# Infinite Fano planes

Let $\kappa$ be an infinite cardinal. Then we call $F\subseteq{\cal P}(\kappa)$ a Fano plane on $\kappa$ if

1. $\bigcup F = \kappa$; $|F|=\kappa$; and $|a| = \kappa$ for all $|a|\in F$,
2. if $a\neq b\in F$ then $|a\cap b|=1$, and
3. if $x\neq y\in\kappa$ there is $a\in F$ such that $x,y\in a$ (note that by 2. above, this $a$ is uniquely determined).

Is there a Fano plane on every infinite cardinal $\kappa$?

• Given your choice of terminology I guess you already thought of this, but I cannot see what is the problem: Let $K$ be a field of cardinality $\kappa$. Fix a bijection between $\kappa$ and the projective plane over $K$. Then let $F$ be (the image under this bijection) the set of lines in the projective plane. Oct 26, 2017 at 10:23
• Oh - thanks, it didn't occur to me to introduce a field. Can you write this down as an answer? Oct 26, 2017 at 11:56

Let $K$ be a field of cardinality $\kappa$. Fix a bijection between $\kappa$ and the projective plane over $K$. Then let $F$ be (the image under this bijection of) the set of lines in the projective plane.

• Maybe you also see a solution to the "uniqueness" sequel of this question: mathoverflow.net/questions/284454/… Oct 26, 2017 at 12:09
• I don't know anything about this topic, but I will guess the answer to the uniqueness question is "no", and maybe one can use non-Desarguesian planes to prove it. Good luck! Oct 26, 2017 at 12:32
• The lack of uniqueness is established (with room to spare) by Alex Kruckman's answer, but here's another, more combinatorial way to prove it. Given the projective plane over a field (as an abstract structure), one can recover the field up to isomorphism. Since there are plenty of non-isomorphic fields of any infinite cardinality (different characteristics, algebraically closed or not), you get plenty of non-isomorphic projective planes by Pooter's construction. Oct 29, 2017 at 17:45

What you call a Fano plane on $\kappa$ is usually simply called a projective plane of cardinality $\kappa$ (with set of points equal to $\kappa$). Pooter's answer uses the fact that any field of size $\kappa$ gives rise to a projective plane of size $\kappa$, which is the easiest way to answer the question. I want to give a more general model-theoretic perspective on the question as well as your follow-up question about uniqueness in the comments.

The projective plane axioms can be written down as a first-order theory $T$, and $T$ has at least one infinite model, so by the Löwenheim-Skolem theorem, it has a model of every infinite cardinality $\kappa$. Moreover, if we let $T_\infty$ be $T$ together with axioms saying the domain is infinite, then $T_\infty$ is not a complete theory (for example, the Desargues axiom and its negation are both consistent with $T_\infty$). So $T_\infty$ has multiple non-isomorphic (even non-elementarily equivalent) models of every infinite cardinality.

To say that a first-order theory $T$ has a unique model up to isomorphism in some uncountable cardinality $\kappa$ is to say that $T$ is uncountably categorical. In a sense, Morley's theorem says that this property is very unusual. It certainly doesn't hold for the theory of projective planes. But it does hold for a completion of this theory, namely the theory of projective planes over algebraically closed fields of characteristic $0$ (indeed, projective planes over fields are bi-interpretable with their base fields, and the theory of algebraically closed fields of characteristic $0$ is uncountably categorical - the same is true for fixed characteristic $p$).

In this recent paper, Gabe Conant and I isolate a particularly model-theoretically interesting class of infinite non-Desarguesian projective planes (the generic planes), which are the models of a complete first-order theory $T_{2,2}$. Among other things, we show that there are continuum-many countable generic planes up to isomorphism.

• Very nice, thanks for your elucidating remarks! Oct 30, 2017 at 7:40