# FOM: Query on 1st-order theory of fields

JoeShipman@aol.com JoeShipman at aol.com
Wed Nov 14 12:07:10 EST 2001

```Can anyone tell me whether the following problem is known to be decidable?

Say that a multivariate polynomial f(x1,...,xn,z) is "solvable over the field K" if, for all k1,...,k_n in K^n, the 1-variable polynomial f(k1,...,k_n,z) either has a zero in K or is a constant polynomial.

Given: multivariate polynomials f_i(x1,x2,...,xn,z), g(x1,x2,...,xn,z) with rational coefficients
Problem: determine whether the solvability of all the f_i over K implies the solvability of g over K for all number fields K.  (Alternatively, for all characteristic 0 fields K.)

Motivating example:  We know from the work of Cardano that the general 4th-degree polynomial can be solved in terms of square and cube roots.  Thus, we have
f_1 = z^2 - x1
f_2 = z^3 - x1
g = x1*(z^4) + x2*(z^3) + x3*(z^2) + x4*z + x5

For any number field (indeed, for any field at all whose characteristic is not 2 or 3) Cardano's explicit formula gives a zero for g in terms of square
and cube roots, which exist because f_1 and f_2 are solvable.  So the answer
in this case is YES.

Second motivating example:
f_1 = z^2 - x1
f_2 = z^3 - x1
f_3 = z^5 - x1
g = x1*(z^5) + x2*(z^4) + x3*(z^3) + x4*(z^2) + x5*z + x6

Because the general quintic is insolvable by radicals, the answer in this case is NO (the field formed by closing Q under the operations of square, cube, and 5th roots does not contain zeros for certain 5th-degree
polynomials).

Third motivating example:
f_1 = z^2 - x1
f_2 = z^3 - x1
f_3 = z^5 - x1
f_4 = z^5 - x1*z - x2
g = x1*(z^5) + x2*(z^4) + x3*(z^3) + x4*(z^2) + x5*z + x6

The work of Tschirnhausen and Bring showed that the general quintic can be brought into either the form x^5 - a = 0 or the form x^5 - x - a = 0 by transformations involving square and cube roots, so the answer in this case is YES.

Is this question decidable in general?  What about the case where there is only one f?  (The first two examples are equivalent to examples with only one
polynomial in the set f_i.  z^6 - x1 is solvable iff z^2 - x1 and z^3 - x1 are; z^30 - x1 is solvable iff z^2 - x1 and z^3 - x1 and z^5 - x1 are.)

-- Joe Shipman

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