On the infinitesimally small

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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

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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
 

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