[FOM] Need reference for results in Field Theory

zahidi@logique.jussieu.fr zahidi at logique.jussieu.fr
Fri Oct 20 04:37:35 EDT 2006

> 2a. I would think so. For instance I would estimate Q to be elementarily
> equivalent to Q(X). This estimate is based on the observation that when
> playing the Fraissé-Ehrenfeucht game between Q and Q(X) (say of length
> n), one can always find for example a very large prime which satisfies the
> same `n-transcendency' as X. Therefore one has a winning strategy in this
> game, implying the two fields are elementarily equivalent. (But I did not
> write this down as a rigourous proof.)

I don't think this is true actually. Q (rational numbers) and Q(x) are not
elementary equivalent. An easy way to see this is as follows:
the Pythagors number p(K) of a field K is defined as the smallest number
of squares needed to represent any number of squares.
It follows from Langrange's theorem that the Pythagors number of Q is 4
(any rational number that is a sum of squares is positive; by Lagrange's
every positive rational number is the sum of four squares).
The Pythagoras number of Q(x) is 5.
It is easy to see that for any natural number n, the statement p=n is an
elementary statement. Hence the statement p=4 is true in Q but not in

> Summing up: I do not think the transcendence degree is of the utmost
> importance. For instance the algebraic closure of A(X) is elementarily
> equivalent to A (classic result, the theory of an algebraic closed field
> of characteristic 0 is Aleph_1 categorical).

Results by Florian Pop show that for large classes of fields transcendence
degree is first-order definable and hence important for elementary
equivalence. E.g. Q(x) is never elementary equivalent to Q(x_1,...,x_n)
(field of rational functions in n variables), for any n>1.

Also, an argument using the Pythagors number can be used to show that R(x)
(where R denotes the reals) is never elementary equivalent to
R(x_1,...,x_n) (field of rational functions in n variables), for any n>1.

Karim Zahidi

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