# On the infinitesimally small

vshiro
Consider R, or rather the equivalence class of fields isomorphic to R, endowed with an order type of the same kind. Let us consider a new object, dx, a quantity such that dx > 0, but for all a > 0 in R, dx < a. (We are considering dx, something outside of R, but without the context of a larger universe set.)
Now let add dx to R, and field-ify it. That is, we are considering the infinite field extension R[dx]. Furthermore we inherit the order of R, but extend it to include dx. The way I think about the order type is that, suppose we can sort of magnify R until we see a, b, and a, b have no more elts of R between them (there is no c with a<c<b)... we can't actually do this, but entertain this idea for a moment. Adding dx to R is like fitting another whole copy of R between a and b: we have between them a < a + k*dx < b for all k in R. But then we magnify this new copy of R until we get to a, b and add another copy of R (dx^2) and so on, and do this for all nonexistent pairs a, b satisfying there is no c st a<c<b...
My question is, what is the "dimension" of R[dx]? It is not 2.. The above thought experiment suggests that it is |Z| (aleph-1), but I don't believe that is true. It may well be |R|, but I suspect the answer is much more sinister...

Note that the standard definition of dimension in field extension doesn't apply.

damgo
What definition of dimension are you wanting to use?

Hurkyl
Staff Emeritus
Gold Member
Now let add dx to R, and field-ify it. That is, we are considering the infinite field extension R[dx]
Do you mean R(dx)? R[dx] is merely a ring in this case, because dx cannot be written as a root of a polynomial in R[x].

My question is, what is the "dimension" of R[dx]?
For R[dx], it's easy. R[dx] is isomorphic to R[x], which is countably infinite dimensional over R; its basis is
{1, x, x^2, x^3, x^4, ... }

R(dx) is isomorphic (ignoring the order) to R(x), the field of rational functions. Finding a vector space basis of R(x) over R is a lot trickier, but we can solve our problem without it:

Consider S = {p(x)/q(x) | p and q are in R[x] and q(x) is not 0}

Clearly S is a spanning set of R(x) (it includes every element of R(x)!!!), and |S| = |R|, so the dimension of the extension can be no more than |R|.

However, consider T = {1/(x - a) | a in R}

linearly independant, so the dimension of the extension can be no less than |T| = |R|.

So, R(dx) is a field extension of R with dimension |R|

Note that the standard definition of dimension in field extension doesn't apply.
Why?

Hurkyl