How many affine subspaces of dimension $2$ does $\mathbb{Z}_2^3$ have?

Question

How many affine subspaces of dimension $2$ does $\mathbb{Z}_2^3$ have? I think in $\mathbb{Z}_2^3$ any 3 points are affine independent, so any affine combination will determine a uniqe plane. So it will be $C_{2^3}^3 = C_8^3 = 56$ unique planes? Am i wrong?

Answer 1

You are correct that any three points will determine an affine plane (because no three points in $\mathbb{Z}_2^3$ are collinear). However, there are multiple sets of three points which determine the same affine plane.

How many? Any plane consists of $4$ points. You can use any $3$ of those points in your counting argument.

So (in your notation) the total number of planes is

$$ \frac{C_8^3}{C_4^3}=\frac{56}{4}=14 $$

Answer 2

Linear $2$-dimensional subspaces in $\mathbb Z_2^3$ are in bijeciton with $1$-dimensional subspaces in the dual $(\mathbb Z_2^3)^\vee\cong\mathbb Z_2^3$. Each nonzero vector in $\mathbb Z_2^3$ gives rise to a distinct $1$-dimensional subspace (since $1$-dimensional subspaces only consist of two vectors), so there are $8-1=7$ of them.

Now, there are two affine subspaces for each linear subspace, so there are $14$ total.