^2+q+1$ comes from :) Could you advise some resource (preferably online, but a book would be fine too) for further reading? – Juris Aug 9 '13 at 7:18
Can you explain why you divide by $(q-1)$ to get number of vector equivalence classes in $U$? – Juris Aug 9 '13 at 7:20
"Can you explain why you divide by $(q−1)$ to get number of vector equivalence classes in $U$?" Each 1-dimensional subspace of $U$ collapses to a single point of the projective plane, and there are $q-1$ non-zero vectors in a 1-dimensional space. – Jyrki Lahtonen Aug 9 '13 at 9:17
There are systematic ways of selecting the subspaces. The simplest one is by "orthogonal" complements: Each $U$ is a solution set of a homogeneous equation $$a_1x_1+a_2x_2+a_3x_3=0,$$ where $(a_1,a_2,a_3)$ is a non-zero vector of coefficients. Here (again!!) $(a_1,a_2,a_3)$ and $\lambda(a_1,a_2,a_3)$ define the same subspace $U$, so it suffices to let $(a_1,a_2,a_3)$ range over a set of representatives of points of the projective plane. This explains, why the number of 2-dimensional subspaces is also $q^2+q+1$. – Jyrki Lahtonen Aug 9 '13 at 9:22

Does this help?

Let F be a field. We consider the three dimensional vector space over F. Let subspaces of dimension 1 be points. Let subspaces of dimension 2 be lines. The incidence relation is subset inclusion (i.e. if a point is a subset of a line, then you say the point is on the line.)

These definitions satisfy the axioms of a projective plane. You should be able to recover the Fano plane by applying this procedure with the field with two elements.

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The Fano plane with 7 points is constructed starting with the field consisting of 2 elements. See Fano plane.

What you seem to be asking for is an *explanation" of the intuition behind the construction of the projective plane. Here it would be helpful to start with a different construction using "points at infinity". This was in fact historically the first construction of the projective plane (the construction using homogeneous coordinates in $\mathbb{R}^3$ came later).

Briefly, what you want to do is start with the usual plane, and add "points at infinity" to it. There is going to be a separate point at infinity for each "direction" in the plane. To define the notion of "direction" rigorously, the usual approach is to take a pencil (i.e., a set) of parallel lines. Each such pencil defines a "direction". The part about "adding" points at infinity means that there is going to be an extra point on each line (namely, the one corresponding to the "direction" of that line). The details can be found for example in the book on projective geometry by Hartshorne.

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Sorry, but you are just repeating the same statement that I don't understand. That's why I asked the question. I see making 3-component vectors with components either 0 or 1. There aren't even all the possible combinations. How do you get to that? How do you know that you need triples? Would you also take triples if you started with 9-element field? Or it's 10-component vectors as there would be 10 points on each line? And where's that multiplication mentioned by Grumpy? I see only addition... – Juris Aug 8 '13 at 18:08
It's also not clear how the corresponding lines are found. If it would be 4 or 5 element field, would it still be the search for the same number of nonzero digits? Or for equal digits? And why this even works? I have read all those wiki articles and still haven't found (or understood) explanation of relations between fields and projective planes. – Juris Aug 8 '13 at 18:12
Fano plane hides the scalar multiplications, because the field of two elements has only one non-zero scalar! – Jyrki Lahtonen Aug 9 '13 at 4:42