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Consider the projective space $P(\mathbb{F}^{n+1}_q)$, the projective space constructed over $\mathbb{F}^{n+1}_q$, where $q$ is prime and $n \in \mathbb{N}$.

How many points does it have? And how many straight lines? I've already figured it out for $q=2$, but I don't know how to generalise it to every prime number $q$.

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  • Where are you stuck ? The underlying field $k$ doesn't change so much the definition of the projective space $P(k^{n+1})$. And for a line in $P(k^{n+1})$ see this answer – reuns Dec 04 '16 at 17:42
  • $\mathbb{F}^{n+1}_q$ has $q^{n+1}-1$ non-zero vectors. A point in a projective space is defined as a 1-dimensional subspace of the underlying vector space. Since there are $q$ scalar multiples of a vector, and every scalar multiple of a vector determines the same 1-dimensional subspace as that vector, there are $\frac{q^{n+1}-1}{q}$ points in the projective space. Is this correct? –  Dec 04 '16 at 21:04
  • Yes of course, my bad! But what about the straight lines in the projective space? –  Dec 04 '16 at 22:42
  • So you have $\frac{(q^{n+1}-1)(q^{n+1}-q)}{(q-1)^2}$ divided by $\binom {\frac{q^{n+1}-1}{(q-1)}} {2}$ lines? –  Dec 05 '16 at 11:49

1 Answers1

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In general, the number of $m$-dimensional projective subspaces in $PG(n,q)$ is $\frac{(q^{n+1}-1)(q^{n+1}-q)\dots(q^{n+1}-q^m)}{(q^{m+1}-1)(q^{m+1}-q)\dots(q^{m+1}-q^m)}.$

If $m=0$, we get that the number of points is in fact $\frac{q^{n+1}-1}{q-1}$,

if $m=1$, we get that the number of lines is $\frac{(q^n+q^{n-1}+\dots+q+1)(q^{n-1}+\dots+q+1)}{q+1}$.

Sz_Z
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