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?
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2But no, the three vectors $(1,0,0)$, $(0,1,0)$ and $(1,1,0)$ are certainly not independent in $\Bbb Z_2^3$. – David C. Ullrich Jun 12 '22 at 23:52
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By planes I mean affine subspaces of dimension 2 – MathLearner Jun 12 '22 at 23:58
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I wanted to say "How many affine subspaces of dimension 2 $\mathbb{Z}_2^3$ have". Sorry for my bad english :( – MathLearner Jun 13 '22 at 00:04
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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
$$ \frac{C_8^3}{C_4^3}=\frac{56}{4}=14 $$
Micah
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Ok. I think I understand. But i have 2 questions.1.How we check that any 3 points aren't colinear 2.Why any plane consists of 4 points and not 3? – MathLearner Jun 13 '22 at 11:14
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A $k$-dimensional subspace (linear or affine) of $\mathbb{Z}_2^n$ has $2^k$ points. So $3$ distinct points aren't collinear because a line only has $2^1=2$ points, and the planes that you're looking for have $2^2=4$ points. – Micah Jun 13 '22 at 12:44
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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.
Kenta S
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