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Statement i read a few times :

if $k$ is a field, then the rational points are Zariski-dense in the projective space $\mathbb{P}^n$. Does anybody could provide a proof of this fact or a reference? (i think it should be only true if $k$ is infinite).

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What are rational points in $\mathbb{P}^n?$ Aren't they all rational? – Igor Rivin Dec 10 '13 at 16:14
I assume that it means that you fix an algebraically closed field extension $K/k$, the Zariski topology on $\mathbb{P}^n(K)$ is the topology whose closed sets are locus of systems of homogeneous equations in $k[X_0,...,X_n]$. This give you the Zariski $k$-toplogy, for which line in $k^{n+1}$ should be dense... If i dont miss something. – Louis La Brocante Dec 10 '13 at 16:23
A $k$-point is a proper closed subset, so if $k$ is finite, the set of $k$-points cannot be Zariski-dense. – user64687 Dec 10 '13 at 16:29
yes i see the point with finite field, if $Z_1$,...,$Z_n$ are the finite elements, they are all solution of 1 polynomial equation, thus there is an open subset not containing all those. – Louis La Brocante Dec 10 '13 at 16:40
@IgorRivin: in this context "rational points" is being used a shorthand for "$k$-rational points", which is a fancy way of saying "points with coordinates in $k$". The reason for this seemingly superfluous terminology is that $\mathbf{P}^n_k$, considered as a scheme, contains lots of other points besides these. – user64687 Dec 10 '13 at 16:48

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