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:

  1. Can we deduce, from some axiomatic geometries, an algebraic structure?

  2. 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.

  • $\begingroup$ Are you familiar with algebraic geometry? A glance at your profile is inconclusive. $\endgroup$ Nov 28, 2015 at 21:23
  • $\begingroup$ No, unfortunately. $\endgroup$
    – Olivier
    Nov 28, 2015 at 21:33
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    $\begingroup$ I recommend you check out Hartshorne's Geometry: Euclid and Beyond, specifically chapter 3. It deals with analytic geometry over general fields, and the geometric consequences of adding certain properties to them, such as an order, the Archimedean property, square roots for elements of the form $1 + a^2$ (Pythagorean fields), or square roots of arbitrary positive elements (Euclidean fields). For example, if a field is Euclidean, then its corresponding Cartesian plane has the property that any circle that has a point inside another circle will intersect it in two points. $\endgroup$
    – Mauro
    Nov 28, 2015 at 23:37

4 Answers 4


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 $F$.

  • $\begingroup$ Interesting! Do you happen to know what the exact requirement on $F$ is in this case? $\endgroup$
    – Marc
    Nov 28, 2015 at 21:55
  • $\begingroup$ Real-closed fields. Another major achievement of Tarski was to give a decision procedure for elementary geometry, later improved by various people including Seidenberg and Cohen. This decision procedure to my mind partly answers your other question about finding "geometric" methods of proof. Probably algebraic methods are, to a computer, more suitable than more geometry-based methods. $\endgroup$ Nov 28, 2015 at 22:07
  • $\begingroup$ But does "categoricity" of Hilbert's axioms justify using coordinate system(which I guess exist in the language of sets) to prove theorems in the language of plane geometry? Can I claim a proof of some statement from Hilbert's axioms by checking it's true in $R^2$? Hilbert's axioms do not consitute a first order theory so it seems I cannot use completeness theorem. $\endgroup$
    – 183orbco3
    Feb 8, 2019 at 14:24

I'll be repeating some stuff that has already been said, but I have my own spin on it.

Can we deduce, from some axiomatic geometries, an algebraic structure?

Yes. As you can find in Hartshorne's book, or in Hilbert's Foundations, the idea is an "algebra of segments" for any ordered Desarguesian plane in which you construct an Archimedian, ordered division ring $D$ with certain geometric operations such that the points and lines in $D\times D$ are coordinatized in exactly the way we're taught for $\Bbb R \times \Bbb R$. So, they are a natural starting point for ordered geometry.

Now, every Archimedian, ordered division ring embeds in $\Bbb R$. Since $\Bbb R$ is also an Archimedian, ordered field, you can see it is the maximal such field for such a geometry. In fact, needing the reals to coordinatize a plane is equivalent to very strong "completeness" of lines in that geometry. This completeness/maximality property makes it very attractive to study geometry with it.

Ordered geometry is great, but reexamination of the ideas shows that division rings in general are exactly what you need to coordinatize Desaguesian planes. They imbue their lines with exactly the translation and scaling behavior one would expect in a geometry according to the Erlangen program.

Are some axiomatic geometries equivalent, in some way, to their more algebraic counterparts?

Let me continue briefly along the lines above. The amazing theorem that a Desaguesian plane is Pappian iff the division ring is commutative has already been mentioned. There are also theorems about equivalence of certain types of fields and constructability criterion in the geometry.

Desarguesian projective planes enjoy the same coorditinization theorem with division rings. Hyperbolic planes require one more unique axiom before they can be coordinatized by a division ring. The division ring you get is an analogue of the "algebra of segments" called the "field of ends". So as far as these theorems go, you have a really strong connection between synthetic geometry and analytic geometry. Geometry with vector spaces over division ring captures a large part, but not the whole of synthetic geometries.

In fact, there are even more general coordinatization theorems for projective planes using 'ternary rings'. As you add more special properties to the projective plane, the ternary ring gets closer to being a division ring.

More posts you may like:

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This question is a little broad, and so this certainly doesn't answer all of it, but hopefully you find it interesting - an example of a situation where a geometric construction is equivalent to an algebraic property.

Given any division ring $R$ (that is, a set with two binary operations, $\oplus$ and $\otimes$, such that $\oplus$ and $\otimes$ are associative and $\oplus$ is commutative, have identities $0$ and $1$ respectively; every element has an $\oplus$-inverse, and every element other than $0$ has an $\otimes$-inverse; and distributivity holds, i.e., $(a\oplus b)\otimes c=(a\otimes c)\oplus (b\otimes c)$), we may consider the projective plane over $R$, $\mathbb{P}^2_R$ (see for example Wikipedia: Projective Plane for the details). This is a sort of abstraction of the usual projective (not Euclidean, but similar) geometry.

It turns out that certain algebraic properties of the division ring $R$ are equivalent to certain geometric facts about the "plane" $\mathbb{P}^2_R$: see for instance Pappus' Theorem. Pappus' theorem holds if and only if $R$ is a commutative division ring ($a\otimes b=b\otimes a$), that is, a field.

There are other examples of this - for instance, Desargue's theorem - and this might be what you have in mind?

One might ask, "But wait! Isn't starting with a division ring already presupposing algebraic structure?" Well, yes, but it turns out that this algebraic structure is "already there," in a sense: given any "plane" $P$ satisfying some basic axioms of projective geometry (perhaps surprisingly, the crucial property is Desargue's theorem), we can realize $P$ as $\mathbb{P}^2_R$ for some division ring $R$. The converse also holds - $\mathbb{P}_R^2$ satisfies these axioms for any division ring $R$ - so this is really an equivalence between geometric and algebraic "base theories." See Chapter 7 of Foundations of Projective Geometry by Robin Hartshorne for more details.

This can lead to neat cross-applications. For example, every finite projective plane which satisfies Desargue's Theorem also satisfies Pappus' Theorem! Why? Because of the purely algebraic fact there are no finite division rings which aren't commutative! (See Wedderburn's little theorem. I'm pretty sure this is a case of nuking a mosquito, but it's cool!)

  • 1
    $\begingroup$ Notice that I'm talking about projective geometry, as opposed to Euclidean. This is because in many ways projective geometry is "tamer." I suspect there are similar connections for Euclidean geometries, but I know much less about this sort of thing. EDIT: See Andre's answer below. $\endgroup$ Nov 28, 2015 at 21:42
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    $\begingroup$ Stillwell's book "The four pillars of geometry" has a nice and elementary exposition of this equivalence between projective geometry plus extra axioms (e.g. Desargeu's Theorem) and algebra. $\endgroup$
    – Lee Mosher
    Nov 28, 2015 at 22:52
  • 1
    $\begingroup$ Stillwell's book is generally awesome. $\endgroup$ Nov 28, 2015 at 23:14
  • $\begingroup$ "such that $\otimes$ is commutative"—is that really required for a division ring? wouldn't that make $R$ a field? $\endgroup$
    – wchargin
    Nov 29, 2015 at 21:42
  • $\begingroup$ @WChargin Derp, of course you're right - fixed! $\endgroup$ Nov 29, 2015 at 23:32

Euclid certainly did not use real numbers. He showed that the diagonal and side of a square have no "common measure", i.e. there is no segment that can be laid end-to-end an integer number of times to get the length of the side and also to get the length of the diagonal, and he knew how to define what it means to say the ratio of lengths of segment $A$ to segment $B$ is the same as the ratio of lengths of segment $C$ to segment $D$ when the latter two are the side and diagonal of a square.

And as mentioned elsewhere, Hilbert also developed Euclidean geometry while not just not using real numbers, but while making a point of avoiding them.


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