By Abraham Robinson (Eds.)

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**Example text**

E. in the present case, in M itself. 6. 9. 9 possesses a solution in an algebraically closed extension M’ of M although it does not possess a solution in N . Let this solution be given by yl=al, y2=a2, ... rra. , am),and the degree of transcendence of M ” over M is finite ( ~ m )Now . no>1, is as small as possible. We may then assume that the degree of transcendence of M,, over M is exactly m,= 1. For if this is not true a priori, then we may always replace M by the algebraic closure of M(bl, ...

V(al-ai)=b We select an element a, E for i = 2 , ... m. 29 yields v(y) = v(z-u,) - v(uo)= b - b = 0. Also, putting ci=(a,-ul)/uo for i = 2 , ... 29, v(y-c4) =v = i = 2 , ... m. 31. , m. 31 is satisfied in M' by y = (c-u,)/uo, and we have to solve it in M . 29, x = u a + % . e. e. the set of all r E F for which v(r)>O. Then the quotient ring FR=R/J is known to be a field. , ym the elements of FR which contain c2, .. , c, respectively. e. , m since w(ci)= 0. 32. In order to prove that this system possesses a solution in FR, it is sufficient to show that FR is infinite.

X, holds in M' and so M' includes a constant a, such that V(ao) holds in M'. Since Xo is admissible, V(uo)also must be admissible, but the pair { M ' ; V ( a o ) )cannot belong to X' since the number of quantifiers in V(a,) is smaller than p0. It follows that V(a0)is decidable in K u N', N' being the diagram,of M ' . M and M' are models of K and K satisfies z, ... A A 2, r> V ( U o ) is deducible from K . We write 2, A ... , a,, all belong to C,’ and hence can appear only in the free argument places of the relations which occur in 2.