# What are the planes of $AG_2(2)$?

I am trying to understand the question (this is Van Lint's Intro to Combinatorics) by first looking at when $r=2$. The points are ordered pairs with each entry in $\mathbb{F}_2$ and all possible linear combinations $x+ty$ where $x,y \in \mathbb{F}_2^2$ and $t\in\mathbb{F}_2$ make a tetrahedron.

My first question is what is a planes of $AG_2(2)$? It would also help to see what they are for $AG_3(2)$.

My confusion is, if $r$ is the number of elements in our field and $2$, the dimension is fixed, wouldn't there only be one plane regardless of the what $r$ we take? but the question mentions planes.

My second question is about what needs to be shown. From my understanding, what we want to do is take any $3$ points from the tetrahedron, and show that it belongs in exactly one block of size $4$ in the Steiner system given.

• I suppose $AG_r(q)$ stands for the affine geometric incidence structure over the $q$ element field, and plane means $2$ dimensional affine subspace. Then, $AG_2(q)$ has only one plane: the whole space. – Berci Mar 30 '18 at 22:24

You have your $r$ and $q$ mixed up; $\mathrm{AG}_{r}(q)$ is affine geometry of dimension $r$ over $\mathbb{F}_{q}$ (commonly denoted $\mathrm{AG}(r,2)$). You only have one plane of $\mathrm{AG}_{2}(2)$, so it's not a very interesting case. When $r \geq 3$ you will have more (14 when $r=3$).