It is almost impossible to find something noone has done before.
But that is a good thing. Otherwise that would mean that I am the
first one doing it, and I never finish anything.

As I don't really know where else to publish this (mathoverflow etc do not seem to have a section for it), here comes a wrong construction of a field of characteristic 1. Still, it is wrong in a somewhat interesting way. Maybe as an exercise for an introductory lecture in mathematical logic or something.

So to make things easy, we work with ZFC. We could probably use weaker systems, but this would just make things harder to express.

Now, let \operatorname{ZFC} be the first-order theory of Zermelo-Fraenkel set theory, and let us assume that \operatorname{ZFC}\not\vdash\bot. We define first-order constants K, 0, 1, +, \cdot with the axioms

  • 0,1\in K
  • +,\cdot\subseteq K^{K^2}
  • (K,+,\cdot,0,1) is a field

Now define 0.a := 0 and (n+1).a := n.a + n, where the first addition is ordinal addition, and the second addition is addition in the field K. Essentially, this defines that n.a means adding a\in K for n times. This is a simple recursive definition, so it is justified according to the recursion theorem. We add the axiom

  • \exists_{0<\alpha<\omega}\forall_{a\in K} \alpha.a = 0 (K has finite characteristic \neq 0)

let us call the resulting theory \operatorname{ZFC}_K. Obviously, \operatorname{ZFC}_K\not\vdash\bot. Now define the axioms A_n := \underbrace{1+\ldots+1}_{n\times}\neq 0, and A :=
\{A_{n+2}\mid n\in \mathbb{N}_0\}, and consider \operatorname{ZFC}_K \cup A.

Every finite subset of \operatorname{ZFC}_K \cup A is satisfiable, because there are fields of arbitrarily large characteristic. Hence, because of the compactness theorem, \operatorname{ZFC}_K \cup A \not\vdash \bot, and hence, there is a model M \models \operatorname{ZFC} with a constant K such that

  • K is a field
  • K has finite characteristic
  • \operatorname{char} K\neq n for all n\in\mathbb{N}_0\backslash\{1\}.

We can "lift" K out of its model M, by setting \overline{K} := \{k\in M\mid M\models k\in K\}, similar for the other constants.

Now comes the exercise: Why is \overline{K} not a field of characteristic 1?

I could actually imagine that, though the actual characteristic of \overline{K} should be 0 (if I am correct), this thing has some properties that one wants from fields with characteristic 1, so it might be interesting to look at (but if it is, there certainly are people who already do this).