Context: I'm taking a course in geometry (we see affine, projective, inversive, etc, geometries) in which our basic structure is a vector space, usually $\mathbb{R}^2$. It is very convenient, and also very useful, since I can then use geometry whenever I have a vector space at hand.
However, some of that structure is superfluous, and I'm afraid that we can prove things that are not true in the more modest axiomatic geometry (say in axiomatic euclidian geometry versus the similar geometry in $\mathbb{R}^3$).
My questions are thus, in the context of plane geometry in particular:
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Can we deduce, from some axiomatic geometries, an algebraic structure?
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Are some axiomatic geometries equivalent, in some way, to their more algebraic counterparts?
(Note that by « more algebraic » geometry, I mean geometry in a vector space. The « more algebraic » counterpart of axiomatic euclidian geometry would be geometry in $R^2$ with the usual lines and points, and where we might restrict in some way the figures that we can build.)
I think it is useful to know when the two approaches intersect, first to be able to use the more powerful tools of algebra while doing axiomatic geometry, and second to aim for greater generality.
Another use for this type of considerations could be in the modelisation of geometry in a computer (for example in an application like Geogebra). Even though exact symbolic calculations are possible, an axiomatic formulation could be of use and maybe more economical, or otherwise we might prefer to do calculations rather than keep track of the axiomatic formulation. One of the two approaches is probably better for the computer, thus the need to be able to switch between them.
Best Answer
Hilbert's Foundations of Geometry did more or less precisely what you are asking for. Starting from an extension of Euclid's Axioms, Hilbert proves that any model of the axioms is isomorphic to $\mathbb{R}^2$ with the usual definition of line.
Later, Tarski gave a first-order axiomatization of plane geometry. Because of built in restrictions in the first-order approach, one cannot get isomorphism with the natural geometry of $\mathbb{R}^2$. But one can get isomorphism to the natural geometry of $F^2$, for some real-closed field $F$.