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Jun 10, 2011 at 5:32 history edited Aaron Meyerowitz CC BY-SA 3.0
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Jun 10, 2011 at 5:32 history made wiki Post Made Community Wiki by Aaron Meyerowitz
Jun 10, 2011 at 4:40 history edited Aaron Meyerowitz CC BY-SA 3.0
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Jun 8, 2011 at 6:16 vote accept chous
Jun 7, 2011 at 21:48 history edited Aaron Meyerowitz CC BY-SA 3.0
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Jun 7, 2011 at 18:40 comment added Douglas Zare $\mathbb{P}^3(F_3)$ has ${40 \choose 2}/{4 \choose 2} = 130$ lines, not $40$, so the construction for $42$ blocks actually uses $132$ blocks. It was a little too good to be true since it would mean any quadruple would intersect a block in $2$ points, and it would be easy to use the $5$th points to make a line redundant.
Jun 7, 2011 at 17:53 comment added Douglas Zare Aaron Meyerowitz's construction for $42$ blocks used $F_3$ rather than $F_4$, which should make it easier. The construction is just like that of $\mathbb{RP}^3$ so that the points are lines passing through the origin in $\mathbb{R}^4$ and the lines of $\mathbb{P}^3$ correspond to (two-dimensional) planes passing through the origin. $\mathbb{P}^3(F_4)$ has $(4^4-1)/(4-1)=85$ points, though.
Jun 7, 2011 at 12:18 comment added chous Thanks a lot! How would you recommend to build the projective 3 space of order 4? I think I'm currently not ready for this task :(.
Jun 7, 2011 at 12:12 vote accept chous
Jun 8, 2011 at 6:15
Jun 7, 2011 at 9:47 history edited Aaron Meyerowitz CC BY-SA 3.0
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Jun 7, 2011 at 9:33 history edited Aaron Meyerowitz CC BY-SA 3.0
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Jun 7, 2011 at 5:59 history edited Aaron Meyerowitz CC BY-SA 3.0
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Jun 7, 2011 at 4:51 history edited Aaron Meyerowitz CC BY-SA 3.0
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Jun 7, 2011 at 4:33 history edited Aaron Meyerowitz CC BY-SA 3.0
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Jun 7, 2011 at 4:21 history answered Aaron Meyerowitz CC BY-SA 3.0