This is a little question inspired from Hartshorne's Geometry, which I've been juggling around for a while.
Suppose that $\Pi$ is the Cartesian plane $F^2$ for some field $F$, with the set of ordered pairs of elements of $F$ being the points and lines those subsets defined by linear equations. Let \Pi' be the associated projective plane. Here \Pi' is just $\Pi$ together with the points at infinity, which are the pencils of sets of parallel lines. A line in \Pi' will be the subset consisting of a line of $\Pi$ plus its unique point at infinity.
Also, let $V=F^3$ be a three-dimensional vector space over $F$. Let $\Sigma$ be the set of $1$-dimensional subspaces of $V$, and call them points. If $W\subseteq V$ is a $2$-dimensional subspace of $V$, then the set of all 'points' contained in $W$ will be called a 'line,' and $\Sigma$ forms a projective plane.
I would like to see how \Pi' and $\Sigma$ are isomorphic. I figure I should construct a bijection from the points of \Pi' to the points of $\Sigma$, however, I don't see a natural mapping. At first I figured if (a,b)\in\Pi', then maybe $(a,b)\mapsto\text{span}(a,b,0)$, but this isn't injective or surjective. Not only that, but I figure if $A$ and $B$ are points in \Pi' on a line, then the images of all points in that line in \Pi' should be in the same $2$-dimensional subspace in $\Sigma$.
I don't quite see why they should be isomorphic. Suppose to lines $l$ and $m$ intersect in \Pi', shouldn't their intersection point map to the intersection of the lines in $\Sigma$? But what if this intersection doesn't even pass through the origin, and hence isn't even a $1$-dimensional subspace? Thank you for any insight on how to solve this.
Edit: For my last question above, I just realized that all the $2$-dimensional subspaces pass through the origin (whoops!), and thus the intersection will contain the origin as well, so that's not an issue.