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The Krasner hyperfield is an algebraic structure of two operations on $K=\{0,1\}$ called $+\colon K\times K\to \mathcal{P}(K)$ and $\cdot\colon K\times K\to K$ with

  • $0+0=0$
  • $0+1=1+0=1$
  • $1+1=\{0,1\}$

and with the same multiplicative structure as $\mathbb{F}_2$. In Proposition 3.1 of this paper of Connes and Consani, it is shown that projective geometries in which every line contains at least 4 points are "essentially the same" as $K$-modules. What I am trying to understand is the way in which this connects to ordinary vector spaces. For instance, if I look at $\mathbb{F}_p^n$ then, as a vector space, it determines a projective geometry (via, if you like, a matroid) which is a module over $K$. But $\mathbb{F}_p^n$ is also just a module over $\mathbb{F}_p$. I believe that $K$ is the terminal hyperfield, so there's a hyperfield morphism $\mathbb{F}_p\to K$. If I take a projective geometry, realized as a module over $K$, what is the pullback along the terminal morphism? Does the resulting $\mathbb{F}_p$ vector space have the same "projective geometry," abstractly, as the combinatorial one we started with?

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A module over $\mathbb{F}_p$ viewed as a hyperfield is different than that of a module of $\mathbb{F}_p$ viewed as a ring. In particular the pullback of a module over the Krasner hyperfield need not be a vector space over $\mathbb{F}_p$. In fact it is never an $\mathbb{F}_p$ vector space since since it must obey the rule $x+x= \{0,x\}$ for all elements $x \neq 0$ in the module over $K$.

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  • $\begingroup$ Ah thanks very much! Do you have any sense of how we are supposed to think about the relationship between vector spaces and combinatorial geometries? I guess there is a functor from vector spaces over a field 𝔽 to K-modules? $\endgroup$ Commented Oct 14, 2024 at 22:36
  • $\begingroup$ But presumably this functor is not -⊗K...? $\endgroup$ Commented Oct 14, 2024 at 22:42
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    $\begingroup$ I guess the sort of functor you would like would send a vector space to its associated projective geometry viewed as a $K$-module. The issue is that this isn't obviously functorial. I think the naive definition is not functorial (I haven't checked carefully though). I don't have a great intuition for these sorts of things myself. $\endgroup$ Commented Oct 14, 2024 at 23:07

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