[FOM] The Field with One Element?

[Harvey Friedman]

On Fri, Jun 17, 2016 at 2:41 PM, Andrius Kulikauskas <ms at ms.lt> wrote:
> Dear Harvey,
>
> Thank you for your invitations in your letter below and also earlier,  "...I
> am trying to get a dialog going on the FOM and in these other forums as to
> "what foundations of mathematics are, ought to be, and what purpose they
> serve"."
> http://www.cs.nyu.edu/pipermail/fom/2016-April/019724.html
>
.>..Would the "field with one element" be such
> issue for you?
> https://ncatlab.org/nlab/show/field+with+one+element.

First tell me what is wrong with the following plan for the field with
1 element?

DEFINITION. A pseudo field is a nonempty set F together with elements
0,1 (possibly the same) and binary functions +,dot, such that the
following holds.
1. x+0 = x.
2. x+y = y+x.
3. x+(y+z) = (x+y)+z.
4. For all x, there exists y such that x+y = 0.
5. x dot 1 = x.
6. x dot y = y dot x.
7. x dot (y dot z) = (x dot y) dot z.
8. For all x not 0, there exists y such that x dot y = 1.
9. x dot (y+z) = x dot y + x dot z.
10. If x dot y = 0 then x = 0 or y = 0.

THEOREM. There is a pseudo field with exactly one element.

THEOREM. The pseudo fields with more than one element are exactly the fields.

Proof: Suppose the pseudo field has 0,1 the same, and there is more
than one element. Let u be other than 0,1.

We now show that u dot 0 = 0. We have u dot (1 + 0) = u dot 1 = u = u
dot 1 + u dot 0 = u + (u dot 0). In particular u = u + (u dot 0). Let
u + v = v + u = 0. Then 0 = v + u = v + (u + (u dot 0)) = (v + u) + (u
dot 0) = 0 + (u dot 0) = u dot 0. So u dot 0 = 0.

But u dot 1 = u. Since 1 = 0, we have 0 = u. This is a contradiction. QED

Harvey Friedman