I have two questions, which are really the same question phrased in two ways:
- Has there been any research on adjoining infinitesimal elements to finite fields?
- Has anyone considered extensions to finite fields, which would stand in relation to finite fields as e.g. the hyperreal numbers stand in relation to the real numbers in nonstandard analysis?
(Note: I found a thing called "hyperfinite sets", but one of these appears to be a set that is finite in a nonstandard model, and I don't think that's quite the same thing since it internally "thinks" it's finite; whereas the fields I have in mind would know internally that they have infinitesimals.)
The background is as follows. Given $\mathbb R$ we can, using compactness, obtain a model containing an infinitesimal element $\varepsilon$, namely one that is greater than zero but smaller than any strictly positive element in $\mathbb R$. We can also concretely build such a model using an ultrafilter construction over a non-principal ultrafilter $\mathcal U$ on $\mathbb N$. An example of an infinitesimal would be $(1/n \mid n\in\mathbb N)_{\mathcal U}$, the equivalence class of a sequence tending to $0$. This much is well-known.
It seems to me plausible that we might carry out a similar ultrafilter construction over a finite field, and this too would contain infinitesimal elements. For example (sketch construction): $$ (0,0,0,\dots)_{\mathcal U} < (1,0,1,0,1,\dots)_{\mathcal U} < (1,1,1,1,\dots)_{\mathcal U}. $$ Has a construction in this spirit been studied, and if so can anyone provide references?
Thank you.
Finite fields have no ordering: no field with positive characteristic can be made into an ordered field. So your premise of adjoining an element "greater than zero but smaller than any strictly positive element" doesn't make sense with finite fields.
At the same time, you can adjoin to any field $K$ an element $\varepsilon$ where $\varepsilon \not= 0$ and $\varepsilon^2 = 0$ by using $K[x]/(x^2)$. There $x \bmod x^2$ has the desired property, but this is certainly not a field anymore. (Look up the term "dual numbers".) More generally, by using $K[x]/(x^m)$ you have adjoined to $K$ a nonzero element $\varepsilon$ where $\varepsilon^m = 0$ and no smaller power of $\varepsilon$ vanishes.