Re: naive question from a non-mathematician


Denis Feldmann wrote:
David C. Ullrich a écrit :
On 27 May 2006 20:42:18 -0700, "Gene Ward Smith"

Moreover, its not a univoque definition : take a non-standard extension
of C with same cardinal (for instance, the quotient of C^N by the
classical* relation induced by a (non-trivial) ultrafilter on N)

OK.

Now I define the real
numbers R as the subextension fixed by this automorphism, which I now
dub "complex conjugation".

This is R* (the same non-standard extension of R)

Not yet, because I'm not using the nonstandard structure on the field.
All I know so far is that it is a subextension of degree two.


Now I define an archimedian absolute value
by |z| = sqrt(z conj(z)).

Why archimedian????

Because, as your example shows, otherwise it won't work!

The identity element 1 comes from field theory. Hence we can say that

|z| < 1+1+...+1

for some sum of 1s, which enforces an archimedian total order on the
subextension, and now we have "C" and "R" as archimedian valued fields.


Or, I could use valuation theory, and choose to start out with a
valuation of height one on my algbraically closed field. That is, let C
be an algebraically closed field with cardinality c, and let v be a
valuation map to a totally ordered group G of height one, meaning G has
only one isolated subgroup (the trivial group.) Then C is a topological
field from the valuation topology, and I get a unique continuous
automorphism of degree two. I use that to define R.

Now C and R are topological fields under the
topology defined by this, and conj is continuous. Now I define Q as the
intersection of all subfields of C (or R.)

There is a much simpler way : Q is the field generated by 1.

True, but less in the spirit of the thing.

Serious question: If you start with a structure satisfying your
definition of C and then do all this, does it follow that the
R you get is complete (in the sense that every nonempty set
bounded above has a least upper bound?) I bet it doesn't.

It does.

Sure : it is not even archimedian :-)

Under that old structure we decided to forget about. Using an
archimedian valuation, it does.

.

Relevant Pages

  • Re: naive question from a non-mathematician
    ... Now I define an archimedian absolute value ... valuation of height one on my algbraically closed field. ... be an algebraically closed field with cardinality c, ... field from the valuation topology, and I get a unique continuous ...
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  • Re: naive question from a non-mathematician
    ... David C. Ullrich a écrit: ... If tyou dont addd more axiomatic properties (like the fact it is archimedian) to your definition of C, it will not necessarily be what you want... ... valuation of height one on my algbraically closed field. ...
    (sci.math)