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Explicitly the elements of $\mathbb{P}^{1}(\mathbb{F}_{3})$ are $[1:0], [0:1], [1:1]$, and $[1:-1]$. Why is this so? How would I do this for $\mathbb{P}^{1}(\mathbb{F}_{4})$? What about general $\mathbb{P}^{1}(\mathbb{F}_{n})$?

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The "brute force" way would be to write out every element of $\mathbb{F}_n \times \mathbb{F}_n$ and start identifying those elements that only differ by non-zero scaling. For $\mathbb{F}_3$ this is practical, but for large $n$ I suppose you should be more clever –  you Feb 28 '12 at 20:44
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up vote 7 down vote accepted

The elements of $\mathbb{P}^n(k)$ are equivalence classes of elements of $k^{n+1}-\{(0,\ldots,0)\}$, under the equivalent relation $$(a_1,\ldots,a_{n+1}) \sim (b_1,\ldots,b_{n+1})\Leftrightarrow \text{there exists }\lambda\in k, \lambda\neq 0\text{ s.t. }b_i=\lambda a_i\text{ for }i=1,\ldots,n+1.$$

More conceptually: the elements of $\mathbb{P^1}(k)$ corresponds to directions of lines through the origin in $k^2$. You have one direction for each possible slope, which are the points of the form $[1:m]$, $m\in k$; plus the "vertical direction" which has no slope, which corresponds to $[0:1]$ (think of $[a:b]$ as giving you the second point on the line, with $(0,0)$ being the first point). So you have the horizontal line $[1:0]$; the line with slope $1$, $[1:1]$; the line with slope $-1$, $[1:-1]$; and the vertical line $[0:1]$.

For $\mathbb{P}^1(\mathbb{F}_4)$, you proceed the same way: if we write $\mathbb{F}_4 = \{0,1,\alpha,\alpha+1\}$, with $\alpha^2=\alpha+1$, then you have four possible slopes for non-vertical lines, plus the vertical line, which gives the points $[1:0]$, $[1:1]$, $[1:\alpha]$; $[1:\alpha+1]$; and $[0:1]$.

If $n=p^a$ is a prime power, then $\mathbb{F}_n$ has $p^a$ possible slopes for nonvertical lines, and one vertical line. So the points of $\mathbb{P}^1(\mathbb{F}_{p^a})$ are $[1:r]$ with $r\in\mathbb{F}_{p^a}$, and $[0:1]$. If $n$ is not a prime power, then there is no field of $n$ elements.

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Set-theoretically $\mathbb P^N(\mathbb F_n)$ is the quotient of $X=\mathbb F_n^{N+1}\setminus 0$ by the free action of $G=\mathbb F_n^*$ acting as homotheties: $\mathbb P^N(\mathbb F_n)=X/G$
Thus
$$|\mathbb P^N(\mathbb F_n)|=|X/G|=|X|/[G|=\frac {n^{N+1}-1}{n-1}$$

As to writing out explicitly elements of $\mathbb P^N(\mathbb F_n)$: take a non-zero element $(a_0,...,a_n) \in \mathbb F_n^{N+1}\setminus 0$ and replace commas by colons to signify equivalence class, that is write $(a_0:\ldots :a_n) \in \mathbb P^N(\mathbb F_n)$ .

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